Class 12 Commerce Maths - Differential Equations

Find the order of the differential equations of all circles in the plane XOY which have their centres on x-axis and have give radius.

Let the centre on x-axis be

$(h,0)$ and given radius be

$a$ so that the equation of the circle is

$\displaystyle \left ( x-h \right )^{2}+y^{2}=a^{2}$
Differentiating,

$\displaystyle 2\left ( x-h \right )+2y\frac{dy}{dx}=0$
or

$\displaystyle \pm \sqrt{\left ( a^{2}-y^{2} \right )}+y\frac{dy}{dx}=0.\left [ by\left( 1 \right ) \right ]$
We have eliminated one variable h and have got a differential equation of first order,

$a$ being constant.

Order of differential equation of $y=mx+c$

Eliminating arbitrary constants m and c from

$\displaystyle y=mx+c.$

$\displaystyle \frac{dy}{dx}=m.$ Differentiating again,

$\displaystyle \frac{d^{2}y}{dx^{2}}=0\cdots (2)$ The two arbitrary constants have been eliminated and we have got a differential equation of second order.In the light of above we can say that if an equation contains n arbitrary constants, the result of elimination these constants will be a differential
equation of nth order. Equation (1)in ths above examples are called the General solution of
differential equation. Hence we can say that by solving a differential equation of nth order, we mean to find a relation between the two variables x,y and n independent arbitrary constants, such that when these constants are eliminated the given differential equation of nth order will be formed.

Order of differential equation of $y=mx$

Eliminate the arbitrary constant m in the given equation

$y=mx...(1)$

$\displaystyle \frac{dy}{dx}=m=\frac{y}{x}$ from given equation

$\displaystyle \therefore y=x\frac{dy}{dx}$ is a differential
equation of 1st order obtained by eliminating one constant m.

Solve the differential equation: $(1+x^2)dy=(1+y^2)dx$ .

Give differential equation is

$(1+x^2)dy=(1+y^2)dx$

$\Rightarrow \displaystyle\int\frac{dy}{1+y^2}=\displaystyle\int\frac{dx}{1+x^2}$

Integrating both the sides

$\displaystyle\int\frac{dy}{1+y^2}=\int\displaystyle\frac{dx}{1+x^2}$

$\displaystyle \tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}C$

$\Rightarrow \displaystyle \tan ^{-1}y-\tan ^{-1}x=\tan ^{-1}C$

$\tan ^{-1}\left(\displaystyle\frac{y-x}{1+xy}\right)=\tan ^{-1}C$

$\Rightarrow \displaystyle\frac{y-x}{1+xy}=C$

Solve : $\dfrac{dy}{dx} = 1 + x + y + xy$

The given equation can be written in the form

$\dfrac{dy}{dx}=(1 + x) + y (1 + x)$

$\Rightarrow \dfrac{dy}{dx} = (1 + x)(1 + y)$

$\Rightarrow \dfrac{dy}{1+y}= (1 + x)dx$
On integrating, we have

$\log (1 + y)= x + \dfrac{x^2}{2} +c$ , which is the required solution.

The integrating factor of the differential equation $\dfrac { dy }{ dx } -y\tan { x } =\cos { x }$ is :

Given differential equation is,

$\cfrac { dy }{ dx } -y\tan { x } =\cos { x }$

Comparing above differential equation with first order linear differential equation,

$\dfrac{dy}{dx}+Py=Q$

We get

$P=-\tan x$ and

$Q=\cos x$

So integrating factor is

$e^{ \int Pdx}=e^{ \int -\tan xdx}=e^{-\ln \sec x}=e^{\ln \cos x}=\cos x$

Remember:

$a^{\log_ax}=x$

Find the particular solution of the differential equation:
$y(1 + \log x) \dfrac {dx}{dy} - x\log x = 0$
when $y = e^{2}$ and $x = e$ .

On rearranging, we have

$\dfrac{dy}{dx}=\dfrac{y(1+\log x)}{x\log x}$

$\therefore \dfrac{dy}{y}=\dfrac{1+\log x}{x\log x}dx$

Integrating on both sides, we get

$\displaystyle \int\dfrac{dy}{y}=\int\dfrac{1+\log x}{x\log x}dx$

$\therefore \displaystyle \log y=\int\dfrac{1+\log x}{x\log x}dx$

Put

$z=x\log x\implies dz=(1+\log x)dx$

$\therefore \displaystyle \log y=\int\dfrac{1}{z}dz$

$\therefore \log y=\log z +k$

$\therefore y= ze^k$

$\therefore y= cx\log x$

Now, substituting

$x=e$ and

$y=e^2$ , we have

$e^2= ce\log e$

$\therefore c=e$

Hence, required particular solution is

$y= e(x\log x)$

Distinguish which one is initial valued ordinary differential equation and boundary valued ordinary differential equation:
i) $y''+2y=e^x, y(\pi)=1; y'(\pi)= 2$
i) $y''+2y=e^x, y(0)=1; y'(1)= 1$

Given the differential equations with the subsidiary conditions are

$y''+2y=e^x, y(\pi)=1; y'(\pi)= 2$

ii)

$y''+2y=e^x, y(0)=1; y'(1)= 1$ .

Between them the first one is initial valued ordinary differential equation as the subsidiary condition is about the point

$\pi$ .

And the second one is boundary valued as the subsidiary conditions are about the points

$0$ and

$1$ .

Solve the following differential equation:
$\dfrac{dy}{dx} = e^{x-y}+ x^2 e^{-y}$

$\Rightarrow \dfrac {dy} {dx} =e^{-y} \left (e^{x} +x^{2}\right)$

$\Rightarrow e^{y} dy=\left (x^{2}+e^{x} \right) dx$

Integrate on both sides

$\Rightarrow e^{y} =e^{x} +\dfrac {x^{3}}{3}+C$ , where

$C$ is a constant

A particle is moving along the X-axis. Its acceleration at time $t$ is proportional to its velocity at that time. Find the differential equation of the motion of the particle.

Let

$x$ be the distance traveled by the particle in time

$t$ .

Now, its velocity is given by

$\dfrac{dx}{dt}$ and acceleration is given by

$\dfrac{d^2x}{dt^2}$ .

According to the problem,

$\dfrac{d^2x}{dt^2}=\dfrac{dx}{dt}$ .

This is the required differential equation.

Solve:
$\cos y\dfrac{dy}{dx} = \sin x$ .

$\cos y\dfrac{dy}{dx} = \sin x$

or,

$\cos y\ dy = \sin x\ dx$

Integrating both sides we get,

or,

$\sin x=-\cos x+c$ [Where

$c$ being integrating constant]

or,

$\sin x+\cos x=c$ .

This is the required solution.

Solve: $(x+y)(dx-dy)=dx+dy$

Given,

$(x+y)(dx-dy)=dx+dy$

$(x+y-1)dx=(x+y+1)dy$

$\therefore \dfrac{dy}{dx}=\dfrac{(x+y-1)}{(x+y+1)}$

integrating on both sides, we get,

$y+c=x-2 \ln |x+y+1|+y+1+c$

$2 \ln |x+y+1|=x+1$

Solve the following differential equation:
$\log \left(\dfrac{dy}{dx}\right)= ax + by$

$\log \left(\dfrac{dy}{dx}\right)= ax + by$

$\Rightarrow \dfrac {dy} {dx} =e^{ax+by} =e^{ax}. e^{by}$

$\Rightarrow e^{-by} dy=e^{ax} dx$

Integrate on both sides

$\Rightarrow \dfrac {e^{-by}} {-by} =\dfrac {e^{ax}} {ax} +C$ , where

$C$ is a constant

Verify $y=cx+\dfrac{1}{x}$ is a solution of the differential equations $y\dfrac{dy}{dx}=x\left(\dfrac{dy}{dx}\right)^2+1$ , where c is arbitrary constant.

since the given equation is

$y\frac { dy }{ dx } =x{ \left( \frac { dy }{ dx } \right) }^{ 2 }+1\\ let,\quad \frac { dy }{ dx } =p\\ \therefore \quad yp=x{ p }^{ 2 }+1\quad \longrightarrow (1)\\$

differentiating w.r.t x

$\frac { dy }{ dx } .p+y\frac { dp }{ dx } ={ p }^{ 2 }+2xp\frac { dp }{ dx } \\ \Rightarrow y\frac { dp }{ dx } -2xp\frac { dp }{ dx } ={ p }^{ 2 }-\frac { dy }{ dx } .p\\ \Rightarrow \frac { dp }{ dx } (y-2xp)=0\quad (since\quad \frac { dy }{ dx } =p)\\ \therefore \frac { dp }{ dx } =0\quad or\quad y-2xp=0\\ \Rightarrow p=constant\quad or\quad p=\frac { y }{ 2x } \\ \because \quad p=\frac { y }{ 2x } \\ \Rightarrow \frac { dy }{ dx } =\frac { y }{ 2x } \\ \Rightarrow \frac { dy }{ y } =\frac { dx }{ 2x } \\ integrating\quad both\quad side\\ \int { \frac { dy }{ y } } =\int { \frac { dx }{ 2x } } \\ \Rightarrow logy=log2x+logC\\ \Rightarrow logy=log2x.C\quad (since\quad loga+logb=loga.b)\\ \therefore y=2xC\\$

hence

$y=Cx+\frac { 1 }{ x }$ is not a solution of the differential equation

Solve:- $\sin x\frac{{dy}}{{dx}} + 3y = \cos x$

$-sinx \cfrac{dy}{dx} + 3y = cosx$

$\cfrac{dy}{dx} - \cfrac{3}{sinx} = -cotx$

$I.F. = e^{\int -\cfrac{dx}{sinx}}$

$I.F. = e^{\int -cosec x dx}$

$\int cosec x dx = -ln|cosecx + cotx|$

$I.F. = e^{ln|cosec x + cotx|}$

$I.F. = cosec x + cot x$

Multiplying the differential equation with

$I.F.$

$y(cosecx + cotx) = -\int cotx(cosec x + cotx)dx$

$y(cosec x+ cotx) = -(cosec x+ cotx + x)$

$y = -(1+\cfrac{x}{cosecx+cotx})$

Find the value of $dy/dx$ ,
${y}^{2}=4ax$

Given,

$y^2=4ax$

differentiating on both sides, we get,

$\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(4ax)$

$2y\dfrac{dy}{dx}=4a$

$y\dfrac{dy}{dx}=2a$

Solve the following Differential equation :
$\frac{dy}{dx}=(y-4x^2)^2$

Given,

$\dfrac{dy}{dx}=(y-4x^2)^2$

integrating on both sides, we have,

$\int \dfrac{dy}{dx}=\int (y-4x^2)^2$

$y+c=\int (y^2-8yx^2+16x^4)dx$

$y+c=y^2x-\dfrac{8yx^3}{3}+\dfrac{16x^5}{5}+c$

$y=y^2x-\dfrac{8yx^3}{3}+\dfrac{16x^5}{5}$

The $D\cdot E$ is $\left( \dfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) ^{ \dfrac { 1 }{ 6 } }\cdot \left( \dfrac { { d }y }{ d{ x } } \right) ^{ \dfrac { 1 }{ 3 } }=5$
Prove that: $\left( \dfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) \cdot \left( \dfrac { { d }y }{ d{ x } } \right) ^{ 2 }=5^{ 6 }$

Given,

$\left ( \dfrac{d^2y}{dx^3} \right )^{\dfrac{1}{6}}\cdot \left ( \dfrac{dy}{dx} \right )^{\dfrac{1}{3}}=5$

cubing on both sides, we get,

$\left ( \dfrac{d^2y}{dx^3} \right )^{\dfrac{3}{6}}\cdot \left ( \dfrac{dy}{dx} \right )^{\dfrac{3}{3}}=5^3$

$\left ( \dfrac{d^2y}{dx^3} \right )^{\dfrac{1}{2}}\cdot \left ( \dfrac{dy}{dx} \right )=5^3$

squaring on both sides, we get,

$\left ( \dfrac{d^2y}{dx^3} \right )^{\dfrac{2}{2}}\cdot \left ( \dfrac{dy}{dx} \right )^2=(5^3)^2$

$\left ( \dfrac{d^2y}{dx^3} \right )\cdot \left ( \dfrac{dy}{dx} \right )^2=5^6$

find the solution of
$(e^{x}+e^{-x})\dfrac {dy}{dx}=e^{x}-e^{-x}$

Given,

$(e^x+e^{-x})\dfrac{dy}{dx}=(e^x-e^{-x})$

$\dfrac{dy}{dx}=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

integrating on both sides, we get,

$\int \dfrac{dy}{dx}=\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

substitute

$u=e^x+e^{-x}\rightarrow du=(e^x-e^{-x})dx$

$y+c=\log (e^x+e^{-x})+c$

$\therefore y=\log (e^x+e^{-x})$

Prove that
$y = 2e^x + 3e^2x$ is not g.s. of differential equation $y^2 - 3y + 2y = 0$

Given,

$y=\left(2e^x+3e^{2x}\right)$

$y^2-3y+2y=0$

$y^2-y=0$

$(\left(2e^x+3e^{2x}\right))^2-\left(2e^x+3e^{2x}\right)=0$

$4e^{2x}+12e^{3x}+9e^{4x}-2e^x-3e^{2x}=0$

$-2e^x+e^{2x}+12e^{3x}+9e^{4x}\neq 0$

Hence proved.

If $e^x+e^y=e^{x+y}$ , prove that $\dfrac{dy}{dx}+e^{y-x}=0$ .

$e^x+e^y = e^{x+y}$

$e^x + e^y \cfrac{dy}{dx} = e^{x+y}(1+\cfrac{dy}{dx})$

$e^x + e^y \cfrac{dy}{dx} = (e^x + e^y)(1+\cfrac{dy}{dx})$

$e^x \cfrac{dy}{dx} + e^y = 0$

$\cfrac{dy}{dx} + e^{y-x} = 0$

The soultion of $3e^x \cos^2 ydx + (1- e^x)\cot ydy = 0$ is

Given,

$3e^x\cos ^2y dx+(1-e^x)\cot y dy=0$

$3e^x\cos ^2y dx=(e^x-1)\cot y dy$

$\dfrac{3e^x}{e^x-1}dx=\dfrac{\cot y}{\cos ^2y}dy$

integrating on both sides, we get,

$\int \dfrac{3e^x}{e^x-1}dx=\int \dfrac{\cot y}{\cos ^2y}dy$

$\int \dfrac{3e^x}{e^x-1}dx=\int \dfrac{\sec y}{\sin y}dy$

$3\log (e^x-1)=\log (\tan y)+c$

$\cos{x}\dfrac{dy}{dx}-\cos{2x}=\cos{3x}$

Given,

$\cos x \dfrac{dy}{dx}-\cos 2x=\cos 3x$

$\cos x \dfrac{dy}{dx}=\cos 3x+\cos 2x$

$\cos x \dfrac{dy}{dx}=4\cos ^3x-3\cos x+2\cos ^2x-1$

$\dfrac{dy}{dx}=4\cos ^2x-3+2\cos x-\sec x$

integrating on both sides, we get,

$y+c=2x+\sin (2x)-3x+2\sin x -\log (\tan x+\sec x)+c$

$\therefore y+x=\sin (2x)+2\sin x-\log (\tan x+\sec x)+c$

Find the particular sol of $D.E. \dfrac {dy}{dx} = y\tan x;\ y = 1\ x = 0$ .

Given,

$\dfrac{dy}{dx}=y\tan x$

$\dfrac{dy}{y}=\tan x dx$

integrating on both sides, we get,

$\int \dfrac{dy}{y}=\int \tan x dx$

$\log y = \log \sec x+c$

put

$y=1,x=0$

$\log 1 = \log \sec 0+c$

$0=0+c\Rightarrow c=0$

Therefore the required equation is,

$\log y = \log \sec x$

Solve $\dfrac{dy}{dx}=x+y$ .

$\displaystyle \frac{dy}{dx} = x+y$

$\displaystyle \frac{dy}{dx}+(-y) = x$

$\displaystyle I-F = e^{\displaystyle \int -1dx} = e^{-x}$

$\displaystyle h-s = ye^{-x} = \int xe^{-x}+c$

$\displaystyle ye^{-x} = -xe^{-x}-e^{-x}$

$\displaystyle ye^{-x} = e^{-x}(-1-x)$

$\displaystyle x+y+1 = 0$

1237350_1134391_ans_c605f4d45c154eb08678cd0a6c5e04e7.jpg

(1+ $y^{2}$ ) dx=2xy dy
By simplifying get the below expression:
$\int \dfrac{2y}{1+y^{2}}dy=\int \dfrac{1}{x}dx$

We have,

$\left( 1+{{y}^{2}} \right)dx=2xydy$

$\Rightarrow \dfrac{dx}{x}=\dfrac{2y}{\left( 1+{{y}^{2}} \right)}dy$

On integrating we get,

$\int{\dfrac{dx}{x}}=\int{\dfrac{2y}{\left( 1+{{y}^{2}} \right)}}dy$

$\int{\dfrac{2y}{\left( 1+{{y}^{2}} \right)}}dy=\int{\dfrac{dx}{x}}$

Hence proved.

Solve the following differentiate:
$x (x - y)dy + y^{2}dx = 0$ .

Given,

$x(x-y)dy+y^2dx=0$

$x(x-y)dy=-y^2dx$

$\dfrac{1}{x^2}\dfrac{dx}{dy}-\dfrac{1}{xy}=-\dfrac{1}{y^2}$

substitute

$-\dfrac{1}{x}=v\rightarrow \dfrac{1}{x^2}dx=dv$

$\dfrac{dv}{dy}+\dfrac{v}{y}=-\dfrac{1}{y^2}$

$I.F=e^{\int \frac{1}{y}}dy=y$

$vy=\int- \dfrac{1}{y^2}ydy+c$

$vy=-\log y +\log c$

$vy=\log \dfrac{c}{y}$

$e^{vy}=\dfrac{c}{y}$

$e^{-\frac{y}{x}}=\dfrac{c}{y}$

Solve: $\dfrac { d y } { d x } = 1 +\dfrac {1}{ y ^ { 2 }}$

$\\(\dfrac{dy}{dx})=1+(\dfrac{1}{y^2})$

$\\or\>(\dfrac{dy}{dx})=(\dfrac{1+y^2}{y})$

$\\or\>\int(\dfrac{y}{1+y^2})dy=\int\>dx$

let $1+y^2=t$

then $2ydy=dt$

$\\\therefore\>\int(\dfrac{1}{t})(\dfrac{dt}{2})=x+c$

$\\(\dfrac{1}{2})\log t=x+c$

$\log(1+y^2)=2x+c_1$

$F(x, y) = x^3 + y^3 + 3x^2y + 3 xy^2$ . Is this homogeneous function.

$F(x,y)$ is a homogeneous function of degree

$n$ if, for any real number

$t$ ,

$F(tx,ty)=t^{n}F(x,y)$

Here,

$F(x,y)=x^{3}+y^{3}+3x^{2}y+3xy^{2}$

$F(tx,ty)=(tx)^{3}+(ty)^{3}+3(tx)^{2}ty+3(tx)(ty)^{2}$

$=t^{3}\left ( x^{3}+y^{3}+3x^{2}y+3xy^{2} \right )$

$=t^{3}\left ( F(x,y) \right )$

So,

$F(x,y)$ is a homogeneous function of degree

$3$

$y-x\dfrac{dy}{dx}=0$

Given,

$y-x\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=\dfrac{y}{x}$

$\dfrac{dy}{y}=\dfrac{dx}{x}$

integrating on both sides, we get,

$\int \dfrac{dy}{y}=\int \dfrac{dx}{x}$

$\log y =\log x+c$

$\dfrac{dy}{dx} + y = x^{-2}$

$\dfrac{dy}{dx}+y=x^{-2}$

$\dfrac{dy}{dx}=x^{-2}-y$

integrating on both sides, we get,

$y+c=-yx-\dfrac{1}{x}+c$

$y+yx=-\dfrac{1}{x}+c$

$y(1+x)+\dfrac{1}{x}=0$

Verify that $y=2\sin{3x}$ is the solution of the D.E $\cfrac{{d}^{2}y}{d{x}^{2}}+9y=0$

$y=2\sin 3x$

$\dfrac{dy}{dx} = 6\cos 3x$

$\dfrac{d^{2}y}{dx^{2}}=-18\sin 3x$

$\dfrac{d^{2}y}{dx^{2}}+9y =-18\sin 3x+9(2\sin 3x)$

$=-18\sin 3x+ 18\sin 3x$

$=0$

Solve
$\cos x \cdot \cos y\ dy-\sin x \cdot \sin y\ dx=0$

Given,

$\cos x \cos y dy-\sin x \sin y dx=0$

$\cos x \cos y dy=\sin x \sin y dx$

$\cot y dy=\tan x dx$

integrating on both sides, we get,

$\int \cot y dy=\int \tan x dx$

$\log \sin y=\log \sec x+c$

Find the Particular solution of the differential equations
${ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0;y=1$ when , $x=1$

Given,

$x^2dy+(xy+y^2)dx=0$

$x^2dy=-(xy+y^2)dx$

$\dfrac{dy}{dx}=\dfrac{-(xy+y^2)}{x^2}$

substitute

$y=vx\rightarrow \dfrac{dy}{dx}=x\dfrac{dv}{dx}+v$

$x\dfrac{dv}{dx}+v=\dfrac{-(x^2v+x^2v^2)}{x^2}$

$x\dfrac{dv}{dx}+v=-(v+v^2)$

$x\dfrac{dv}{dx}=-v-v^2-v$

$\dfrac{dv}{v^2+2v}=-\dfrac{dx}{x}$

integrating on both sides, we get,

$\int \dfrac{dv}{v^2+2v}=-\int \dfrac{dx}{x}$

$\log \sqrt {\dfrac{v}{v+2}}=-\log x+\log c$

$\log \dfrac{x\sqrt v}{\sqrt {v+2}}=\log c$

$\dfrac{x\sqrt v}{\sqrt {v+2}}=c$

re-substitute

$y=vx$

$x^2y=c^2(y+2x)$

$y=1,when,x=1$

$1^2(1)=c^2(1+2(1))$

$c^2=\dfrac{1}{3}$

$\therefore x^2y=\dfrac{1}{3}(y+2x)$

$y+2x=3x^2y$

Define particular solution of a differential equation

$Particluar\,solution: -$ The solution of a differential equation obtained by assigning particular values to the arbitrary constants in the general solution

$.$

Solve the following differential equations.
$y - x \dfrac{dy}{dx} = 0$

1324241_1196971_ans_ebf502b7ab984c28a32b61fbd907d1da.jpg

Find $\dfrac {dy}{dx}$
$\left( x ^ { 3 } - 2 y ^ { 3 } \right) d x + 3 x y ^ { 2 } d y = 0$

Consider the given equation.

$(x^3-2y^3)dx+3xy^2dy=0$

$(x^3-2y^3)dx=-3xy^2dy$

$\dfrac{dy}{dx}=-\dfrac{x^3-2y^3}{3xy^2}$

$\dfrac{dy}{dx}=\dfrac{2y^3-x^3}{3xy^2}$

Hence, this is the answer.

Consider the differential equation,
$\dfrac{dy}{dx}+\sin y=\cos x$ .
Find the order of the differential equation.

Given the differential equation,

$\dfrac{dy}{dx}+\sin y=\cos x$ .

Due to the presence of the highest order derivative term

$\dfrac{dy}{dx}$ which is of first order.

So the given differential equation is of order

$1$ .

Find solution of $e ^ { x } \tan y dx+ \left( 1 - e ^ { x } \right) \sec ^ { 2 } y d y = 0$ .

We have,

${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$

Now,

${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$

$\Rightarrow {{e}^{x}}\tan ydx=-\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy$

$\Rightarrow {{e}^{x}}\tan ydx=\left( {{e}^{x}}-1 \right){{\sec }^{2}}ydy$

$\Rightarrow \dfrac{{{e}^{x}}}{\left( {{e}^{x}}-1 \right)}dx=\dfrac{{{\sec }^{2}}y}{\tan y}dy$

On integration and we get,

$\int{\dfrac{{{e}^{x}}}{\left( {{e}^{x}}-1 \right)}}dx=\int{\dfrac{{{\sec }^{2}}y}{\tan y}}dy$

$\Rightarrow \log \left( {{e}^{x}}-1 \right)=\log \tan y+\log C$

Hence, this is the answer.

Find general solution of
$\dfrac {dy}{dx} = 1 + x + y + xy$

1340347_1300850_ans_68df7bc9ed9047119ad25896cfc3553c.jpg

Solve the differential equation by variable separable method: $\left( { e }^{ y }+1 \right) \cos { x } dx+{ e }^{ y }\sin { x } dy=0.$

Consider given the deferential equation,

$\left( {{e}^{y}}+1 \right)\cos xdx+{{e}^{y}}\sin xdy=0$

$\left( {{e}^{y}}+1 \right)\cos xdx=-{{e}^{y}}\sin xdy$

$\dfrac{\cos xdx}{\sin x}=-\dfrac{{{e}^{y}}}{{{e}^{y}}+1}dy$

$\cot x.dx=-\dfrac{{{e}^{y}}}{{{e}^{y}}+1}dy$

Taking integration both sides,

$\int{\cot x.dx=-\int{\dfrac{{{e}^{y}}}{{{e}^{y}}+1}dy}}$

$\log \sin x=-\log \left( {{e}^{y}}+1 \right)+\log C$

$\log \sin x=\log \dfrac{C}{\left( {{e}^{y}}+1 \right)}$

$\sin x=\dfrac{C}{\left( {{e}^{y}}+1 \right)}$

$\left( {{e}^{y}}+1 \right)\sin x=C$

Hence, this is the answer.

Solve:
$\dfrac{dy}{dx}=1+y^2$ .

Given the differential equation,

$\dfrac{dy}{dx}=1+y^2$

or,

$\dfrac{dy}{1+y^2}=dx$

Now integrating we get,

$\tan^{-1}y=x+c$

or,

$y=\tan (x+c)$ . [ Where

$c$ is integrating constant]

Find an expression for y given
$\dfrac{dy}{dx}=7x^{5}$

$\frac{dy}{dx}= 7x^{5}\Rightarrow dy = 7x^{5}dx$

Integrating both sides, we have

$\int dy= \int 7x^{5}dx\Rightarrow y=\frac{7x^{6}}{6}+c$

1115709_1201988_ans_32c6ae8b6f024c588aa2b00f2c3dc408.jpg

Solve :
$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$

The given differential equation is

$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$

$\dfrac{dy}{1+y^2}=(1+x^2)dx$

Integrating both sides, we get,

$\int \dfrac{dy}{1+y^2}=\int (1+x^2)dx$

$tan^{-1}y=x+\dfrac{x^3}{3}+C$

This is the required solution.

If $( 2 + \sin x ) \dfrac { d y } { d x } + ( y + 1 ) \cos x =0$ and $y ( 0 ) = 1,$ then $y \left( \dfrac { \pi } { 2 } \right)$ is equal to :

We have,

$\begin{array}{l} \left( { 2+\sin x } \right) \cfrac { { dy } }{ { dx } } +y\cos x=-\cos x \\ \cfrac { { dy } }{ { dx } } +y.\left( { \cfrac { { \cos x } }{ { 2+\sin x } } } \right) =\cfrac { { -\cos x } }{ { \left( { 2+\sin x } \right) } } \\ I.F={ e^{ \int { \cfrac { { \cos x } }{ { 2+\sin x } } dx } } } \\ ={ e^{ \log \left( { 2+\sin x } \right) } } \\ \Rightarrow \left( { 2+\sin x } \right) \\ y.\left( { 2+\sin x } \right) =-\int { \cos x } dx+C \\ y\left( { 2+\sin x } \right) =-\sin x+C \\ When\, \, x=0,\, \, \, y=1\, \, then \\ 2=C \\ y\left( { 2+\sin x } \right) =2-\sin x \\ then\, \, x=\cfrac { \pi }{ 2 } \\ y\times 3=2-1 \\ y=\cfrac { 1 }{ 3 } \end{array}$

Hence, this is the answer.

Find the general solution of the following differential equation.
$\dfrac{{dy}}{{dx}} = 1 + x + y + xy$

1340376_1324138_ans_58108f3f38884283bd95fcf6d901070f.jpg

Solve
$\dfrac { dy }{ dx } ={ x }^{ 2 }y+y$

$\dfrac{dy}{dx}=x^2y+y$

$\Rightarrow \dfrac{dy}{dx}=(x^2+1)y$

$\Rightarrow \dfrac{dy}{y}=(x^2+1)dx$

Integrating on both sides we get

$log y=\dfrac{x^3}{3}+x+c$

$\Rightarrow 3 log y=x^3+3x+3c$

$\Rightarrow log y^3=x^3+3x+c_1$

where

$c_1=3c$ .

1190067_1330161_ans_12aa19c2f0d84d22be1028cf012ae6a9.JPG

Solve :
$(x+y)^2\dfrac{dy}{dx}=a^2$

Let

$x+y=y=v$ . Then,

$1+\dfrac{dy}{dx}=\dfrac{dv}{dx}$

$\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$

Therefore,

$v^2(\dfrac{dv}{dx}-1)=a^2$

$v^2\dfrac{dv}{dx}=a^2+v^2$

$v^2 dv = (a^2 +v^2)dx$

$\dfrac{v^2}{v^2+a^2}dv=dx$

$(1-\dfrac{a^2}{v^2+a^2})dv=dx$

$\int dv -a^2 \int \dfrac{1}{v^2+a^2}dv = \int dx+C$

$v-a \tan^{-1}(\dfrac{v}{a}) = x+C$

$(x+y)-a \tan^{-1}(\dfrac{x+y}{a})=x+C$

Find the particular solution of the differential equation $\left( 1 + x ^ { 2 } \right) \dfrac { d y } { d x } + 2 x y = \dfrac { 1 } { 1 + x ^ { 2 } }$ given that at $x = 1 , y = 0$

We have,

$\left( { 1+{ x^{ 2 } } } \right) \cfrac { { dy } }{ { dx } } +2xy=\cfrac { 1 }{ { 1+{ x^{ 2 } } } } \\ \cfrac { { dy } }{ { dx } } +\cfrac { { 2x } }{ { 1+{ x^{ 2 } } } } y=\cfrac { 1 }{ { { { \left( { 1+{ x^{ 2 } } } \right) }^{ 2 } } } }$

Therefore,

$y.{ e^{ \int { \cfrac { { 2x } }{ { 1+{ x^{ 2 } } } } dx } } }=\int { { e^{ \int { \cfrac { { 2x } }{ { 1+{ x^{ 2 } } } } dx } } } } \times \cfrac { 1 }{ { \left( { 1+{ x^{ 2 } } } \right) } } dx \\ y\left( { 1+{ x^{ 2 } } } \right) ={ \tan ^{ -1 } }\left( x \right) +C \\ 0=\cfrac { \pi }{ 4 } +C \\ C=\cfrac { { -\pi } }{ 4 }$

Therefore,

$y\left( { 1+{ x^{ 2 } } } \right) ={ \tan ^{ -1 } }x-\cfrac { \pi }{ 4 }$

Hence, this is the answer.

Solve :
$\sin^{-1}\left (\dfrac{dy}{dx}\right )=x+y$

$sin^{-1}(\dfrac{dy}{dx})=x+y$

$\dfrac{dy}{dx}=\sin (x+y)$

Let

$x+y=v$ . Then,

$1+\dfrac{dy}{dx}=\dfrac{dv}{dx}$

$\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$

Therefore,

$\dfrac{dv}{dx}-1=\sin v$

$\dfrac{dv}{dx}=1+\sin v$

$\dfrac{dv}{1+\sin v}=dx$

$\int \dfrac{1}{1+\sin v} = \int dx$

$\int dx= \int \dfrac{1-\sin v}{1-sin^2v}dv$

$\int dx = \int \dfrac{1-\sin v}{cos^2v}dv$

$\int dx = \int (sec^2v-\tan v \sec v)dv$

$x=\tan v -\sec v +C$

$x=\tan (x+y) -\sec (x+y)+C$ , which is the required solution.

Find general solution of D.E
$\dfrac{{dy}}{{dx}} = {x^2}y + y$

We have,

$\dfrac{{dy}}{{dx}} = {x^2}y + y$

$\dfrac{{dy}}{{dx}} = (x^2 + 1)y$

$\dfrac{{dy}}{{y}} = (x^2 + 1)dx$

On taking integration both sides, we get

$\int\dfrac{{dy}}{{y}} =\int (x^2 + 1)dx$

$\ln y=\dfrac{x^3}{3}+x+C$

Hence, this is the answer.

Solve the differential equation:
$\dfrac{d^2y}{dx^2}-y=0$ .

Given the differential equation is

$\dfrac{d^2y}{dx^2}-y=0$ .

Let

$y=Ae^{mx}$ for

$A\ne 0$ be a trial solution for the given differential equation.

Then we get from the differential equation is

$m^2-1=0$ [ Auxiliary equation]

or,

$m=\pm 1$ .

So the general solution will be

$y=c_1e^x+c_2e^{-x}$ . [ Where

$c_1$ and

$c_2$ are arbitrary constants]

Solve :
$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$

$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$

$\dfrac{dy}{1+y^2}=(1+x^2)dx$

$\int \dfrac{dy}{1+y^2}=\int (1+x^2)dx$

$tan^{-1}y = \int dx + \int x^2 dx$

$tan^{-1}y=x+\dfrac{x^3}{3}+C$

Find the general solution of $\dfrac{dy}{dx} = e^{x + y}$ .

$\Rightarrow \dfrac {dy} {dx} =e^{x}. e^{y}$

$\Rightarrow e^{-y} dy =e^{x} dx$

Integrate on both sides

$\Rightarrow \dfrac {e^{-y}} {-1}=e^{x}+c$

$\Rightarrow e^{x} +e^{-y} +c=0$ is general solution

Define a ordinary differential equation.

A differential equation involving derivative w.r.t. a single independent variable is called an ordinary differential equation.

Solve the differential equation :
$y \log y dx-xdy=0$

$y\log y dx-xdy=0$

$y\log y dx=xdy$

$\dfrac{dy}{y\log y}=\dfrac{dx}{x}$

Integrating both sides,

$\int \dfrac{dy}{ylogy}=\int \dfrac{dx}{x}$ ......(1)

Let

$\log y=t$

$\dfrac{1}{y}=\dfrac{dt}{dy}$

$\dfrac{1}{y}dy=dt$

Substituting this value in equation 1, we get,

$\int \dfrac{dt}{t} =\int \dfrac{dx}{x}$

$\log t = \log x + \log C$

$log(\log y)=\log Cx$

$\log y = Cx$

$y=e^{Cx}$

This is the required general solution of the given differential equation.

Solve :
$\dfrac{dy}{dx}=e^{x+y}$

$\dfrac{dy}{dx}=e^{x+y} = e^x. e^y$

$\dfrac{dy}{e^y}=e^xdx$

$e^{-y}dy=e^xdx$

$\int e^{-y}dy=\int e^xdx$

$-e^{-y}=e^x+C$

$e^x+e^{-y}=-C$

$e^x+e^{-y}=k$

$(k=-C)$

Solve the following differential equation.
$\dfrac{dy}{dx}=x-1$ .

Given,

$\dfrac{dy}{dx}=x-1$ .

Integrating on both sides

$\displaystyle \int \dfrac {dy}{dx}=\int x-1\ dx$

$y=x^2-x +c$

Solve:
$\dfrac{{dy}}{{dx}} = \sin x - x$

Given the differential equation,

$\dfrac{{dy}}{{dx}} = \sin x - x$

Now integrating both sides we get,

$y=-\cos x-\dfrac{x^2}{2}+c$ . [ Where

$c$ is integrating constant]

Find the solution of $\dfrac{dy}{dx}=2^{y-x}$ .

$\dfrac{dy}{dx}=2^{y-x}$

$\dfrac{dy}{dx}=\dfrac{2^y}{2^x}$

$\dfrac{dy}{2^y}=\dfrac{dx}{2^x}$

Integrating both sides, we get,

$\int 2^{-y}dy=\int 2^{-x}dx$

$\dfrac{-2^{-y}}{\log 2}=\dfrac{-2^{-x}}{\log 2}+C$

$-2^{-y}+2^{-x}=C \log 2$

$2^{-x}-2^{-y}=k$ [Where k= C log 2]

Define a differential equation.

An equation involving derivatives or differentials of a function with respect to one or more independent variable is called differential equation.

Solve $\dfrac{dy}{dx}=e^{x+y}$

The given differential equation is :

$\dfrac{dy}{dx}=e^{x+y}$

$\dfrac{dy}{dx}=e^x . e^y$

$dy=e^x e^y \ dx$

$e^{-y} \ dy=e^x \ dx$

$\int e^{-y} \ dy = \int e^x \ dx$

$-e^{-y}=e^x+C$ , which is the required solution.

Solve the following differential equation:
$y^2dy-x^2dx=0$ .

$y^2dy-x^2dx=0$ .

$\Rightarrow y^2dy=x^2dx$

Integrating on both sides

$\displaystyle \int y^2 dy=\int x^2 dx$

$\dfrac{y^3}3=\dfrac {x^3}3+c$

$\Rightarrow x^3-y^3=C$

For the following differential equation, find the general solution.
$\dfrac{dy}{dx}=\sin x$

$\dfrac{dy}{dx}=\sin x\\$ .

Integrating both sides, we get

$\displaystyle \int dy=\int \sin x dx$

Using,

$\displaystyle\int \cos \ x=\sin x+C$ , we get

$y=-\cos x +c$

Solve the following differential equation.
$\dfrac{dy}{dx}=e^x$

$\dfrac{dy}{dx}=e^x$ .

Integrating on both sides

$\displaystyle \int \dfrac {dy}{dx}=\int e^x dx$

$\Rightarrow y=e^x +c$ .................Using

$\left [\int e^x \ dx=e^x+C\right ]$

For the following differential equation, find the general solution.
$x dy=dx$ .

$x dy=dx$ .

Separating the variables, we get

$dy =\dfrac 1x dx$

Integrating both sides, we get

$\displaystyle \int dy=\int \dfrac 1x dx$

$\Rightarrow y=\log x +c$ ................

$\left [\because \int\dfrac {dx}{x}=\log x+c\right ]$

Solve the differential equation :
$\dfrac{dy}{dx}=3x$

$\dfrac{dy}{dx}=3x$

Integrating on both sides

$\displaystyle \int dy=\int 3xdx$

$y=\dfrac{3x^2}2+c$

Solve the following differential equation.
$\dfrac{dy}{dx}+2x=e^{3x}$ .

$\dfrac{dy}{dx}+2x=e^{3x}$ .

Integrating on both sides

$\displaystyle \int dy +\int 2x dx =\int e^{3x}dx$

Using

$\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+C$ and

$\left (\int e^{ax} \ dx=\dfrac {e^{ax}}{a}+C\right )$

$\Rightarrow y+x^2=\dfrac 13 e^{3x}+c$

Solve the following differential equation.
$\dfrac{dy}{dx}+1=\sin x$ .

$\dfrac{dy}{dx}+1=\sin x$

$\Rightarrow \dfrac {dy}{dx}=\sin x -1$

Integrating both sides, we get

$\Rightarrow \displaystyle \int dy=\int \sin x-1 dx$

$\left [\because \int \sin x=-\cos \ x+C\right ]$

$\Rightarrow y=-\cos x -x$

$\Rightarrow x+y+\cos x =0$ .

Show that $y=e^x(A\cos x+B\sin x)$ is the solution of the differential equation $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

We have,

$y=e^x(A\cos x+B\sin x)$ ……(i)

$\Rightarrow \dfrac{dy}{dx}=e^x(A\cos x+B\cos x)+e^x(-A\sin x+B\cos x)$

$\Rightarrow \dfrac{dy}{dx}=y+e^x(-A\sin x+B\cos x)$ …….(ii)

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)+e^x(-A\cos x-B\sin x)$

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)-y$ …….(iii)

Subtracting (ii) from (iii), we get

$\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=\dfrac{dy}{dx}-2y\Rightarrow \dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

This is the given differential equation. So,

$y=e^x(A\cos x+B\sin x)$ satisfies the differential equation. Hence, it is the solution of the given differential equation.

Answer the following question in one word or one sentence or as per exact requirement of the question.
How many arbitrary constants are there in the general solution of the differential equation of order $3$ .

No. of arbitrary constant in a general solution of differential equation is equal to the order of differential equations .

So, For a differential equation of order

$3$ , there will be three

$(3)$ arbitrary constant in its general solution .

Solve the following differential eqauton.
$y^2dy-x^2dx=0$ .

Given:

$y^2dy=x^2dx$

Integrating on both sides

$\displaystyle \int y^2 dy=\int x^2 dx$

$\Rightarrow \dfrac{y^3}3=\dfrac {x^3}3+c$

$\Rightarrow x^3-y^3=c$

Find the area of the triangle with vertices at the points :
$(-1,-8),(-2,-3) \ and \ (3,2)$

Given Points are

$(-1,-8),(-2,-3),(3,2)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

$\therefore$ Area of triangle

$:-$

$=\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|-1(-3-2)-2(2+8)+3(-8+3) |$

$=\dfrac 12|5-20-15|$

$=15$

Write the order of the differential equation associated with the primitive $y=C_1+C_2e^x+C_3e^{-2x+C_4}$ , where $C_1, C_2, C_3, C_4$ are arbitrary constants.

$3$

Show that the function $y=A\cos x+B\sin x$ is a solution of the differential equation $\dfrac{d^2y}{dx^2}+y=0$ .

We have,

$y=A\cos x+B\sin x$

$\Rightarrow \dfrac{dy}{dx}=-A\sin x+B\cos x$ and

$\dfrac{d^2y}{dx^2}=-A\cos x-B\sin x$

$\Rightarrow \dfrac{d^2y}{dx^2}=-y$ or,

$\dfrac{d^2y}{dx^2}+y=0$

This is the given differential equation. So,

$y=A\cos x+B\sin x$ satisfies the given differential equation. Hence,

$y=A\cos x+B\sin x$ is the solution of the given differential equation.

Solve the following systems of linear equations
$x+y= 5$
$y+z = 3$
$x+z = 4$

$x+y= 5\cdots(1)$

$y+z = 3\cdots(2)$

$x+z = 4\cdots(3)$

$(1)+(2)$

$x+2y+z=8$

$x+z=4$

$\implies 2y=4$

$y=2\implies x=3,z=1$

Solve $\dfrac{\ln \left ( \sec x+\tan x \right )}{\cos x}dx=\dfrac{\ln \left ( \sec y+\tan y \right )}{\cos y}dy$

Given differential equation is

$\displaystyle \frac{\ln \left ( \sec x+\tan x \right )}{\cos x}dx=\frac{\ln \left ( \sec y+\tan y \right )}{\cos y}dy$

Integrating we get,

$\displaystyle \int \frac{\ln \left ( \sec x+\tan x \right )}{\cos x}dx=\int\frac{\ln \left ( \sec y+\tan y \right )}{\cos y}dy$

Put

$z=\ln { \left( \sec x+\tan x \right) }$ in LHS

$\Rightarrow \cfrac{dz}{dx}=\cfrac { \sec x.\tan x+\sec ^{ 2 } x }{ \sec x+\tan x }$

$\Rightarrow \cfrac{dz}{dx}=\cfrac { \sec x(\tan x+\sec x) }{ \sec x+\tan x }$

$\Rightarrow dz=\cfrac { dx }{ \cos x }$

Put

$t=\ln { \left( \sec y+\tan y \right) }$ in RHS

$\Rightarrow \cfrac{dt}{dx}=\cfrac { \sec y.\tan y+\sec ^{ 2 } y }{ \sec x+\tan x }$

$\Rightarrow \cfrac{dt}{dy}=\cfrac { \sec y(\tan y+\sec y) }{ \sec y+\tan y }$

$\Rightarrow dt=\cfrac { dy }{ \cos y }$

So, the differential equation becomes

$\int z\, dz=\int t \, dt$

$\Rightarrow \displaystyle \frac { { z }^{ 2 } }{ 2 } =\frac { { t }^{ 2 } }{ 2 } +C$

$\Rightarrow z^2=t^2+c$

$\Rightarrow \ln^2 \left ( \sec x+\tan x \right )=\ln^2 \left ( \sec y+\tan y \right )+c$

$f(x, y) = \sqrt{x^{2} + 2xy + 3y^{2}}$
If homogeneous enter 1 else enter 0.

Consider,

$f(x,y)=\sqrt { x^{ 2 }+2xy+3y^{ 2 } }$

$\Rightarrow f(tx,ty)=\sqrt { \left( tx \right) ^{ 2 }+2\left( tx \right) \left( ty \right) +3\left( ty \right) ^{ 2 } }$

$=\sqrt { { t }^{ 2 }x^{ 2 }+{ t }^{ 2 }2xy+{ t }^{ 2 }3y^{ 2 } }$

$=t\sqrt { x^{ 2 }+2xy+3y^{ 2 } }$

$={ t }f(x,y)$

Hence this is homogeneous function of degree 1

Hence the answer is !.

If $y_1\, \&\, y_2$ be solutions of the differential equation $\displaystyle \frac{dy}{dx}\, +\, Py\, +\, Q$ , where $P$ & $Q$ are functions of $x$ alone and $y_2\, =\, y_1 z$ , then $z\, =\, 1\, -\, a\, e^{-\displaystyle {\int \frac{Q}{y_1} dx}}$ , ' $a$ ' being an arbitrary constant. If you think this is true write 1 otherwise write 0.

$\displaystyle {\frac{dy_1}{dx}\, +\, Py_1\, =\, Q,\, \frac{dy_2}{dx}\, +\, Py_2\, =\, Q}$
Substitue

$y_2\, =\, y_1\, z$

$\Rightarrow\, \displaystyle {\frac{dy_2}{dx}\, =\, y_1\, \frac{dz}{dx}\, +\, z \frac{dy_1}{dx}}$

$\Rightarrow\, y_1\, \displaystyle {\frac{dz}{dx}\, +\, z\, \frac{dy_1}{dx}\, =\, Q\, -\, Py_2}$

$\Rightarrow\, y_1\, \displaystyle {\frac{dz}{dx}\, +\, z\, \frac{dy_1}{dx}\, +\, Py_1\, z\, =\, Q}$

$\Rightarrow\, y_1\, \displaystyle {\frac{dz}{dx}\, +\, z\, Q\, =\, Q\, \Rightarrow\, y_1\, \frac{dz}{dx}\, =\, Q(1\, -\, z)}$

$\Rightarrow\, \displaystyle {\int \frac{dz}{1\, -\, z}\, =\, \int \frac{Q}{y_1} dx}$

$\Rightarrow\, l n\, \left |z\, -\, 1 \right |\, =\, - \displaystyle {\int \frac{Q}{y_1} dx\, +\, \lambda}$

$\Rightarrow\, z\, =\, 1\, +\, a\, e^{-\, \int \dfrac{Q}{y_1}dx}$

For each of the differential equations given, find a particular solution satisfying the given condition.
1. $\cfrac{dy}{dx}+2y\,\tan\,x=\sin x:y=0$ when $x=\cfrac{\pi}{3}$
$(1+x^2)\cfrac{dy}{dx}+2xy=\cfrac{1}{1+x^2};y=0$ when $x=1$
$\cfrac{dy}{dx}-3\,y\cot\,x=\sin\,2x; y=2$ when $x=\cfrac{\pi}{2}$

$\dfrac{dy}{dx}+2y\tan x = \sin x;\, y=0$ , when

$x=\dfrac{\pi}{3}$

$4(x) = \int_e 2\tan x.dx = 2ln(x) = \sec^2x$

$\sec^2x(y^1+2\tan xy)=\sin x.sec^2x$

$(\sec^2x)^1=\sin x.\sec^2x$

$\sec62x.y = \int x.\sec^2x.dx$

$\int \sin x.\dfrac{1}{\cos^2x}.dx$

$\cos x = +$

$\sin x.dx = -d+$

$=\int \dfrac{-dt}{+2} = \dfrac{1}{+} = \dfrac{1}{\cos x}+c$

$y=\cos x+c \cot^2x$

$y(x=\dfrac{\pi}{3})=0$

$o=\cos(\dfrac{\pi}{3})+c\cos^2(\dfrac{\pi}{3})$

$o=\dfrac{1}{2}+c.\dfrac{1}{4}$

$c=2$

$y(x) = \cos (x) + \cos^2x$

$(1+x^2) \dfrac{dy}{dx}+2xy = \dfrac{1}{1+x^2}; y(x=1) = 0$

$\dfrac{dy}{dx} = \dfrac{2x}{1+x62}y=\dfrac{1}{(1+x^2)^2}$

$4(x) = \int_e\dfrac{2x}{1+x^2} .dx$

$=\int_e\dfrac{dt}{1+1} =eln (1+1)$

$=(1+x^2)$

$(1+x^2) (y^1+\dfrac{2x}{1+x^2}y)=\dfrac{1}{(1+x^2)^2}.(1+x^2)$

$((1+x^2)y)^1=\dfrac{1}{1+x^2}$

$(1+x^2)y=\int \dfrac{1}{1+x^2}.dx$

$(1+x^2)y = \tan (x)+c$

$y = \dfrac{\tan (x)}{1+n^2} + \dfrac{c}{1+x^2}$

$y(n=1)=o \Rightarrow 0=\dfrac{\dfrac{\pi}{2}}{2}+\dfrac{c}{2}$

$c=-\dfrac{\pi}{4}$

$y=\dfrac{\tan (x)}{1+x^2} - \dfrac{\pi}{4(1+x^2)}$

$\dfrac{dy}{dx} - 3y \cot x = \sin x; (y(x=\dfrac{x}{2})=2$

$4(x) = \int_e-3\cot x. dx = -3\log |\sin x| = \dfrac{1}{\sin ^3x}$

$\dfrac{1}{\sin^2x}(y^2-3y\cot x)=\dfrac{\sin$2x}{\sin^3x}$

$\left[\dfrac{y}{\sin^3z}\right]^1=\dfrac{\sin x^2x}{\sin^3x}$

$\dfrac{y}{\sin^3x} = \int \dfrac{2\cos x}{\sin x}.dx$

$\sin x = + \Rightarrow \cos x. dx = d+$

$\int \dfrac{2dt}{+2}$

$\dfrac{-2}{+} = \dfrac{-2}{\sin x}=c$

$y=-2\sin^2x+c\sin^3x$

$y(x=\dfrac{\pi}{2})=2$

$2=-2+c$

$c=4$

$y(x) = -2\sin^2x+4\sin^3x$

Find a particular solution of the differential equation $(x-y)(dx+dy)=dx-dy$ , given that $y=-1$ when $x=0$ .

let

$t=(x-y)\\ dt=dx-dy$

$\Rightarrow dx+dy = dx-dy+2dy = dt+2dy$
Put this in given equation

$t(dt+2dy)=dt\\ dt+2dy=\dfrac{dt}{t}\\ 2dy=\dfrac{dt}{t}-dt$

Integrate both sides

$2y=\ln t-t+c\\ 2y=\ln(x-y)-(x-y)+c$
Now

$y=-1 \;\; x=0$

$-2=\ln(0+1)-(0+1)+c\\ \Rightarrow c=-1\\$
Equation is

$2y=\ln(x-y)-(x-y)-1$

Find the equation of a curve passing through the point (0,-2) given that at any point (x, y) on the curve, the product of the slope of its tangent and $y$ coordinate of the point is equal to the $x$ coordinate of the point.

$\textbf{Step 1: Forming the differential equation}$

$\text{Slope of the tangent at any point }(x,y)\text{ on a curve is given by }\dfrac{dy}{dx}$

$\text{Given that product of slope tangent and y coordinate gives the x coordinate}$

$y\dfrac{dy}{dx}=x$

$\implies ydy=xdx$

$\textbf{Step 2: Solving the differential equation to find the equation of the curve}$

$\text{Integrating both sides we get}$

$\int{ydy}=\int{xdx}$

$\dfrac{y^2}{2}=\dfrac{x^2}{2}+c\;\;.........(1)$

$\text{The curve passes through }(0,-2)$

$\implies\dfrac{(-2)^2}{2}=c$

$\implies c=2$

$\therefore\text{Equation of required curve is }\dfrac{y^2}{2}=\dfrac{x^2}{2}+2$

$\implies y^2=x^2+4$

$\textbf{Hence option C is correct}$

Find the particular solution of the differential equation
$(1+e^{2x})dy+(1+y^2)e^xdx=0$ , given that $y=1$ when $x=0$ .

$(1+e^{2x})dy+(1+y^2)e^xdx=0\\ \dfrac{dy}{(1+y^2)}=- \dfrac{e^x dx}{(1+e^{2x})}\\ \tan^{-1}y=- \dfrac{e^x dx}{(1+e^{2x})}$
Let

$t=e^x\;\; dt=e^xdx\\ \tan^{-1}y=- \dfrac{dt}{(1+t^2)}\\ \tan^{-1}y=- \tan^{-1}t+c\\ \tan^{-1}y=- \tan^{-1}(e^x)+c$
When

$x=0$ and

$y=1$ ,

$\tan^{-1}1=- \tan^{-1}(e^0)+c\\ \dfrac{\pi}{4}=- \dfrac{\pi}{4}+c\\ c=\dfrac{\pi}{2}$

$\tan^{-1}y=- \tan^{-1}(e^x)+ \dfrac{\pi}{2}$

The general solution of the differential equation $\dfrac{dy}{dx}=e^{x+y}$ is

This equation can be written as

$\dfrac{dy}{dx}={e}^{x}{e}^{y}$

${e}^{-y}dy={e}^{x}dx$

Integrating both the sides, we get

$-{e}^{-y}={e}^{x}+c$

${e}^{-y}+{e}^{x}=C$

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

$y = e^x + 1$ : $y''- y = 0$
$y = x^2 + 2x + C$ : $y'- 2x -2 = 0$
$y = \cos \,x + C$ : $y' + \sin \,x = 0$
$y = \sqrt{1+ x^2}$ : $y' = \cfrac{xy}{1+x^2}$
$y = Ax$ : $xy = y (x \neq 0)$
$y = x\, \sin\, x$ : $xy = y + x \sqrt{x^2 y^2} (x \neq 0 \,and\, x > y \,or\, x < y)$
$xy = \log \,y + C$ : $y'=\cfrac{y^2}{1-xy}(xy\neq 1)$
$y-\cos \,y =x$ : $(y\, \sin \, y+\cos\, y+x)y'=y$
$x+y=\tan^{-1}y$ : $y^2y'+y^2+1=0$
$y=\sqrt{a^2-x^2} x \epsilon (-a, a)$ : $x+y\cfrac{dy}{dx} = 0(y\neq 0)$

1) Given, $y = e^x + 1$ ; $y''- y = 0$

For ex. $y’’-y=0 \Rightarrow \dfrac{d^2y}{dx^2} – y=0$

$\Rightarrow \dfrac{d^2y}{dx^2} = y$ …(1)

For this type of equation solunation is

$y=e^x+c$ $C \in R$

And it satisfies the above eqn.

So, $y=e^x+1$

2) $y'- 2x -2 = 0$

$\Rightarrow \dfrac{dy}{dx} = 2x+2$

$\Rightarrow dy=(2x+2)dx$

Integrating on both sides, we get

$\int dy = \int (2x+2)dx$

$y=x^2+2x+c, C\in R$

3) $y' + \sin \,x = 0$

$\Rightarrow \dfrac{dy}{dx} + \sin =0$

$\Rightarrow dy = \sin dx$

Integrating on both side

$\int dy = \int \sin x \,dx$

$\Rightarrow y=+\cos x+c$ $\left\{\int \sin x\, dx = \cot x\right\}$

4) $y' = \cfrac{xy}{1+x^2}$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{xy}{1=x^2}$

$\Rightarrow \dfrac{dy}{y} = \dfrac{x\, dx}{1+x^2}$

$\Rightarrow \int \dfrac{dy}{y} = \int \dfrac{xdx}{1+x^2}$

Let $x^2+1=t$

$2xdx=dt \pm xdx=\dfrac{dt}{2}$

$\Rightarrow \ln(y) = \dfrac{1}{2}$ $\displaystyle \int \dfrac{dt}{(1pt)}$

$\Rightarrow \ln (y) = \dfrac{1}{2} [ \ln (1+2)] + c, \,c \in R$

$\Rightarrow$ taking $c=0$

$\ln (y) = \displaystyle \dfrac{1}{2} \ln (1+1)=\ln (1+1)^{1/2}$

$y=(1+1)^{1/2}$

$y=\sqrt{1+x^2}$

5) $y = Ax$ ; $xy = y (x \neq 0)$

$x\dfrac{dy}{dx} = y\Rightarrow \dfrac{dy}{y} = \dfrac{dx}{x} \Rightarrow \int \dfrac{dy}{y} = \int \dfrac{dx}{x}$

$\ln (y) = \ln (x) + \ln (A) \Rightarrow \ln (y) = \ln (Ax), A \in R$

$\Rightarrow y=Ax$

9) $x+y=\tan^{-1}y$ ; $y^2y'+y^2+1=0$

$y^2y^1 = -(1+y^2)$

$y^1=\dfrac{-(1+y^2)}{y^2}$

$\dfrac{y^2y^1}{1+y^2}=(-1)$

$\displaystyle \int \dfrac{y^2}{1+y^2}dy = \int –dx$

$\Rightarrow \int \dfrac{1+y^2-1}{1+y^2}dy = -\int dx$

$=\displaystyle \int dy - \int \dfrac{1}{1+y^2}dy = -\int dx$

$\Rightarrow y - \tan ^{-1} (y) = -x+c$

$\Rightarrow x+y = \tan ^{-1} (y) + c$

10) $y=\sqrt{a^2-x^2} x \epsilon (-a, a)$ ; $x+y\cfrac{dy}{dx} = 0(y\neq 0)$

$x+y\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = \dfrac{-x}{y}$

$\int y dy = \int –x \, dx$

$\dfrac{y^2}{2} = \int –x dx$

$y^2 = a-x^2$

$y=\sqrt{a^2-x^2}, x\epsilon (-a, a)$

Find the general solution of the differential equation $\dfrac {dy}{dx} + \sqrt {\dfrac {1 - y^{2}}{1 - x^{2}}} = 0$ .

Given,

$\cfrac {dy}{dx}+\sqrt {\cfrac {1-y^2}{1-x^2}}=0$

$\cfrac{dy}{\sqrt{1-{y}^{2}}}=\cfrac{-dx}{\sqrt{1-{x}^{2}}}$

Integrating both sides, we get

${\sin}^{-1}y={\cos}^{-1}x+c$

Find a particular solution of the differential equation $(x+1)\cfrac{dy}{dx}=2\,e^{-y}-1$ , given that $y=0$ when $x=0$ .

$(x+1)\cfrac{dy}{dx}=2e^{-y}-1\\ \cfrac{dy}{2e^{-y}-1}= \cfrac{dx}{x+1}\\ \cfrac{dy}{2e^{-y}-1}=\ln(x+1)+c$
Let

$t=2e^{-y}-1$
So,

$dt=-2e^{-y}dy$
Putting these in obtained equation

$\cfrac{-dt}{t(t+1)}=\ln(x+1)+c\\ \ln(t)-\ln(t+1)=\ln(x+1)+c\\ \ln(2e^{-y}-1)-\ln(2e^{-y})=\ln(x+1)+c$
Putting

$x=0,\;\;y=0$

$\ln(1)-\ln2=\ln1+c\\ c=-\ln2$
Particular solution is

$\ln(2e^{-y}-1)-\ln(2e^{-y})=\ln(x+1)-\ln2$

Solve the differential equation:
$(\tan^{-1}y-x)dy = (1+y^2)dx$ .

On rearranging the equation becomes:

$(\tan ^{ -1 }{ y } -x)dy=(1+{ y }^{ 2 })dx\\ \Rightarrow \dfrac { dx }{ dy } +\dfrac { x }{ 1+{ y }^{ 2 } } =\dfrac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } }$ .

This is in the form

$\dfrac { dx }{ dy } +P(y).x=Q(y)$ , so it is a non homogeneous linear differential equation of the first order.

the integrating factor is:

$I.F.={ e }^{ \int { \dfrac { dy }{ 1+{ y }^{ 2 } } } }={ e }^{ \tan ^{ -1 }{ y } }$

Multiplying both sides of the equation with integrating factor, we get:

${ e }^{ \tan ^{ -1 }{ y } }(\dfrac { dx }{ dy } +\dfrac { x }{ 1+{ y }^{ 2 } } )={ e }^{ \tan ^{ -1 }{ y } }\dfrac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } \\ \Rightarrow { e }^{ \tan ^{ -1 }{ y } }\dfrac { dx }{ dy } +{ e }^{ \tan ^{ -1 }{ y } }\dfrac { x }{ 1+{ y }^{ 2 } } ={ e }^{ \tan ^{ -1 }{ y } }\dfrac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } \\ \Rightarrow \dfrac { d }{ dy } (x{ e }^{ \tan ^{ -1 }{ y } })={ e }^{ \tan ^{ -1 }{ y } }\dfrac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } }$

Integrating by parts, by taking

$\tan ^{ -1 }{ y }$ as first function and

$\dfrac { { e }^{ \tan ^{ -1 }{ y } } }{ 1+{ y }^{ 2 } }$ as second function we get:

$\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=\int { { e }^{ \tan ^{ -1 }{ y } }\dfrac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } dy } \\=\tan ^{ -1 }{ y } \int { \dfrac { { e }^{ \tan ^{ -1 }{ y } } }{ 1+{ y }^{ 2 } } dy } -\int d(\tan ^{ -1 }{ y }) { (\int { \dfrac { { e }^{ \tan ^{ -1 }{ y } } }{ 1+{ y }^{ 2 } } dy } } ) =\tan ^{ -1 }{ y } .{ e }^{ \tan ^{ -1 }{ y } }-\int \dfrac { dy }{ 1+{ y }^{ 2 } } { (\int { \dfrac { { e }^{ \tan ^{ -1 }{ y } } }{ 1+{ y }^{ 2 } } dy } ) } \\ \Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=\tan ^{ -1 }{ y } .{ e }^{ \tan ^{ -1 }{ y } }-{ e }^{ \tan ^{ -1 }{ y } }+C\Rightarrow x=\tan ^{ -1 }{ y } -1+C{ e }^{ -\tan ^{ -1 }{ y } }$

Find a particular solution of the differential equation $\dfrac{dy}{dx}+y\, \cot\,x=4x\, \text{cosec } x(x\neq 0)$ , given that $y = 0$ when $x=\dfrac{\pi}{2}$

$\dfrac{dy}{dx}+y\cot x = 4x\text{cosec }x$

$= e^{ \int {\cot xdx}}= \sin x$

$y\times \sin x = \displaystyle \int {4x\sin x.\text{cosec }xdx}$

$y\times \sin x = \displaystyle \int {4x}dx$

(since

$\sin x.\text{cosec }x=1$ )

$y\times \sin x = 2.x^2 +c$

Now

$y=0\;\; x=\dfrac{\pi}{2}$

$0=2.\cfrac{\pi^2}{4}+c$

$c=-\cfrac{\pi^2}{2}$

Particular solution is

$y\times \sin x = 2x^2 -\dfrac{\pi^2}{2}$

If $y = e^{m.{\cos}^{-1}x}$ , prove that :
$(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=m^2y$

Given :

$y=e^{{m\cos}^{-1}x}$ ...

$(i)$

Differentiating

$(i)$ , w.r.t.

$x$ , we get

$\dfrac{dy}{dx}=e^{{m\cos}^{-1}x}.m\begin{pmatrix}\dfrac{-1}{\sqrt{1-x^2}}\end{pmatrix}$

$\Rightarrow \sqrt{1-x^2}\dfrac{dy}{dx}=-my$ [Using

$(i)$ ]

$\Rightarrow (1-x^2)\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}^2=m^2y^2$

Differentiating again w.r.t.

$x$ , we get

$(1-x^2)2.\dfrac{dy}{dx}\dfrac{d^2y}{dx^2}+\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}^2.(-2x)=m^2 2y\dfrac{dy}{dx}$

$\Rightarrow (1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=m^2y$

Solve the differential equation:
$\log \left (\dfrac{dy}{dx} \right ) = 2x - 3y$

$\log \left ( \dfrac{dy}{dx} \right ) = 2x - 3y$

$\Rightarrow \dfrac{dy}{dx} = e^{2x - 3y}$

$= \dfrac{dy}{dx} = \dfrac{e^{2x}}{e^{3y}}$

$\Rightarrow e^{3y} dy = e^{2x} dx$

Integrating both sides

$\displaystyle\int e^{3y} dy = \int e^{2x} dx$

$\dfrac{e^{3y}}{3} = \dfrac{e^{2x}}{2} + c$

Find general solution of differential equation $\dfrac {dy}{dx} = \dfrac {1 + y^{2}}{1 + x^{2}}$

$\cfrac{dy}{dx} = \cfrac{1 + y^2}{1 + x^2}$

$\therefore \cfrac{dy}{1 + y^2} = \cfrac{dx}{1 + x^2}$

Integrating, we get

$\displaystyle \int \cfrac{dy}{1 + y^2} = \int \cfrac{dx}{1 + x^2}$

$\therefore tan^{-1}y = tan^{-1}x + c$

Find general solution of differential equation $\dfrac {dy}{dx} + y = 1 (y\neq 1)$ .

Comparing given differential equation with first order linear equation,

$\dfrac{dy}{dx}+Py=Q$

$P=1$ and

$Q=1$

So integrating factor,

$I.F. =e^{\int 1 dx}=e^x$

Hence solution is given by,

$y\cdot I.F=\int q\cdot I.F. dx$

$\displaystyle \Rightarrow ye^x=\int e^xdx=e^x+C$

$\Rightarrow y=1+Ce^{-x}$

Solve the differential equation: $y \dfrac {dy}{dx} = \dfrac {x}{e^{y}}$

Given,

$y \dfrac {dy}{dx} = \dfrac {x}{e^{y}}$

$\Rightarrow \displaystyle \int y\cdot e^{y}dy = \int x\ dx$

$\Rightarrow y\cdot e^{y} -\displaystyle \int 1\cdot e^{y}dy = \dfrac {x^{2}}{2} + c$

$\Rightarrow y \cdot e^{y} - e^{y} = \cfrac {x^{2}}{2} + c$

$\Rightarrow (y - 1) e^{y} = \cfrac {x^{2}}{2} + c$ .

Find the general solution of differential equation $\cfrac { dy }{ dx } +\sqrt { \cfrac { 1-{ y }^{ 2 } }{ 1-{ x }^{ 2 } } } =0$ ; $x\ne 1$ .

Given differential equation can be written as,

$\dfrac{dy}{dx}=-\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

$\Rightarrow \dfrac{dy}{\sqrt{1-y^2}}=-\dfrac{dx}{\sqrt{1-x^2}}$

Now integrate both sides,

$\Rightarrow \sin^{-1}y=-\sin^{-1}x+C$

$\Rightarrow \sin^{-1}x+\sin^{-1}y=C$

Find $y = (sin^{-1} x)^2$ , then show that $(1- x^2) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} = 2$ .

$y = (\sin^{-1}x)^2$

Differentiating both sides, we get

$y' = 2\sin^{-1}x\cdot\cfrac{1}{\sqrt{1-x^2}}$

Differentiating both sides once more, we get

$y'' = 2\cdot\cfrac{1}{\sqrt{1-x^2}}\cdot\cfrac{1}{\sqrt{1-x^2}} + 2\sin^{-1}x\cdot\left(-\cfrac12\right)\cdot\cfrac1{(1-x^2)\sqrt{1-x^2}}\cdot(-2x)$

$\Rightarrow y'' = \cfrac{2}{{1-x^2}}+ \cfrac{2x\sin^{-1}x}{(1-x^2)\sqrt{1-x^2}}$

$\Rightarrow (1-x^2)y'' = 2+ \cfrac{2x\sin^{-1}x}{\sqrt{1-x^2}}$

$\Rightarrow (1-x^2)y'' = 2+ xy'$

$\Rightarrow (1-x^2)y'' -xy'-2 = 0$

$\Rightarrow (1- x^2) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2 = 0$

Find the general solution of the differential equation $\dfrac{dy}{dx} - \dfrac{y}{x} = 0$ .

$\cfrac{dy}{dx} - \cfrac{y}{x} = 0$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{y}{x}$

$\therefore \cfrac{dy}{y} = \cfrac{dx}{x}$

Integrating both sides,

$\log y = \log x + c$ where c is an arbitrary constant.

$\Rightarrow \log y = \log kx$ ,

$k$ being greater than zero.

$\Rightarrow y = kx$ is the general solution

Solve the differential equation $x\ dy = y\ dx + \sqrt {x^{2} + y^{2}}dx$

Re-arranging the differential equation gives us

$\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\sqrt{y^{2}+x^{2}}}{x}$

$\dfrac{dy}{dx}=\dfrac{y}{x}+\sqrt{\dfrac{y^{2}+x^{2}}{x^{2}}}$
Let

$y=vx$ , then

$\dfrac{dy}{dx}=v+\dfrac{xdv}{dx}$
Hence

$v+\dfrac{xdv}{dx}=v+\sqrt{1+v^{2}}$

$\dfrac{xdv}{dx}=\sqrt{1+v^{2}}$

$\int \dfrac{dv}{\sqrt{1+v^{2}}}=\int \dfrac{dx}{x}$

$ln(v+\sqrt{1+v^{2}})=lnx+C$
Or

$ln(y+\sqrt{x^{2}+y^{2}})-lnx=lnx+c$

$ln(y+\sqrt{x^{2}+y^{2}})=2lnx+c$

Solve: $(x^2 + y^2)dx + (2xy)dy = 0$

Let

$v = \cfrac{y}{x}$

$\therefore y = vx$ and

$dy = xdv + vdx$

$\therefore (x^2 + y^2)dx - 2xydy = 0$ becomes

$(x^2 + v^2x^2)dx - 2vx^2(xdv + vdx) = 0$

$\therefore (1 + v^2)dx = 2v(xdv + vdx)$

$\therefore (1 - v^2)dx = 2vxdv$

$\therefore \cfrac{dx}{x} = \cfrac{2vdv}{1 - v^2}$

Integrating both sides, we have

$\log x = -\log (1 - v^2) + c$ where c is an arbitrary constant

$\therefore \log x = \log \left ( \cfrac{k}{1 - v^2} \right )$ where k is a constant greater than zero

$\therefore k = x(1 - v^2)$

$\therefore k = x \left (1 - \cfrac{y^2}{x^2} \right )$

i.e.

$x^2 - y^2 = kx$

Find a particular solution of the differential equation $\left( 1+{ x }^{ 2 } \right) dy+2xydx=\cot { x } dx$ , given that $y=0$ if $x=\cfrac { \pi }{ 2 }$

$\dfrac{dy}{dx}+\dfrac{2x}{1+x^{2}}y=\dfrac{cotx}{1+x^{2}}$

$I.F=e^{\int \frac{2x}{1+x^{2}}.dx}$

$=e^{ln(1+x^{2})}$

$=1+x^{2}$
Hence, multiplying the entire equation by

$1+x^{2}$ , gives us

$(1+x^{2})\dfrac{dy}{dx}+2xy=cotx$

$(1+x^{2})dy+2xy.dx=cotxdx$

$d((1+x^{2})y)=cotx.dx$

$\int d(y+x^{2}y)=\int cotx dx$

$y(1+x^{2})=ln|sinx|+C$
At

$x=\dfrac{\pi}{2}$ ,

$y=0$ .
Hence

$ln(1)+C=0$
Or

$C=0$
Therefore the particular solution of the above differential equation is

$y=\dfrac{ln|sinx|}{1+x^{2}}$

Solve the differential equation: $(xy^{2} + x)dx + (yx^{2} + y)dy = 0$

$(xy^{2}+x)dx=-(yx^{2}+y)dy$

$x(y^{2}+1)dx=-y(x^{2}+1)dy$

$\dfrac{xdx}{x^{2}+1}=\dfrac{-ydy}{y^{2}+1}$

$\dfrac{2xdx}{x^{2}+1}=\dfrac{-2ydy}{y^{2}+1}$

$d(\ln(1+x^{2}))=d(-\ln(1+y^{2}))$

$\displaystyle \int d(\ln(1+x^{2}))=\int d(-\ln(1+y^{2}))$

$\ln(1+x^{2})=-\ln(1+y^2)+c$
Or

$\ln(1+x^{2})+\ln(1+y^{2})=c$
Or

$(1+x^{2})(1+y^{2})=e^{c}$
Or

$x^{2}y^{2}+x^{2}+y^{2}+1=C$

If $y=\sin(2\sin^{-1}x)$ , then prove that: $\displaystyle\frac{dy}{dx}=2\sqrt{\displaystyle\frac{1-y^2}{1-x^2}}$ .

Given that

$y=\sin (2\sin ^{-1}x)$

$\Rightarrow \sin ^{-1}y=2\sin ^{-1}x$

Differentiation with respect to

$x$ .

$\therefore \displaystyle\frac{1}{\sqrt{1-y}}\frac{dy}{dx}=2\frac{1}{\sqrt{1-x^2}}$

$\displaystyle\frac{dy}{dx}=2\sqrt{\displaystyle\frac{1-y^2}{1-x^2}}$ .

$[hence \quad proved]$

Solve the Differential Equation, $(1 + x) y dx + (1 - y) x dy = 0$ .

Differential equation is

$(1 + x) y dx + (1 - y) x dy = 0$

$\Rightarrow (1 + x)ydx = -(1 - y)dx$

$\Rightarrow \left (\dfrac {1 - x}{x}\right )dx = -\left (\dfrac {1 - y}{y}\right )dy$

$\Rightarrow \left (\dfrac {1}{x} + 1\right ) dx = \left (\dfrac {-1}{y} + 1\right ) dy$ .

On integrating both sides

$\therefore \displaystyle \int \left (\dfrac {1}{x} + 1\right ) dx = \displaystyle \int \left (\dfrac {-1}{y} + 1\right ) dy$

$\Rightarrow \displaystyle \int \dfrac {1}{x} dx + \displaystyle \int 1 dx = -\displaystyle \int \dfrac {1}{y} dy + \displaystyle \int 1 dy$

$\Rightarrow \log_{e}x + x = -\log_{e} y + y + c$

$\Rightarrow \log_{e}x + \log_{e}y = y - x + c$

$\Rightarrow \log_{e}(x .y) = y - x + c$

$\Rightarrow x . y = e^{y - x + c}$

This is required solution.

Solve the differential equation $\left( 1+{ x }^{ 2 } \right) \dfrac { dy }{ dx } +y={ e }^{ \tan ^{ -1 }{ x } }$ .

$(1+x^{2})\dfrac{dy}{dx}+y=e^{tan^{-1}x}$

$\dfrac{dy}{dx}+\dfrac{y}{1+x^{2}}=\dfrac{e^{tan^{-1}x}}{1+x^{2}}$
Hence

$IF=e^{\int\frac{1}{1+x^{2}}dx}$

$=e^{tan^{-1}x}$
Therefore

$e^{tan^{-1}x}dy+\dfrac{e^{tan^{-1}x}y}{1+x^{2}}dx=\dfrac{e^{2tan^{-1}x}}{1+x^{2}}dx$

$d(e^{tan^{-1}x}.y)=d(\dfrac{e^{2tan^{-1}x}}{2})$
Integrating both sides

$\int d(e^{tan^{-1}x}.y)=\int d(\dfrac{e^{2tan^{-1}x}}{2})$

$e^{tan^{-1}x}.y=0.5e^{2tan^{-1}x}+c$
Or

$2y=e^{tan^{-1}x}+2ce^{-tan^{-1}x}$

Find the general solution of the differential equation
$e^{x} \tan y dx + (1 - e^{x}) \sec^{2} y\ dy = 0$ .

Given differential equation is

$e^x \tan y\ dx + (1-e^x) {\sec}^2 y\ dy=0$

$\implies (1-e^x){\sec}^2y\ dy=-e^x \tan y\ dx$

$\implies \dfrac{{\sec}^2 y}{\tan y}dy=-\dfrac{e^x}{(1-e^x)}dx$

Taking integration on both the sides we get,

$\displaystyle \int \dfrac{{\sec}^2 y}{\tan y}dy=\int \dfrac{-e^x}{(1-e^x)}dx$

$\implies \log(\tan y)=\log(1-e^x)$

$\implies \tan y=1-e^x$

$\implies y=\tan^{-1}(1-e^x)$ is the general solution of given DE.

Solve: $\dfrac {dy}{dx} =1-xy +y -x$

$\dfrac {dy}{dx} = 1-xy + y -x$

$\dfrac {dy}{dx} = \left ( 1+y \right )\left ( 1-x \right )$

$\dfrac {dy}{1+y} = \left ( 1-x \right )dx$

Upon Integrating both side we get,

$\ln \left | 1+y \right |= x-\dfrac{x^{2}}{2} + C$

$y = -1 + e^ \left ( x-\dfrac{x^{2}}{2} + C\right )$

Solve: $\displaystyle \frac{dy}{dx} = \cos^3 x \sin^2 x + x \sqrt{2x + 1}$

$\dfrac{\mathrm{d} y}{\mathrm{d} x} = \cos ^{3}\sin ^{2}+\sqrt[x]{2x+1}$

$\Rightarrow dy = (\cos ^{3}\sin ^{2}) dx +(\sqrt[x]{2x+1}) dx$

$\Rightarrow \int dy = \int \cos. \cos^{2}\sin ^{2}dx +\int\sqrt[x]{2x+1}dx$

$= \int \sin ^{2}(1-\sin ^{2})\cos dx +\int\sqrt[x]{2x+1}dx$

{.................( $I_{1}$ )................} {...............( $I_{2}$ )..........}

For $I_{1} \rightarrow$ take $\sin = u\Rightarrow \cos dx =du$

$\therefore I_{1} =\int u^{2}(1-u^{2})du = \int u^{2}du-\int u^{4}du$

$\Rightarrow I_{1} =\dfrac{\sin ^{3}}{3} - \dfrac{\sin ^{5}}{5} + c_{1}$

$I_{2} \rightarrow$ take $2x+1 =t^{2}\Rightarrow 2dx =tdt$

$\Rightarrow I_{2} = \int \dfrac{(t^{2}-1)}{2}.t.tdt= \dfrac{1}{2}\int (t^{4}-t^{2})dt$

$\Rightarrow I_{2} = \dfrac{1}{2}[\dfrac{(2x+1)^{5/2}}{5} - \dfrac{(2x+1)^{3/2}}{3} + c_{2}$

$\therefore y = I_{1} + I_{2}$

Find the particular solution of the differential equation $\displaystyle\dfrac{dy}{dx}+2y\tan x=\sin x$ , given that $y=0$ when $x=\displaystyle\dfrac{\pi}{3}$ .

Given the equation is

$\dfrac{dy}{dx} + 2y tanx = sinx$

Multiplying by

$sec^{2} x$ on both sides

$\dfrac{dy}{dy} sec^{2}(x) + 2y sec^{2} (x) tanx = sinx sec^{2} (x)$

This can be written as

$\dfrac{d(y sec^{2} (x) )}{ dx} = tan(x) sec (x)$

Thus, integrating on both the sides, we get,

$\Rightarrow y sec^{2} (x) = sec(x) + c$ is the general solution

On substituting the values,

$y=0$ when

$x=\dfrac{\pi}{3}$ we get

$c=2$

Thus the particular solution on substituting the values is

$y sec^{2} (x) = sec(x) -2$

Solve the differential equation $\cfrac { dy }{ dx } =\cfrac { y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x }$

$y =x\tan{t}$ ........

$(1)$

$dy = x\sec^2tdt +\tan tdx$

Given equation can be written as

$xdy - ydx- \sqrt{x^2+y^2}dx = 0$

Substituting

$y$ and

$dy$ in above equation we get

$x(\tan tdx +x\sec^2tdt) - (x\tan t + x\sec t)dx=0$

$x^2\sec^2tdt = x \sec tdx$

$x\sec tdt = dx$

$dx -x\sec tdt = 0$

$\dfrac{dx}{x} -\sec tdt = 0$

$ln|x|-ln|\sec t + \tan t | = ln c$

$\dfrac{x}{ \sec t + \tan t} = c$

Rationalizing we get

$\dfrac{x( \sec t - \tan t )}{ \sec ^2t - \tan ^2t} = c$

Since,

$\sec ^2t - \tan ^2t = 1$

$\Rightarrow x \sec t = \sqrt{x^2(1+ \tan ^2t)}$

$= \sqrt{x^2 + (x \tan t)^2}$

$= \sqrt{x^2 + y^2}$ ......

$(2)$

Hence,

$x \sec t - x \tan t =c$

From (1) and (2)

$\sqrt{x^2 + y^2} - y =c$

which is the solution to given differential equation.

Solve the differential equation:
$(1 + e^{2x})dy + (1 + y^{2})e^{x}dx = 0$ .

$(1+e^{2x})dy+(1+y^2)e^x dx=0$

separating

$y$ and

$x$

$\Rightarrow \dfrac{dy}{-(1+y^2)}=\dfrac{e^x}{1+e^{2x}}dx$

Integrating both side

$\displaystyle -\int\dfrac{dy}{1+y^2}=\int\dfrac{e^x dx}{1+e^{2x}}$

Let

$p=e^x$

$\Rightarrow \dfrac{dp}{dx}=e^x\Rightarrow dx=\dfrac{dp}{e^x}$

$\displaystyle \therefore -\int \dfrac{dy}{1+y^2}=\int e^x \dfrac{dp}{e^2(1+e^{2x})}=\int \dfrac{dp}{1+e^{2x}}=\int \dfrac{dp}{1+p^2}$

$(\therefore p=e^x,p^2=e^{2x})$

$\displaystyle \Rightarrow -\int \dfrac{dy}{1+y^2}=\int \dfrac{dp}{1+p^2}$

$\Rightarrow -\tan^{-1}(y)=\tan^{-1}(p)+c$

$\Rightarrow -\tan^{-1}(y)=\tan^{-1}(e^x)+c$

$\Rightarrow \boxed{\tan^{-1}(y)=-[\tan^{-1}(e^x)+c]}$

Solve the differential equation ${ (x+y) }^{ 2 }\cfrac { dy }{ dx } ={ a }^{ 2 }\quad$

Put

$x+y=t\\\Rightarrow 1+\dfrac{dy}{dx}=\dfrac{dt}{dx}$

$\Rightarrow \dfrac{dt}{dx}=\dfrac{a^2}{t^2}+1=\dfrac{a^2+t^2}{t^2}$

$\Rightarrow \displaystyle \int \dfrac{t^2}{a^2+t^2}dt=\int dx$

$\Rightarrow t-a\tan ^{-1}\dfrac{t}{a}=x+c$ , where

$c$ is any constant.

$\Rightarrow y-a\tan ^{-1}\dfrac{x+y}{a}=c$

Find the general solution of the differential equation $x\dfrac {dy}{dx} + 2y = x^{2}\log x$ .

Given:

$x\dfrac {dy}{dx} + 2y = x^{2}\log x$

$\dfrac {dy}{dx} + 2 \dfrac {1}{x}y = x\log x$
This is in the form

$\dfrac {dy}{dx} + Py = Q$
Here

$P = \dfrac {1}{x}; Q = x\log x$

$\therefore$ Integrating factor

$= IF = e^{\int Pdx} = e^{\displaystyle \int \dfrac {2}{x}dx} = e^{2\log x} = e^{\log x^{2}}$

$IF = x^{2}$

$\therefore$ Solution is

$y(IF) = \int Q(IF) + C$

$\therefore yx^{2} = \int x\log x. x^{2} + C$

$\therefore yx^{2} = \left [\log x \cdot \dfrac {x^{4}}{4} - \displaystyle \int \dfrac {x^{4}}{4}\cdot \dfrac {1}{x}dx\right ] + C$

$yx^{2} = \log x . \dfrac {x^{4}}{4} - \dfrac {1}{4} . \dfrac{x^{4}}{4} + C$

$\therefore yx^{2} = \log x . \dfrac {x^{4}}{4} - \dfrac {x^{4}}{16} + C$ .

Solve the differential equation $\cos ^{ 2 }{ x } \cfrac { dy }{ dx } +y=\tan { x }$

$\cos^{2}x\dfrac{dy}{dx}+y=\tan x$

$\dfrac{dy}{dx}+\dfrac{y}{\cos^{2}x}=\dfrac{\tan x}{\cos^{2}x}$ (1)

This is of the form :

$\dfrac{dy}{dx}+P(x)y=Q(x)$

where

$P(x)=\dfrac{1}{\cos^{2}x}=\sec^{2}x$

$Q(x)=\tan x\sec^{2}x$

Integrating factor:

$e^{\int\sec^{2} xdx}=e^{\tan x}$

Mutiplying (1) by Integrating Factor:

$e^{\tan x}\dfrac{dy}{dx}+y\sec^{2}e^{\tan x}=\tan x\sec^{2}e^{\tan x}$

$\dfrac{d(ye^{\tan x})}{dx}=\tan x\sec^{2}e^{\tan x}$

Integerating both sides,

$y\tan x=\int \tan x\sec^{2}e^{\tan x}dx$

Taking RHS

Let

$\tan x =t$

$\sec^{2}xdx=dt$

$\int \tan x\sec^{2}e^{\tan x}dx=\int te^{t}dt=t\int e^{t}+\int e^{t}=(t+1)e^{t}$

$(t+1)e^{t}=(\tan x+1)e^{\tan x}$

Hence

$y\tan x=(\tan x+1)e^{\tan x}+c$ where c is constant of integration

$y=\dfrac{(\tan x+1)e^{\tan x}}{\tan x}+c{\tan x}$

Find the order of the differential equation of the family of all the circles with their centres at the origin.

$\rightarrow$ For the family of circle general equation is

$(x-h)^2 + (y-k)^2 = r^2$

Here,

$(h,k) = (0,0)$ as their centres are at origin.

$\cdot$ For circle

$1$ lets assume radius

$r_1$ and we know centre is

$(0,0).$

$\rightarrow$ Let us assume another circle

$2$ with radius

$r_2$ and which has center

$(0,0)$

$\cdot$ Similarly many concentric circles will be formed.

(Refer the image)

For circle

$1$ ,

$x^2 + y^2 = r_1^2$

For circle

$2 , x^2 + y^2 = r_2^2$

$\rightarrow$ The differential equation we will find after differentiating any above equation w.r.t.

$x$ so we get :

$\therefore x^2 + y^2 = r_1^2$

$\therefore 2x + 2y = \dfrac {dy}{dx} = 0$

$\therefore \dfrac {dy}{dx} = \left( \dfrac {-x}{y} \right)$

Therefore, the order of differential equation is

$1.$

1166904_879612_ans_6de08126debf42dcbf4a91427bb3b61e.png

Show that $3e^x \tan y \ dx + (1 - e^x) \sec^2 y \ dy = 0$

For showing the above result, basically we have to find the solution of the given expression, for doing so we have to simplify the above equation in a such a way that function of x and y are form in either side of the equation and then integrate both sides to have the final solutions.

$3e^x \tan y \ dx + (1 - e^x) \sec^2 y \ dy = 0$ ....(1)

Now let consider

$e^{ x-1}=t$ ,...(2)

$e^{ x }dx=dt..$ .(3)

puting 2 and 3 in (1)

$\dfrac{3dt}{t}=\dfrac{1}{\cos ^{2}y*tany} dy$

$\dfrac{3dt}{t}=\dfrac{1}{\cos ^{2}y*\dfrac{siny}{cosy}} dy$

$\dfrac{3dt}{t}=\dfrac{1}{\sin y\cos y} dy$

multiply and divide by 2

$\dfrac{3dt}{t}=\dfrac{2}{2\sin y \cos y} dy$

now we know that

$2\sin y \cos y=2\sin2y$

$\dfrac{3dt}{t}=\dfrac{2}{\sin2y} dy$

$\dfrac{3dt}{t}=2\csc2 y dy$

$\int \dfrac{3dt}{t}=2\int \csc2 y dy$

integrate both sides,

$3\ln t=\ln (\csc 2y-\cot 2y)+c$ .....(4)

putting the value of t in equation (4)

$3\ln ({e^{ x-1}})=\ln (\csc 2y-\cot 2y)+c$

This is the required solution for the given expression . and which eventually showed the required expression.

The general solution of differential equation $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}$ is

$\cfrac{dy}{dx}=\cfrac{x+y}{x-y}$

Put

$y=vx$

$\cfrac{dy}{dx}=v+x\cfrac{dv}{dx}=\cfrac{1+v}{1-v}$

$x\cfrac{dv}{dx}=\cfrac{1+v}{1-v}-v=\cfrac{1+v-v+v^{2}}{1-v}=\cfrac{1+v^{2}}{1-v}$

$\cfrac{1-v}{1+v^{2}}dv=\cfrac{dx}{x}$

$\cfrac{v-1}{1+v^{2}}dv=\cfrac{-dx}{x}$

By integrating, we get

$\log(1+v^{2})-\tan^{-1}v=-\log x-\log C$

$\log(Cx(1+v^{2}))=\tan^{-1}v$

$e^{\tan ^ {-1}(\cfrac{y}{x})}=Cx(1+(\cfrac{y}{x})^{2})$

$xe^{\tan ^{-1}(\cfrac{y}{x})}=C(x^{2}+y^{2})$

Form the different equations of all concentric circles at the origin.

$(x-h)^{2}+(y-k)^{2}=r^{2}$

$Centre=(h,k), R=r$

We have centre as

$(0,0)$

Equation will be

$x^{2}+y^{2}=r^{2}$

Differentiate with respect to x

$2x+2y\cfrac{dy}{dx}=0$

$\cfrac{dy}{dx}=-\cfrac{x}{y}$

Solve $x\cos ydy=(xe^x In x+e^x)dx$

$x\cos { y } dy=(x{ e }^{ x }\ln { x } +{ e }^{ x })dx\\ \Rightarrow x\cos { y } dy=x({ e }^{ x }\ln { x } +\cfrac { { e }^{ x } }{ x } )dx\\ \Rightarrow \cos { y } dy=({ e }^{ x }\ln { x } +\cfrac { { e }^{ x } }{ x } )dx$

Integrating both sides we get

$\Rightarrow \int { \cos { y } dy } =\int { ({ e }^{ x }\ln { x } +\cfrac { { e }^{ x } }{ x } )dx } \\ \Rightarrow \int { \cos { y } dy } =\int { ({ e }^{ x }\ln { x } )dx } +\int { (\cfrac { { e }^{ x } }{ x } )dx } \\ \Rightarrow \int { \cos { y } dy } =\ln { x } \int { { e }^{ x }dx } -\int { { e }^{ x }.\cfrac { 1 }{ x } dx } +\int { \cfrac { { e }^{ x } }{ x } dx } \\ \Rightarrow \int { \cos { y } dy } =\ln { x } \int { { e }^{ x }dx } \\ \Rightarrow \sin { y } ={ e }^{ x }\ln { x } +e\\ \Rightarrow y=\sin ^{ -1 }{ ({ e }^{ x }\ln { x } +e) }$

Solve:
$\dfrac{dy}{dx}=\dfrac{y+x}{y-x}$ .

Now,

$\dfrac{dy}{dx}=\dfrac{y+x}{y-x}$

or,

$y\ dy-x\ dy=y\ dx+x\ dx$

or,

$y\ dy-x\ dx=x\ dy+y\ dx$

or,

$y\ dy-x\ dx= d(xy)$

Now integrating both sides we get,

$\dfrac{y^2}{2}-\dfrac{x^2}{2}=xy+\dfrac{c}{2}$ [ Where

$c$ is integrating constant]

or,

$y^2-x^2=2xy+c$ .

Solve: $x\left( x-1 \right) \cfrac { dy }{ dx } -\left( x-2 \right) y={ x }^{ 3 }\left( 2x-1 \right)$

Given,

$x(x-1)\dfrac{dy}{dx}-(x-2)y=x^3(2x-1)$

$\dfrac{dy}{dx}-\dfrac{(x-2)}{x(x-1)}y=\dfrac {x^2(2x-1)}{(x-1)}$

$\dfrac{dy}{dx}+Py=Q$

$P=-\dfrac{(x-2)}{x(x-1)}\rightarrow I.F=e^{\int P dx}=e^{\int -\frac{(x-2)}{x(x-1)} dx}=e^{\log \frac{x-1}{x^2}}=\dfrac{x-1}{x^2}$

$Q=\dfrac {x^2(2x-1)}{(x-1)}$

$y \times I.F =\int Q \times I.F dx$

$y \times \left ( \dfrac{x-1}{x^2} \right )=\int \dfrac {x^2(2x-1)}{(x-1)} \times \dfrac{x-1}{x^2} dx$

$y\left ( \dfrac{x-1}{x^2} \right )=\int (2x-1)dx$

$y\left ( \dfrac{x-1}{x^2} \right )=x(x-1)+c$

$y=x^3+C$

Solve:
$y-x\dfrac{dy}{dx}=a\left(y^2+\dfrac{dy}{dx}\right)$

$y-x\dfrac{dy}{dx}=a\left(y^2+\dfrac{dy}{dx}\right)$

$\Rightarrow y-ay^2=(x+1)\dfrac{dy}{dx}$

$\Rightarrow \dfrac{dx}{x+1}=\dfrac{dy}{y-ay^2}$

$\Rightarrow In (x+1)=\displaystyle \int \dfrac{dy}{y(1-ay)}+C$

$\Rightarrow In(x+1)=\displaystyle \int \dfrac{dy}{y}+\displaystyle \int \dfrac{ady}{1-ay}+C$

$\Rightarrow In (x+1)In y +a(-1)In (1-ay)$

$\Rightarrow (x+1)=In \dfrac{y}{(1-ay)^a}+C$

$\Rightarrow x+1=\dfrac{Ky}{(1-ay)^a}$

Solve:
$\dfrac{dy}{dx}=\dfrac{x^2+2}{y}$ .

Now,

$\dfrac{dy}{dx}=\dfrac{x^2+2}{y}$

or,

$y\ dy=x^2\ dx+2\ dx$

Now integrating we have,

$\dfrac{y^2}{2}=\dfrac{x^3}{3}+2x+c$ [ Where

$c$ is integrating constant].

Solve:
$\dfrac{dy}{dx}=\dfrac{x-y}{x+1}$ .

Now,

$\dfrac{dy}{dx}=\dfrac{x-y}{x+1}$

or,

$x\ dy+dy=x\ dx-y\ dx$

or,

$x\ dy+y\ dx+dy=x\ dx$

or,

$d(xy)+dy=x\ dx$

Now integrating we get,

$xy+y=\dfrac{x^2}{2}+c$ [ Where

$c$ is integrating constant]

Solve: $\frac{dy}{dx} = \frac{x(2y -x)}{x(2y +x)}$ , if $y =1$ when $x= 1$

Given,

$\dfrac{dy}{dx}=\dfrac{x(2y-x)}{x(2y+x)}$

$\dfrac{dy}{dx}=\dfrac{(2y-x)}{(2y+x)}$

substitute

$y=vx\rightarrow \dfrac{dy}{dx}=x\dfrac{dv}{dx}+v$

$x\dfrac{dv}{dx}+v=\dfrac{(2vx-x)}{(2vx+x)}$

$x\dfrac{dv}{dx}=\dfrac{(2v-1)}{(2v+1)}-v$

$x\dfrac{dv}{dx}=\dfrac{(-2v^2+v-1)}{(2v+1)}$

$\dfrac{(2v+1)}{(-2v^2+v-1)}dv=\dfrac{1}{x}dx$

integrating on both sides, we get,

$\int \dfrac{(2v+1)}{(-2v^2+v-1)}dv=\int \dfrac{1}{x}dx$

$-\dfrac{1}{2}v^2+v-\log (2v+1)+\dfrac{5}{8}=\log x+c$

$-\dfrac{1}{2}\left ( \dfrac{y}{x} \right )^2+\dfrac{y}{x}-\log \left ( 2\dfrac{y}{x}+1 \right )+\dfrac{5}{8}=\log x+c$

put

$y=1=x$ , we get,

$-\dfrac{1}{2}\left ( \dfrac{1}{1} \right )^2+\dfrac{1}{1}-\log \left ( 2\dfrac{1}{1}+1 \right )+\dfrac{5}{8}=\log 1+c$

$-\dfrac{1}{2}+1+\dfrac{5}{8}=c$

$\therefore c=\dfrac{9}{8}$

$-\dfrac{1}{2}\left ( \dfrac{y}{x} \right )^2+\dfrac{y}{x}-\log \left ( 2\dfrac{y}{x}+1 \right )+\dfrac{5}{8}=\log x+\dfrac{9}{8}$

$\therefore -\dfrac{1}{2}\left ( \dfrac{y}{x} \right )^2+\dfrac{y}{x}-\log \left ( 2\dfrac{y}{x}+1 \right )=\log x+\dfrac{1}{2}$

A man, $2m$ tall, walks ar the rate of $1\frac{2}{3}m/s$ towards a street light which is $5\frac{1}{3}m$ above the ground. At what rates is the tip of his shadow moving ? At what is the length of the shadow changing when he is $3\dfrac{1}{3}m$ from the base of the light?

Let

$AB$ be the lamp-post. Let at any time

$t,$ the man

$CD$ be at a distance

$x$ meters from the lamp-post, and

$y$ meters be the length of his shadow

$DE$ . It is given that

$\dfrac{dx}{dt}=1\dfrac{2}{3}m/s$

From figure it is clear

$\triangle ABE$ and

$\triangle CDE$ are similar

$\therefore$

$\dfrac{AB}{CD}=\dfrac{BE}{DE}=\dfrac{AE}{CE}$ ----- ( 1 )

$AB=5\dfrac{1}{3}m$ and

$CD=2m$ [ Given ]

Let

$BD=x$ and

$DE=y$

$\Rightarrow$

$AE=x+y$

Now, substituting the values we get,

$\dfrac{5\dfrac{1}{3}}{2}=\dfrac{x+y}{y}$

$\Rightarrow$

$\dfrac{16}{6}=\dfrac{x+y}{y}$

$\Rightarrow$

$16y=6x+6y$

$\Rightarrow$

$10y=6x$ ------ ( 2 )

Differentiating w.r.t we get,

$10\dfrac{dy}{dt}=6\dfrac{dx}{dt},$ but

$\dfrac{dx}{dt}=1\dfrac{2}{3}=\dfrac{5}{3}$

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{6}{10}\times \dfrac{5}{3}$

$\Rightarrow$

$\dfrac{dy}{dx}=1$

Hence, the shadow increases at the rate of

$1m/s.$

$AB=5\dfrac{1}{3}m$ and

$CD=2m$

To find the rate at which the tip of the shadow

$E$ moves,

we have to find the rate at which

$BE=x+y$ moves.

Let

$BE=x+y=p$

From ( 1 ),

$\dfrac{AB}{CD}=\dfrac{BE}{DE}$

$\Rightarrow$

$\dfrac{8}{3}=\dfrac{x+y}{y}=\dfrac{p}{y}$

$\Rightarrow$

$\dfrac{8}{3}y=p$

Differentiating w.r.t

$t$ bothe sides we get,

$\Rightarrow$

$\dfrac{8}{3}\dfrac{dy}{dt}=\dfrac{dp}{dt}$

But

$\dfrac{dy}{dt}=1$

$\therefore$

$\dfrac{dp}{dt}=\dfrac{8}{3}$

The rate at which the tip of the shadow moves is

$\dfrac{8}{3}m/s$

1339523_1048626_ans_03b42aa7fe2c4a2fb829f9fc6b4f568d.png

The solution of the differential equation $x\dfrac{{dy}}{{dx}} = - \dfrac{y}{2} - \dfrac{{\sin 2x}}{{2y}}$ is given by

The differential equation can be rewritten as

$2y\dfrac{dy}{dx}x=-y^2-\sin2x$

$\Rightarrow y^2+2y\dfrac{dy}{dx}=-\sin2x$

$\Rightarrow \dfrac{d}{dx}(xy^2)=\dfrac{d}{dx}(\cos^2x)$

$\Rightarrow xy^2=\cos^2x+c$

$x\sqrt {1+y} +y\sqrt {1+x}=0$ , prove that $\dfrac { dy }{ dx } =\dfrac { -1 }{ { \left( 1+x \right) }^{ 2 } }$

$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$

$\Rightarrow x\sqrt {1 + y} = - y\sqrt {1 + x}$

$\Rightarrow {x^2}(1 + x) = {y^2}(1 + x)$

$\Rightarrow {x^2} + {x^2}y = {y^2} + x{y^2}$

$\Rightarrow {x^2} - {y^2} = x{y^2} - {x^2}y$

$\Rightarrow \left( {x - y} \right)\left( {x + y} \right) = xy\left( {y - x} \right)$

$\Rightarrow \left( {x + y} \right)\left( {x - y} \right) = - xy\left( {x - y} \right)$

$\Rightarrow x + y = - xy$

$\Rightarrow x + xy + y = 0$

$\Rightarrow y\left( {x + 1} \right) = - x$

$\Rightarrow y = \cfrac{{ - x}}{{x + 1}}$

$\Rightarrow \cfrac{{dy}}{{dx}} = - \left[ {\cfrac{{1.\left( {x + 1} \right) - x \times 1}}{{{{\left( {x + 1} \right)}^2}}}} \right]$

$\Rightarrow \cfrac{{dy}}{{dx}} = - \left[ {\cfrac{{x + 1 - x}}{{{{\left( {x + 1} \right)}^2}}}} \right]$

$\Rightarrow \cfrac{{dy}}{{dx}} = -\cfrac{1}{{{{\left( {x + 1} \right)}^2}}}$

Solve $e^{y}(x+1)=1$

$e^y(x+1)=1$

$\Rightarrow e^y=\dfrac{1}{x+1}$ .....

$(1)$

Differentiate on both sides

$\Rightarrow e^y\dfrac{dy}{dx}=\dfrac{-1}{(x+1)^2}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{e^y(x+1)^2}$

Substituting the value of

$y$ from

$(1)$ in the above equation, we get

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-(x+1)}{(x+1)^2}$

$Therefore, \Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{x+1}$

Find the solution of $\dfrac{dy}{dx} = 1 +y$ .

Given the differential equation is

$\dfrac{dy}{dx} = 1 +y$

or,

$\dfrac{dy}{1+y}=dx$

Now integrating both sides we get,

$\log|(1+y)|=x-\log|c|$ [ Where

$c$ is integrating constant]

or,

$1+y=ce^x$

or,

$y=ce^x-1$ .

$y - x\frac{{dy}}{{dx}} = 5\left( {{y^2} + \frac{{dy}}{{dx}}} \right)$

$y-5y^2$ =

$\cfrac{dy}{dx}(5+x)$

$\cfrac{dx}{5+x} = \cfrac{dy}{y-5y^2}$

$\cfrac{dx}{5+x} = -\cfrac{dy}{y^2(5-\cfrac{1}{y})}$

Integrating both side we get

$ln|5+x| = -ln(|\cfrac{1}{y} - 5|) + c$

$\dfrac{dy}{dx} + \dfrac{y}{x} = xy^2\sin x$

$\dfrac{dy}{dx} + \dfrac{y}{x} = xy^2 \sin x$

$\dfrac{1}{y^2} \dfrac{dy}{dx} + \dfrac{1}{xy} = x \sin x$

Let

$\dfrac{1}{y} = t$

$\dfrac{-1}{y^2} \dfrac{dy}{dx} = \dfrac{dt}{dx}$

$-\dfrac{dt}{dx} + \dfrac{t}{x} = x \sin x$

$\dfrac{dt}{dx} + \left(\dfrac{-1}{x}\right) t = x \sin x$

$I.f = e^{-\int \dfrac{1}{x} dx}$

$= e^{-ln x}$

$= \dfrac{1}{x}$

$\dfrac{t}{x} = \displaystyle \int x \sin x. \dfrac{1}{x} dx$

$\dfrac{t}{x} = -\cos x + c$

$\dfrac{1}{xy} = -\cos x + c$

$\dfrac{1}{xy} + \cos x = c$

Solve:
$x^{e}+e^{x}+e^{e}$

Consider given the expression,

Let,

$y={{x}^{c}}+{{e}^{x}}+{{e}^{e}}$

Differentiate with respect to x,

$\dfrac{d}{dx}y=\dfrac{d}{dx}\left( {{x}^{c}}+{{e}^{x}}+{{e}^{e}} \right)$

$=c.{{x}^{c-1}}+{{e}^{x}}+0$

$\dfrac{dy}{dx}=c.{{x}^{c-1}}+{{e}^{x}}$

Hence, this is the answer.

Find the general solution of the differential equation $\left( {1 + {y^2}} \right)+\left( {x - {e^{\text{tan} ^{ - 1}y}}} \right)\frac{{dy}}{{dx}} = 0$

Given,

$(1+y^2)(x-e^{\tan ^{-1}y})\dfrac{dy}{dx}=0$

$(1+y^2)=(e^{\tan ^{-1}y}-x)\dfrac{dy}{dx}$

$\dfrac{1}{1+y^2}=\dfrac{1}{e^{\tan ^{-1}y}-x}\dfrac{dx}{dy}$

$\dfrac{e^{\tan ^{-1}y}}{1+y^2}-\dfrac{x}{1+y^2}=\dfrac{dx}{dy}$

$x \times I.F = \int Q\times I.F dy$

$I.F=e^{\int \frac{1}{1+y^2}dy}=e^{\tan ^{-1}y}$

$x\times e^{\tan ^{-1}y}=\int \dfrac {e^{\tan {-1}y}}{1+y^2}e^{\tan ^{-1}y}dy$

substitute

$\tan {-1}y=t$

$xe^t=\int e^t\times e^tdt$

$xe^t=\int e^{2t}dt$

$xe^t=\dfrac{e^{2t}}{2}+c$

$x=\dfrac{e^{t}}{2}+ce^{-t}$

$x=\dfrac{e^{\tan {-1}y}}{2}+ce^{-\tan {-1}y}$

Solve $(x^{3} + x^{2} + x + 1)\dfrac {dy}{dx} = 2x^{2} + x$ if the value of $y = 1$ when $x = 0$ .

Given,

$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$

$\dfrac{dy}{dx}=\dfrac{2x^2+x}{x^3+x^2+x+1}$

$\dfrac{dy}{dx}=\dfrac{2x^2+x}{(x+1)(x^2+1)}$

$\dfrac{2x^2+x}{(x+1)(x^2+1)}=\dfrac{A}{x+1}+\dfrac{B}{x^2+1}$

$2x^2+x=A(x^2+1)+(Bx+C)(x+1)$

when

$x=-1\rightarrow A=\dfrac{1}{2}$

when

$x=0\rightarrow C=-\dfrac{1}{2}$

when

$x=1\rightarrow B=\dfrac{3}{2}$

$\dfrac{2x^2+x}{(x+1)(x^2+1)}=\dfrac{1}{2(x+1)}+\dfrac{3x-1}{2(x^2+1)}$

$\therefore \dfrac{dy}{dx}=\dfrac{1}{2(x+1)}+\dfrac{3x-1}{2(x^2+1)}$

Integrating on both sides, we get,

$\int \dfrac{dy}{dx}=\int \dfrac{1}{2(x+1)}+\int \dfrac{3x-1}{2(x^2+1)}$

$y=\dfrac{1}{2}\log (x+1)+\dfrac{3}{4}\log (x^2+1)-\dfrac{1}{2}\tan ^{-1}x+c$

$x=0,y=1$

$1=\dfrac{1}{2}\log (0+1)+\dfrac{3}{4}\log (0^2+1)-\dfrac{1}{2}\tan ^{-1}0+c$

$1=c$

$\therefore y=\dfrac{1}{2}\log (x+1)+\dfrac{3}{4}\log (x^2+1)-\dfrac{1}{2}\tan ^{-1}x+1$

Find the order and degree of
$y''' + y^2 + e^{y^1} = 0$

$y^m + y^2 + e^{y^1} = 0$

$y^m + y^2 + e^{\left(\dfrac{dy}{dx}\right)}$

Order

$=$ degree

$=$

$1$

Integrate for y : $(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$ .

Given,

$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$

$dy=\dfrac{2x^2+x}{x^3+x^2+x+1}dx$

integrating on both sides, we get,

$\int dy=\int \dfrac{2x^2+x}{x^3+x^2+x+1}dx$

$y=\int \dfrac{2x^2+x}{x^3+x^2+x+1}dx$

$y=\int \dfrac{2x^2+x}{(x+1)(x^2+1)}dx$

$\dfrac{2x^2+x}{(x+1)(x^2+1)}dx=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+1}$

$2x^2+x=A(x^2+1)+(Bx+C)=(x+1)$

by putting

$x=-1$ we get,

$A=\dfrac{1}{2}$

by putting

$x=1$ we get,

$B=\dfrac{3}{2}$

by putting

$x=0$ we get,

$C=-\dfrac{1}{2}$

$\therefore y=\dfrac{1}{2(x+1)}+\dfrac{3x-1}{2(x^2+1)}$

$\Rightarrow y=\dfrac{1}{2}\log (x+1)+\dfrac{3}{4}\log (x^2+1)-\dfrac{1}{2}\tan ^{-1}x+c$

Solve: $ydx-xdy=xydx$ .

Given,

$ydx-xdy=xydx$

$\dfrac{dx}{x}-\dfrac{dy}{y}=dx$

integrating on both sides, we get,

$\log x -\log y =x +\log c$

$\log x -\log y =\log e^x+\log c$

$\log \left ( \dfrac{x}{y} \right )=\log (ce^x)$

$\therefore \dfrac{x}{y}=ce^x$

$x^2(y+1)dx+y^2(x-1)dy=0$ .

${ x }^{ 2 }(y+1)dx+{ y }^{ 2 }(x-1)dy=0\\ \Rightarrow { x }^{ 2 }(y+1)dx=-{ y }^{ 2 }(x-1)dy\\ \Rightarrow (x+\cfrac { x }{ x-1 } )dx=(-y+\cfrac { y }{ y+1 } )dy\\ \Rightarrow \int { (x+\cfrac { x }{ x-1 } )dx } =\int { (-y+\cfrac { y }{ y+1 } )dy } \\ \Rightarrow \int { (x+1+\cfrac { 1 }{ x-1 } )dx } =\int { (-y+1-\cfrac { 1 }{ y+1 } )dy } \\ \Rightarrow \cfrac { { x }^{ 2 } }{ 2 } +x+\log { (x-1) } =\cfrac { -{ y }^{ 2 } }{ 2 } +y-\log { (y+1) } +C$

If $y=y(x)$ and $\cfrac { 2+\sin { x } }{ y+1 } \left( \cfrac { dy }{ dx } \right) =-\cos { x } ,y(0)=1$ , then $y\left( \cfrac { \pi }{ 2 } \right)$ equals:

$\cfrac { dy }{ y+1 } =-\cfrac { \cos { x } }{ 2+\sin { x } } dx$

$2+\sin { x } =y$

$\cos { x } dx=dt$

$\Rightarrow \log { \left( y+1 \right) } =-\int { \cfrac { dt }{ t } }$

$\log { \left( y+1 \right) } =-\log { \left( t \right) } +C$

$\Rightarrow \log { \left( y+1 \right) } =-\log { \left( 2+\sin { x } \right) } +C$

$\Rightarrow \log { \left( y+1 \right) } =\log { \left( \cfrac { 1 }{ 2+\sin { x } } \right) } +C$

when

$x=0,y=1$

$\Rightarrow \log { \left( 2 \right) } =\log { \left( \cfrac { 1 }{ 2 } \right) } +C$

$\Rightarrow C=-2\log { \left( \cfrac { 1 }{ 2 } \right) } =2\log { 2 }$

$\therefore$ when

$x=\cfrac { \pi }{ 2 }$

$\Rightarrow \log { \left( y+1 \right) } =\log { \left( \cfrac { 1 }{ 2+1 } \right) } +2\log { 2 }$

$\Rightarrow \log { \left( y+1 \right) } =\log { \left( \cfrac { 4 }{ 3 } \right) }$

$\Rightarrow \log { \left( y+1 \right) } -\log { \left( \cfrac { 4 }{ 3 } \right) } =0$

$\Rightarrow \log { \left( \cfrac { 3 }{ 4 } \left( y+1 \right) \right) } =0$

$\Rightarrow$ Taking anti log

$\Rightarrow \cfrac { 3 }{ 4 } \left( y+1 \right) =1$

$\Rightarrow 3y+3=4$

$\Rightarrow y=\cfrac { 1 }{ 3 }$

$\therefore y\left( \cfrac { \pi }{ 2 } \right) =\cfrac { 1 }{ 3 } =\cfrac { 1 }{ 3 }$

The solution of the equation $(2y-1)dx-(2x+3)dy=0$ is

Given

$(2y-1)dx-(2x+3)dy=0$

Separating the variables we get

$\dfrac{dy}{(2y-1)}=\dfrac{dx}{2x+3}$

On integration, we get

$\int{\dfrac{dy}{(2y-1)}}=\int{\dfrac{dx}{2x+3}}$

$\Rightarrow \dfrac{1}{2}\log(2y-1)=\dfrac{1}{2}\log(2x+3)+\log c$

$\Rightarrow \log(2y-1)=\log 2c(2x+3)$

$\Rightarrow \dfrac{2y-1}{2x+3}=2c$

$\Rightarrow \dfrac{2x+3}{2y-1}=\dfrac{1}{2c}=k$

where

$k=\dfrac{1}{2c}$

Hence,the solution of the equation is

$\dfrac{2x+3}{2y-1}=k$

Solve : $\dfrac{dy}{dx} + \dfrac{3x^2}{1 + x^3} y = \dfrac{1 + x^2}{1 + x^3}$

Given,

$\dfrac{dy}{dx}+\dfrac {3x^2}{1+x^3}y=\dfrac{1+x^2}{1+x^3}$

$\dfrac{dy}{dx}+Py=Q$

$P=\dfrac {3x^2}{1+x^3},Q=\dfrac{1+x^2}{1+x^3}$

$I.F=e^{\int Pdx}=e^{\int \frac {3x^2}{1+x^3}dx}=e^{\log |1+x^3|}=1+x^3$

$y \times I.F = \int Q\times I.F dx$

$y \times(1+x^3)=\int \dfrac{1+x^2}{1+x^3} \times (1+x^3)dx$

$y(1+x^3)=\int (1+x^2)dx$

$y(1+x^3)=x+\dfrac{x^3}{3}+c$

Solve: $(1+y^2)\dfrac{dx}{dy}+x =e^{-\tan^{-1}}y$

Given,

$(1+y^2)\dfrac{dx}{dy}+x=e^{-\tan ^{-1}y}$

$\dfrac{dx}{dy}+\dfrac{1}{1+y^2}x=\dfrac{e^{-\tan ^{-1}y}}{1+y^2}$

$\dfrac{dx}{dy}+Px=Q$

$P=\dfrac{1}{1+y^2}\rightarrow I.F=e^{\int P dy}=e^{\int \frac{1}{1+y^2} dy}\Rightarrow I.F=e^{\tan ^{-1}y}$

$Q=\dfrac{e^{-\tan ^{-1}y}}{1+y^2}$

$x \times I.F = \int Q \times I.F dy$

$x \times e^{\tan ^{-1}y}=\int \dfrac{e^{-\tan ^{-1}y}}{1+y^2} e^{\tan ^{-1}y} dy$

$x e^{\tan ^{-1}y}=\int \dfrac{1}{1+y^2} dy$

$x e^{\tan ^{-1}y}=e^{\tan ^{-1}y}+c$

$\dfrac{{dy}}{{dx}} = {x^2}\left( {x - 2} \right)$
Given $y=2$ ; when $x=0$ .

We have,

$\dfrac{dy}{dx}={{x}^{2}}\left( x-2 \right)$

$\dfrac{dy}{dx}={{x}^{3}}-2{{x}^{2}}$

$dy=\left( {{x}^{3}}-2{{x}^{2}} \right)dx$

On taking integral both sides, we get

$\int{dy}=\int{\left( {{x}^{3}}-2{{x}^{2}} \right)dx}$

$y=\dfrac{{{x}^{4}}}{4}-\dfrac{2{{x}^{3}}}{3}+c$

At $x=0,y=2$

So,

$2=0-0+c$

$c=2$

Therefore, the general solution

$y=\dfrac{{{x}^{4}}}{4}-\dfrac{2{{x}^{3}}}{3}+2$

Hence, this is the answer.

Solve the differential equation $({e^x} + 1)ydy + (y + 1)dx = 0$

We have,

$\left( {{e}^{x}}+1 \right)ydy+\left( y+1 \right)dx=0$

$\left( {{e}^{x}}+1 \right)ydy=-\left( y+1 \right)dx$

$\dfrac{y}{\left( y+1 \right)}dy=-\dfrac{1}{\left( {{e}^{x}}+1 \right)}dx$

On taking integration both sides, we get

$\int{\dfrac{y}{\left( y+1 \right)}dy}=-\int{\dfrac{1}{\left( {{e}^{x}}+1 \right)}dx}$

$\int{\dfrac{y+1-1}{\left( y+1 \right)}dy}=-\int{\dfrac{1}{\left( {{e}^{x}}+1 \right)}dx}$

$\int{1dy}-\int{\dfrac{1}{\left( y+1 \right)}dy}=-\int{\dfrac{{{e}^{x}}}{{{e}^{x}}\left( {{e}^{x}}+1 \right)}dx}$ ……. (1)

Let $I=-\int{\dfrac{{{e}^{x}}}{{{e}^{x}}\left( {{e}^{x}}+1 \right)}dx}$

Let $t={{e}^{x}}$

$dt={{e}^{x}}dx$

Therefore,

$I=-\int{\dfrac{1}{t\left( t+1 \right)}dt}$

$I=-\int{\left( \dfrac{1}{t}-\dfrac{1}{t+1} \right)dt}$

$I=-\left( \ln \left( t \right)-\ln \left( t+1 \right) \right)$

$I=\ln \left( t+1 \right)-\ln \left( t \right)$

$I=\ln \left( {{e}^{x}}+1 \right)-\ln \left( {{e}^{x}} \right)$

$I=\ln \left( \dfrac{{{e}^{x}}+1}{{{e}^{x}}} \right)$

From equation (1),

$\int{1dy}-\int{\dfrac{1}{\left( y+1 \right)}dy}=\ln \left( \dfrac{{{e}^{x}}+1}{{{e}^{x}}} \right)$

$y-\ln \left( y+1 \right)=\ln \left( \dfrac{{{e}^{x}}+1}{{{e}^{x}}} \right)+C$

Hence, this is the answer.

Solve: $\dfrac{{dy}}{{dx}} + 2y = \sin x$

We have,

$\dfrac{dy}{dx}+2y=\sin x$ ……. (1)

Since, $P=2,Q=\sin x$

Integrating factor,

$I.F={{e}^{\int{Pdx}}}$

$I.F={{e}^{\int{2dx}}}$

$I.F={{e}^{2x}}$

Therefore, the general solution

$y\times I.F=\int{Q\times I.Fdx}$

$y\times {{e}^{2x}}=\int{\sin x\times {{e}^{2x}}dx}$

We know that

$\int{{{e}^{ax}}\sin bxdx}=\dfrac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\sin bx-b\cos bx \right)+C$

Thus,

$y\times {{e}^{2x}}=\dfrac{{{e}^{2x}}}{{{2}^{2}}+{{1}^{2}}}\left( 2\sin x-\cos x \right)+C$

$y\times {{e}^{2x}}=\dfrac{{{e}^{2x}}}{5}\left( 2\sin x-\cos x \right)+C$

Hence, this is the answer.

The solution of $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + xy = 5x$

Consider the given equation.

$\left( 1+{{x}^{2}} \right)\dfrac{dy}{dx}+xy=5x$

$\dfrac{dy}{dx}+\dfrac{xy}{\left( 1+{{x}^{2}} \right)}=\dfrac{5x}{\left( 1+{{x}^{2}} \right)}$

We know that the general equation

$\dfrac{dy}{dx}+Py=Q$

So,

$P=\dfrac{x}{1+{{x}^{2}}},Q=\dfrac{5x}{1+{{x}^{2}}}$

We know that the integrating factor,

$I.F={{e}^{\int{Pdx}}}$

$I.F={{e}^{\int{\dfrac{x}{1+{{x}^{2}}}dx}}}$

Let $t=1+{{x}^{2}}$

$\dfrac{dt}{dx}=0+2x$

$\dfrac{dt}{2}=xdx$

Therefore,

$I.F={{e}^{\dfrac{1}{2}\int{\dfrac{1}{t}dt}}}$

$I.F={{e}^{\dfrac{1}{2}\left( \ln t \right)}}$

$I.F={{e}^{\dfrac{1}{2}\left( \ln \left( 1+{{x}^{2}} \right) \right)}}$

$I.F={{e}^{\left( \ln \left( \sqrt{1+{{x}^{2}}} \right) \right)}}$

$I.F=\sqrt{1+{{x}^{2}}}$

We know that the general solution,

$y\times I.F=\int{Q\left( I.F \right)}dx+C$

$y\times \sqrt{1+{{x}^{2}}}=\int{\dfrac{5x}{1+{{x}^{2}}}\times \left( \sqrt{1+{{x}^{2}}} \right)}dx+C$

$y\times \sqrt{1+{{x}^{2}}}=5\int{\dfrac{x}{\sqrt{1+{{x}^{2}}}}}dx+C$

Let $u=1+{{x}^{2}}$

$\dfrac{du}{dx}=0+2x$

$\dfrac{du}{2}=xdx$

Therefore,

$y\times \sqrt{1+{{x}^{2}}}=\dfrac{5}{2}\int{\dfrac{du}{\sqrt{u}}}+C$

$y\times \sqrt{1+{{x}^{2}}}=\dfrac{5}{2}\left( 2\sqrt{u} \right)+C$

On putting the value of $u$ , we get

$y\times \sqrt{1+{{x}^{2}}}=5\sqrt{1+{{x}^{2}}}+C$

$y-5=\dfrac{C}{\sqrt{1+{{x}^{2}}}}$

Hence, this is the answer.

If $dx + dy = (x+y)(dx-dy) \Rightarrow y+ \log (x+y)=x+C$ is the solution of differential equation

Given,

$dx+dy=(x+y)(dx-dy)$

$-(x+y-1)dx=-(x+y+1)dy$

$\dfrac{dy}{dx}=\dfrac{x+y-1}{x+y+1}$

substitute

$x+y=v\rightarrow 1+\dfrac{dy}{dx}=\dfrac{dv}{dx}$

$\dfrac{dv}{dx}-1=\dfrac{v-1}{v+1}$

$\dfrac{dv}{dx}=\dfrac{2v}{v+1}$

$\dfrac{v+1}{2v}dv=dx$

integrating on both sides, we get,

$\int \dfrac{v+1}{2v}dv=\int dx$

$\dfrac{1}{2}(v+\log v)=x+c$

$x+y+\log (x+y)=2x+C$

$y+\log (x+y)=x+C$

Show that the function $y=e^{3x}(A+Bx)$ is a solution of the differential equation $\dfrac{d^2y}{dx^2}-6\dfrac{dy}{dx}+9y=0$ .

given differential equation is

$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -6\cfrac{dy}{dx}+9y=0....(1)$

$y={e}^{3x}(A+Bx)$

diiferetiating

$=3A{e}^{3x}+B(3x+1){e}^{3x}$

differentiating again

$=9A{e}^{3x}+3B(3x+1){e}^{3x}+3B{e}^{3x}$

Substituting in (1)

$9A{e}^{3x}+9Bx{e}^{3x}+6B{e}^{3x}-18A{e}^{3x}-18Bx{e}^{3x}-6B{e}^{3x}+9A{e}^{3x}+9Bx{e}^{3x}=0$

$0=0$

$\therefore$

$y={e}^{3x}(A+Bx)$ is a solution of (1)

Show that the function $y=Ax+\dfrac{B}{x}$ is a solution of the differential equation $x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}-y=0$ .

given

$y=Ax+\cfrac{B}{x}...(1)$

Differntiating on both sides

$\Rightarrow$

$\cfrac{dy}{dx}=A+B(-\cfrac{1}{{x}^{2}})$

$\cfrac{dy}{dx}=A-\cfrac{B}{{x}^{2}}....(2)$

Differentiating

$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\cfrac{2B}{{x}^{3}}$

$B=\cfrac{{x}^{3}}{2}\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } ....(3)$

Substituting in (2)

$\cfrac{dy}{dx}=A-\cfrac{x}{2}\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$

$A=\cfrac{dy}{dx}+\cfrac{x}{2}\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } ....(4)$

Substitute (3) and (4) in (1)

$y=(\cfrac{dy}{dx}+\cfrac{x}{2}\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } )x+(\cfrac{{x}^{3}}{2}\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } )\cfrac{1}{x}$

${x}^{2}\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +x\cfrac{dy}{dx}-y=0$

Hence proved

$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x; y=1$ when $x=0$ .

Particulation solution

$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x\Rightarrow dy=\dfrac{2x^2+x}{x^3+x^2+x+1}dx$

Integrating both sides w.r.t x

$\displaystyle y=\int \dfrac{2x^2+x}{(x+1)(x^2+1)}dx$

$x=-1$ as a solution of given

$eqn:x^3+x^2+x+1$

as:

$(-1)^3+(-1)^2+(-1)+1=0$

$\therefore (x+1)$ is a factor.

Now, we have to integrating using partial functions

$\displaystyle \int \dfrac{2x^2+x}{(x+1)(x^2+1)}=\dfrac{A}{(x+1)}+\dfrac{Bx+c}{x+1}=\dfrac{A(x^2+1)(Bx+c)(x+1)}{(x+1)(x^2+1)}$

$\Rightarrow 2x^2+x=A(x^2+1)(Bx+c)(x+1)$

Putting

$x=-1$

$2(-1)^2-1=A((-1)^2+1)(B(-1)+c)(-1+1)$

$2-1=A(2)+(-B+c)(0)$

$1=2A\Rightarrow \dfrac{1}{2}$

Putting

$x=1$

$2(1)+1=A(12+1)+(B(1)+c)(1+1)$

$3=2A+2B+2c$

$3=3\times \dfrac{1}{2}+2B+2(\dfrac{-1}{2})$

$3=2B$

$B=\dfrac{3}{2}$

Putting

$x=0$

$0=A(0+1)+(B(0)+c)(0+1)$

$0=A+c(1)\Rightarrow A=-c\Rightarrow \boxed{C=\dfrac{-1}{2}}$

So, the partial equation become

$\dfrac{2x^2+x}{(x+1)(x^2+1)}=\dfrac{1}{2(x+1)}=\dfrac{\dfrac{3}{2}x-\dfrac{1}{2}}{x^2+1}=\dfrac{1}{2(x+1)}+\dfrac{3x-1}{2(x^2+1)}$

Now main equation become

$\displaystyle y=\int \dfrac{2x^2+x}{(x+1)(x^2+1)}dx$

$\displaystyle y=\int \dfrac{1}{2(x+1)}dx+\int \dfrac{3x}{2(x^2+1)}dx-\int \dfrac{1}{2(x^2+1)}dx$

$\displaystyle y=\dfrac{1}{2}\log (x+1)+\int \dfrac{3x}{2(x^1+1)}dx-\dfrac{1}{2}\tan^{-1}x$ ...(1)

Integrating

$\displaystyle \int\dfrac{3x}{2(x^2+1)}dx$

Put

$x^1+1=z$

$2x=\dfrac{dz}{dx}$

$\Rightarrow dx=\dfrac{dz}{zx}$

$\displaystyle \int \dfrac{3x}{2(x^2+1)}dx=\frac{3}{2}\int \dfrac{x}{z}x\dfrac{dz}{zx}=\int \frac{3}{4}\dfrac{dz}{z}=\dfrac{3}{4}\log |z|+c$ d

Substituting value of

$z$ we get,

$\displaystyle \int \dfrac{3x}{2(x^2+1)}dx=\dfrac{3}{4}\log (x^2+1)+c$

Now, from(1)

$y=\dfrac{1}{2}\log(x+1)+\int \dfrac{3x}{2(x^2+1)}dx\dfrac{-1}{2}tan^{-1}x$

$y=\dfrac{1}{2}\log(x+1)+\dfrac{3}{4}\log(x^2+1)-\dfrac{-1}{2}tan^{-1}+c$

Put

$x=0$ and

$y=1$

$1=\dfrac{1}{2}\log(0+1)+\dfrac{3}{4}\log(0+1)\dfrac{-1}{2}tan^{-1}(0)+c$

$1=\dfrac{1}{2}\log(1)+\dfrac{3}{4}\log(1)\dfrac{-1}{2}tan^{-1}+c$

$1=0+0-0+c$

$1=c$

$(\therefore \log1=0\,and\,tan=0)$

Putting value of

$c$ in (1)

$y=\dfrac{1}{2}\log(x+1)+\dfrac{3}{4}\log(x^2+1)\dfrac{1}{2}tan^{-1}x+1$

$y=\dfrac{2}{4}\log(x+1)+\dfrac{3}{4}\log(x^2+1)\dfrac{-1}{2}tan^{-1}x+1$

$y=\dfrac{1}{4}\log(x+1)^2+\dfrac{1}{4}\log(x^2+1)^3\dfrac{-1}{2}tan^{-1}x+1$

$\therefore$

$(a\log x=logx^a)$

$y=\dfrac{1}{2}[\log(X+1)^2+\log(X^2+1)^3\frac{-1}{2}tan^{-1}x+1$

$\Rightarrow \boxed{y=\dfrac{1}{4}\log [(x+1)^2(x^2+1)^3]\dfrac{-1}{2}tan^{-1}x+1}$

$(\therefore \log a+\log b=\log ab)$

Solve:
$\left( {2xy + {y^2}} \right)dx + \left( {{x^2} + xy} \right)dy = 0,y\left( 1 \right) = 1$

$(2xy+y^2)dx+(x^2+xy)dy=0, y(1)=1$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{2xy+y^2}{x^2+xy}=0$

Let

$y=vx$

$\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{y(2x+y)}{x(x+y)}=0$

$v+x\dfrac{dv}{dx}+v\left(\dfrac{2x+vx}{x+vx}\right)=0$

$\Rightarrow v+x\dfrac{dv}{dx}+v\left(\dfrac{2+v}{1+v}\right)=0$

$\Rightarrow x\dfrac{dv}{dx}+\dfrac{v^2+2v+v^2+v}{1+v}=0$

$\Rightarrow x\dfrac{dv}{dx}+\dfrac{2v^2+3v}{1+v}=0$

$\Rightarrow \dfrac{1+v}{2v^2+3v}dv+\dfrac{dx}{x}=0$

$\Rightarrow \dfrac{1+v}{v(2v+3)}dv+\dfrac{dx}{x}=0$

$\Rightarrow \left(\dfrac{1}{v(2v+3)}+\dfrac{v}{v(2v+3)}\right)^{dv}+\dfrac{dx}{x}=0$

$\Rightarrow \left\{\dfrac{1}{3}\left(\dfrac{1}{v}-\dfrac{2}{2v+3}\right)+\dfrac{1}{(2v+3)}\right\}dv+\dfrac{dx}{x}=0$

$\Rightarrow \left\{\dfrac{1}{3v}+\dfrac{1}{3(2v+3)}\right\}dv+\dfrac{dx}{x}=0$

Integrating both sides

$\dfrac{1}{3}\left\{log v+\dfrac{1}{2}log (2v+3)\right\}+log x=log c$

$\Rightarrow \dfrac{1}{3}\left\{log\left(\dfrac{y}{x}\right)+log\left(\dfrac{2y}{x}+3\right)^{\dfrac{1}{2}}\right\}+log x=log c$

$\Rightarrow log\left\{x\times \left(\dfrac{y}{x}\times \sqrt{\dfrac{2y}{x}+3}\right)^{\dfrac{1}{3}}\right\}=log c$

$\Rightarrow x^{\dfrac{1}{2}}y^{\dfrac{1}{3}}(\sqrt{2y+3x})^{\dfrac{1}{3}}=c$

$\Rightarrow x^{\dfrac{3}{2}}y(\sqrt{2y+3x})=c^3=c_1$

$\Rightarrow y(\sqrt{2y+3x})=c_1x^{-\dfrac{3}{2}}$

$x=1, y=1$

$\Rightarrow \sqrt{2+3}=4$

$\Rightarrow 4=\sqrt{5}$

$\therefore y(\sqrt{2y+3x})=\sqrt{5}x^{-\dfrac{3}{2}}$ .

For each of the differential equations, find the general solution:
1. $\dfrac{dy}{dx} = \dfrac{1- cos x}{1 + cos x}$
2. $\dfrac{dy}{dx} = \sqrt{4 - y^2} (-2 < y < 2)$

$(1)$

$\dfrac{dy}{dx}=\dfrac{1-\cos x}{1+\cos x}$

$\dfrac{dy}{dx}=\dfrac{1-1+2\cos ^{2}\dfrac{x}{2}}{1+2\sin^{2}\dfrac{x}{2}-1}$

$\dfrac{dy}{dx}=\cot^{2}\dfrac{x}{2}$

$dy=\cot^{2}\dfrac{x}{2}dx$

$\displaystyle\int dy=\int \left(\csc^{2}\dfrac{x}{2}-1\right)dx$

$\boxed{y=-2\cot\dfrac{x}{2}-x+c}$

$(2)$

$\dfrac{dy}{dx}=\sqrt{4-y^{2}}$

$\displaystyle\int\dfrac{dy}{\sqrt{4-y^{2}}}=\int dx$

$\boxed{\sin^{-1}\left(\dfrac{y}{2}\right)=x+c}$

Solve the differential equation
$xy\dfrac { dy }{ dx } =\dfrac { 1+{ y }^{ 2 } }{ 1+{ x }^{ 2 } } \left( 1+x+{ x }^{ 2 } \right)$

Given,

$xy\dfrac{dy}{dx}=\dfrac{1+y^2}{1+x^2}(1+x+x^2)$

$\dfrac{y}{1+y^2}dy=\dfrac{1+x+x^2}{x(1+x^2)}dx$

$\dfrac{1}{2}\displaystyle \int \dfrac{2y}{1+y^2}dy=\displaystyle \int \dfrac{1}{x}dx+\displaystyle \int \dfrac{1}{1+x^2}dx$

$\dfrac{1}{2} \log |1+y^2|=\log x+\tan^{-1}x+c$

Verify that $xy = \log y + c$ is a solution of the differential equation $(xy - 1) \dfrac {dy}{dx} + y^{2} = 0$ .

Given

$xy=log y+c$

Differentiate on both sides

$\Rightarrow xy^{'} +y=\dfrac {1}{y}y^{'}$

$\Rightarrow y^{'} =\dfrac {y^{2}}{1-xy}$

We will substitute

$y^{'}$ in given DE and we will verify

$\Rightarrow (xy-1) \dfrac {y^{2}}{1-xy}+y^{2}$

$\Rightarrow - y^{2}+y^{2}=0=RHS$

Given solution is correct for above differential equation

Solve:
$\dfrac{\mathrm{d} y}{\mathrm{d} x}=1+e^{x-y}$

$\cfrac{dy}{dx} = 1 + {e}^{x-y}$

$\cfrac{d}{dx} \left( x - y \right) = 1 + {e}^{x - y}$

Let

$z = x - y \Rightarrow z' = 1 - y'$

$y' = 1 - {e}^{x - y}$

$1 - y' = {e}^{x - y}$

$z' = {e}^{z}$

$\cfrac{dz}{{e}^{z}} = dx$

${e}^{-z} \; dz = dx$

Integrating both sides, we have

$\int{{e}^{-z} \; dz} = \int{dx}$

$- {e}^{-z} = x + {C}_{0}$

${e}^{-z} = C - x \; \left( \because C = {C}_{0} \left( \text{Constant} \right) \right)$

${e}^{y - x} = C - x$

Now taking

$\log$ both sides, we have

$\left( y - x \right) \log{e} = \log{\left( C - x \right)}$

$y = \log{\left( C - x \right)} + x$

Hence

$y = \log{\left( C - x \right)} + x$ .

Solve $(x+y)^{2}\dfrac {dy}{dx}=a^{2}$

Let

$z=x+y$

$\dfrac{dz}{dx}=1+\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{dz}{dx}-1$

Now,

$z^2\left(\dfrac{dz}{dx}-1\right)=a^2$

$z^2\dfrac{dz}{dx}-z^2=a^2$

$z^2\dfrac{dz}{dx}=z^2+a^2$

$\dfrac{z^2dz}{(z^2+a^2)}=dx$

On integration

$\int{\dfrac{z^2dz}{(z^2+a^2)}}=\int{dx}...........(1)$

$\int{\dfrac{z^2dz}{(z^2+a^2)}}=\int{\dfrac{(z^2+a^2-a^2)dz}{(z^2+a^2)}}$

$=\int{dz-a^2\dfrac{1}{2}\tan^{-1}\dfrac{z}{a}}$

$=z-a\tan^{-1}\dfrac{z}{a}$

Now, (1) become

$z-a\tan^{-1}\dfrac{z}{a}=x+c$

$(x+y)-a\tan^{-1}\dfrac{(x+y)}{a}=x+c$

$y-a\tan^{-1}\dfrac{(x+y)}{a}=c$

Solve the following differential equation :
$(1+y+x^2y)dx+(x+x^3)dy=0$

$(1+y+x^2y)dx+(x+x^3)dy=0$

$(x+x^3)dy=-(1+y+x^2y)dx$

$\dfrac{dy}{dx}=\dfrac{-[1+y(1+x^2)]}{x+x^3}$

$\dfrac{dy}{dx}=\dfrac{-[1+y(1+x^2)]}{x(1+x^2)}$

$\dfrac{dy}{dx}+\dfrac{y}{x}=-\dfrac{1}{x(1+x^2)}$

This is a linear differential equation of the form

$\dfrac{dy}{dx}+Py=Q$

where,

$P=\dfrac{1}{x}$ and

$Q=-\dfrac{1}{x(1+x^2)}$

Now,

$IF=e^{\int Pdx} = e^{\log x} = x$

Hence, its solution is given by

$yx=\int \dfrac{-1}{x(1+x^2)}. x dx +C$

$yx=\int \dfrac{-1}{1+x^2}dx+C$

$yx=\cot^{-1}x+C$

$y=\dfrac{\cot^{-1}x+C}{x}$ which is the required solution.

Find $\dfrac {dy}{dx}$ if $y = x^{2} - 2^{\sin x}$ .

Given,

$y=x^2-2^{\sin x}$

differentiating on both sides, we get,

$\dfrac{d}{dx}\left(2^{\sin \left(x\right)}\right)$

$=\dfrac{d}{dx}\left(e^{\sin \left(x\right)\ln \left(2\right)}\right)$

$=\ln \left(2\right)\cdot 2^{\sin \left(x\right)}\cos \left(x\right)$

$\dfrac{dy}{dx}=2x-\ln \left(2\right)\cdot 2^{\sin \left(x\right)}\cos \left(x\right)$

$\dfrac{dy}{dx} + \dfrac{y}{(1-x)} = 1 - x$

$\displaystyle \frac{dy}{dx}+\frac{y}{1-x} = 1-x$

$\displaystyle I.F e^{\int pdx} = e^{\int \dfrac{1}{1-x}}dx = e^{log|1-x|}$

$\displaystyle = \frac{1}{1-x}$

G.1

$\displaystyle = \frac{y}{1-x} = \int (1-x)\frac{1}{(1-x)}dx$

$\displaystyle \frac{y}{1-x} = x+c$

$\displaystyle y = x(1-x)$

1238584_1134647_ans_15612ed77c0840bb96182824f08433f8.jpg

$x(x-1) \dfrac{dy}{dx} - (x-2)y = x^2 (2x - 1)$

Given,

$x(x-1)\dfrac{dy}{dx}-(x-2)y=x^2(2x-1)$

$\dfrac{dy}{dx}-\dfrac{x-2}{x(x-1)}y=\dfrac{x^2(2x-1)}{x(x-1)}$

$\dfrac{dy}{dx}+\left (\dfrac{2-x}{x(x-1)} \right )y=\dfrac{x^2(2x-1)}{x(x-1)}$

$\dfrac{dy}{dx}+Py=Q$

$P=\dfrac{2-x}{x(x-1)} ,Q=\dfrac{x^2(2x-1)}{x(x-1)}$

$I.F=e^{\int Pdx}$

$I.F=e^{\int \frac{2-x}{x(x-1)}dx}=e^{\log (x-1)-2\log x}=e^{\log \frac{x-1}{x^2}}=\dfrac{x-1}{x^2}$

$y\times I.F=\int Q\times I.F\times dx$

$y\times \dfrac{x-1}{x^2}=\int \dfrac{x^2(2x-1)}{x(x-1)} \times \dfrac{x-1}{x^2}dx$

$y\dfrac{x-1}{x^2}=\int \dfrac{2x-1}{x}dx$

$y\dfrac{x-1}{x^2}=2x-\log x+c$

$y(x-1)=2x^3-x^2\log x+cx^2$

Solve: $(1 +x) \dfrac{dy}{dx} - xy = 1 - x$

$\displaystyle (1 +x) \dfrac{dy}{dx} - xy = 1 - x$

$\displaystyle \frac{dy}{dx}+(\frac{-x}{1+x})y = \frac{1-x}{1+x}$

$\displaystyle \int pdx = \int \frac{-x}{1+x} = -\int \frac{1+x-1}{1+x}dx$

$\displaystyle = -\int 1-\frac{1}{1+x}dx$

$\displaystyle = -x+log|1+x|dx$

$\displaystyle I.F e^{\int pdx} = e^{log(1+x)-x} = \frac{1+x}{e^{x}}$

$\displaystyle 4.5 y .\frac{1+x}{e^{x}} = \int \frac{1-x}{e^{x}}$

$\displaystyle y\frac{(1+x)}{e^{x}} = \int e^{-x}(1-x)$

$\displaystyle y(1+x) = e^{x}c^{-x}(+x)$

$\displaystyle y (1+x) = x$

1237364_1138012_ans_a91d424e14544fa4aae09e712c8f1e30.jpg

$(x+y+1)\dfrac{dy}{dx} = 1$

$(x+y+1) = \dfrac{dx}{dy}$

Put

$x+y = t$

$\dfrac{dx}{dy}+1 = \dfrac{dt}{dy}$

$\Rightarrow (t+1) = \dfrac{dt}{dy}-1$

$\Rightarrow dy = \dfrac{1}{t+2}dt$

$\Rightarrow y = log|1+2|$

$\Rightarrow y = log|x+y+2|$

1240061_1134666_ans_a42c1229cb4a4d76b461e7eff90844de.jpg

$x(1-x^2) dy + (2x^2y - y -ax^3) dx = 0$

1374381_1134731_ans_55dd44fc75824f228cd5c8286c3144f1.jpg

Solve:
$x\dfrac{dy}{dx}=y-x\tan\dfrac{y}{x}$

Given

$x\dfrac {dy}{dx} =y-x \tan \dfrac {y}{x} \Rightarrow \tan \dfrac {y}{x} =\dfrac {ydx-xdy}{dx} ---(1)$

Now,

$\dfrac {d}{dx} \left (\dfrac {y}{x}\right) =\dfrac {1}{x^2} (ydx-xdy)----(2)$

Suubstituting

$(2)$ in

$(1),\ x\tan \dfrac {y}{x} =\dfrac {x^2d}{dx} \left (\dfrac {y}{x} \right)$

$\Rightarrow \ \dfrac {dx}{x}=\dfrac {d/y/x}{\tan (y/x)}$

Integrating both side

$\displaystyle \int \dfrac {dx}{x} =\displaystyle \int \dfrac {d(y/x)}{\tan (y/x)}$

$\Rightarrow \ \log x=\log \sin \left (\dfrac {y}{x}\right) +\log c$

$\boxed {y=\sin^{-1} \left (\dfrac {x}{y}\right)}$

Show that the differential equation is homogenous & solve it $ydx+xlog\left(\dfrac{y}{x}\right)dy-2xdy=0$ .

1111215_1137433_ans_791808f175eb44af8a894f4178170b7e.jpeg

Solve the following differential equations
(i) $\dfrac{dy}{dx}+y=e^{-x}$

$\displaystyle \frac{dy}{dx}+y = e^{-x}$

$\displaystyle I.F e^{\displaystyle\int dx} = e^{x}$

G-1

$\displaystyle ye^{x} = \int e^{-x}e^{x}$

$\displaystyle ye^{x} =\int dx$

$\displaystyle ye^{x}=x+c$

1237352_1134497_ans_cd4e2c70d24a41debd31b8ccfc524334.jpg

Solve $\dfrac{dy}{dx}+\dfrac{2y}{x}=e^x$ .

$\displaystyle \frac{dy}{dx}+(\dfrac{2}{x})y = e^{x}$

$\displaystyle I.F = e^{\displaystyle \int \dfrac{2}{x}}dx = e^{2logx} = x^{2}$

$\displaystyle yx^{2} = \int e^{x}x^{2}$

$\displaystyle x^{2}y = x^{2}e^{x}-\int 2xe^{x}$

$\displaystyle = x^{2}e^{x}-2xe^{x}+2e^{x}+c$

1237353_1134551_ans_39474e3a0cbf400989b3bb1f7ec19fea.jpg

$2\ xy\dfrac {dy}{dx}={x}^{2}+{y}^{2}$

Let

$y=vx$

$\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

Substituting the above in

$2xy\dfrac{dy}{dx}={x}^{2}+{y}^{2}$

$\Rightarrow 2x\left(vx\right)\left(v+x\dfrac{dv}{dx}\right)={x}^{2}+{\left(vx\right)}^{2}$

$\Rightarrow 2v{x}^{2}\left(v+x\dfrac{dv}{dx}\right)={x}^{2}\left(1+{v}^{2}\right)$

$\Rightarrow 2v\left(v+x\dfrac{dv}{dx}\right)=\left(1+{v}^{2}\right)$

$\Rightarrow 2{v}^{2}+2vx\dfrac{dv}{dx}-1-{v}^{2}=0$

$\Rightarrow {v}^{2}-1=-2vx\dfrac{dv}{dx}$

$\Rightarrow \dfrac{2vdv}{1-{v}^{2}}=\dfrac{dx}{x}$ on re-arranging

$\Rightarrow \int{\dfrac{2vdv}{1-{v}^{2}}}=\int{\dfrac{dx}{x}}$

Let

$t=1-{v}^{2}\Rightarrow dt=-2vdv$

$\Rightarrow \int{\dfrac{-dt}{t}}=\int{\dfrac{dx}{x}}$

$\Rightarrow-\log{t}=\log{x}+c$ where

$c$ is the constant of integration.

$\Rightarrow -\log{\left(1-{v}^{2}\right)}-\log{x}=c$ where

$t=1-{v}^{2}$

$\Rightarrow -\log{\left(1-{\left(\dfrac{y}{x}\right)}^{2}\right)}-\log{x}=c$ where

$v=\dfrac{y}{x}$

$\Rightarrow \log{\left(\dfrac{{x}^{2}-{y}^{2}}{{x}^{2}}\right)}+\log{x}=-c$ using the identity

$\log{a}+\log{b}=\log{ab}$

$\Rightarrow\left[x\left(\dfrac{{x}^{2}-{y}^{2}}{{x}^{2}}\right)\right]={10}^{-c}=k$ where

$k$ is the constant of integration

$\Rightarrow \dfrac{{x}^{2}-{y}^{2}}{x}=k$

$\Rightarrow {x}^{2}-{y}^{2}=kx$ is the required solution.

Find the particular solution of the differential equation $\left( x-y \right) \frac { dy }{ dx } =\left( x+2y \right)$ , Given that $y=0$ when $x=1$ .

1112434_1138939_ans_fd9883353337465ba6178629c0c7be63.jpeg

Solve the following differential equation:
$x\cos { ydy } =\left( { xe }^{ x }\log { x+{ e }^{ x } } \right) dx$

$x\, cos\, y \,dy = (xe^x logx+e^x) dx$

$\displaystyle \int cos\,y\, dy = \int \dfrac{xe^{x}logx+e^{x}}{x}dx$

$\displaystyle = \int e^{x}logx dx+\int \dfrac{e^{x}dx}{x}$

$\displaystyle = \int e^{x} logx dx$

$\displaystyle = e^{x}logx = \int \dfrac{e^{x}}{x}dx$

$\displaystyle siny = e^{x} log x - \int \dfrac{e^x}{x}dx+\int \dfrac{e^x}{x} dx$

$\therefore e^x logx = siny+c$

1115877_1139680_ans_071b1dbfcafc42479e4a84e33412785f.jpeg

$x\dfrac {dy}{dx}=x+y$

$x\dfrac{dy}{dx}= x+y$

Dividing x both sides

$\Rightarrow \dfrac{dy}{dx}-\dfrac{y}{x}= \dfrac{x}{x}$

$\Rightarrow \dfrac{dy}{dx}-\dfrac{1}{x}.y=1$

From of

$\dfrac{dy}{dx}+P_{y}=Q$

$P=\dfrac{-1}{x},Q=1$

$\therefore 1F= e^{\displaystyle -\dfrac{1}{2}d}$

$=e^{-log x}$

$=(\dfrac{1}{x})$

$\displaystyle \therefore (y)(1F)= \int Q.(1F)dx+c$

$\displaystyle y. \frac{1}{x} = \int 1(\dfrac{1}{x})dx+c$

$\dfrac{y}{x}= log\,x+e$

$y = xlogx+cx$

$y= x\, log x+c_{2}$

1147066_1144274_ans_f79044663ca34a9a96f7b668f26995e7.jpg

Solve:
$\left( { 3x }^{ 2 }+{ y }^{ 2 } \right) dy+\left( { x }^{ 2 }+{ 3y }^{ 2 } \right) dx=0$

$(3x^2 + y^2) dy + (x^2 + 3y^2) dx = 0$

$\dfrac{dy}{dx} = -\dfrac{(x^2 + 3y^2)}{(3x^2 + y^2)}$

$y = vx$

$\dfrac{dy}{dx} = v + x \dfrac{dv}{dx}$

$v + x \dfrac{dv}{dx} = -\dfrac{(x^2 + 3v^2 x^2)}{(3x^2 + v^2 x^2)}$

$= -\left(\dfrac{1 + 3v^2}{3 + v^2} \right)$

$x \dfrac{dv}{dx} = - \dfrac{(1 + 3v^2)}{(3 + v^2)} - v$

$x \dfrac{dv}{dx} = -\dfrac{1 - 3v^2 - 3v - 3v^3}{3 + v^2}$

$x \dfrac{dv}{dx} = -\dfrac{(1 + v)^3}{(3 + v^2)}$

$\displaystyle \int \dfrac{3 + v^2}{(1 + v)^3} dx = \int \dfrac{dx}{x}$

$\displaystyle \int \dfrac{3 + v^2}{(1 + v)^3} dv + \int \dfrac{v^2}{(1 + v)^3} dv = \int \dfrac{dx}{x}$

$3. \dfrac{(1 + v)^{(-3 + 1)}}{(-3 + 1)} \,\,\,\, \dfrac{v^3}{(1 + v)^3} - \displaystyle \int \dfrac{v^2 [-v + 2]}{(1 + v)^4} dv \,\,\, \int - \dfrac{v^3}{(1 + v)^4} dv + \int \dfrac{2v^2 }{(1 _ v)^4} dv$

$\Rightarrow ln |x + 1| + \dfrac{3}{x + 1} - \dfrac{3}{2 (x+ 1)^2} + \dfrac{1}{3 (x + 1)^3} + 2 \left[ \dfrac{-x^2}{3(1 + x)^3} + \dfrac{-2x - 1}{3(x + 1)^2} \right]$

$\Rightarrow -ln |x + 1| - \dfrac{3}{x + 1} + \dfrac{3}{2(x + 1)^2} + \dfrac{3}{2 (x + 1)^2} - \dfrac{1}{3 ( x+ 1)^3} - \dfrac{2x^2}{3 (1 + x)^3} + \dfrac{2(-2 x - 1)}{3(x + 1)^2}$

$\dfrac{x^3}{(1 + x)^3} + ln (x + 1) + \dfrac{3}{x + 1} - \dfrac{3}{2 (x + 1)^2} + \dfrac{1}{3 (x + 1)^3} + \dfrac{2x^2}{3(1 + x)^3} - \dfrac{2 (-2x + 1)}{3(x + 1)^2}$

Replace x by (v) in all above equation By mistake written x, its (v) ,

So, we get : { and replace v by

$\dfrac{y}{x}$ }

$\begin{Bmatrix} \left(\dfrac{y}{x} \right)^3 \dfrac{1}{(1 + y/x)^3} + ln \left(\dfrac{y}{x} + 1 \right) + \dfrac{3}{(y/x + 1)} - \dfrac{3}{2 (y/x + 1)^2} \\ + \dfrac{1}{3(y/x + 1)^3} + \dfrac{2(y/x)}{3(1 + y/x)^3} + \dfrac{2(2(y/x) -1)}{3(y/x + 1)^2}\\ = \log \, x + \log \, c \end{Bmatrix}$

Solve the following differential equation.
$\dfrac{dy}{dx}=e^{3x-2y}+x^{2}e^{-2y}$

1133910_1140533_ans_1e7399417f76418eba5604a47a6e7f9d.jpg

Show that $y=e^{m\sin^{-1}x}$ is a solution of the differential equation $(1-x^{2})y_{2}-xy_{1}-m^{2}y=0$ .

$\displaystyle y = e^{m\,sin^{1}x}$

$\displaystyle {y}' = m\,sin^{1}x$

$e^{m\,sin^{-1}x}$

$-\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle {y}''= \frac{(\sqrt{1-x^{2}})(\frac{m^{2}e^{m\,sin^{-1}x}}{\sqrt{1-x^{2}}})-me^{m\,sin^{-1}}(\frac{1}{2})(1-x^{2})^{-1/2}(-2x)}{(\sqrt{1-x^{2}})^{2}}$

$\displaystyle = \frac{m^{2}e^{m\,sin^{-1}x}+\frac{m\,x\,e^{sin^{-1}x}}{\sqrt{1-x^{2}}}}{1-x^{2}}$

$\displaystyle = \frac{m^{2}y+n{y}'}{1-x^{2}}$

$\displaystyle (1-x^{2}){y}''= m^{2}y+x{y}'$

$\displaystyle (1-x^{2}){y}''-m^{2}y-xy = 0$

1159097_1140221_ans_e86d8739746a483e894ef3f729abb068.jpg

Find solution for $DE \Rightarrow \left (\dfrac {dy}{dx}\right )^{2} - y \dfrac {dy}{dx} + x = 0$ .

Given,

$\left ( \dfrac{dy}{dx} \right )^2-y\dfrac{dy}{dx}+x=0$

substitute

$\dfrac{dy}{dx}=p$

$p^2-py+x=0$

$\Rightarrow x=py-p^2$ .....(1)

using Clairaut's equation, differentiate both sides w.r.t p

$\dfrac{dx}{dp}=y-2p$

to find solution put

$\dfrac{dx}{dp}=0$

$0=y-2p$

$p=\dfrac{y}{2}$

substitute value of p in equation (1)

$\Rightarrow x=\left ( \dfrac{y}{2} \right )y-\left ( \dfrac{y}{2} \right )^2$

$x=\dfrac{y^2}{2}-\dfrac{y^2}{4}$

$\therefore x=\dfrac{y^2}{4}$

Is the required equation.

Solve the following differential equation:
$4\frac { dy }{ dx } +8y={ 5e }^{ -3x }$

$4\dfrac{dy}{dx}+8y = 5e^{-3x}$

$\dfrac{dy}{dx}+2y = \dfrac{5}{4}e^{-3x}$

$m+2 = 0$

$m = -2$

$A.E \Rightarrow y(x) = C_1 e^{-2x}$

$P.I = \dfrac{5e^{-3x}}{4(D+2)} \Rightarrow \dfrac{5e^{-3x}}{4(-3+2)} = \dfrac{-5e^{-3x}}{4}$

C.S

$y(x) = C_1 e^{-2x}-\dfrac{5}{4}e^{-3x}$

1112401_1139703_ans_5174e551ae05446c876c2473ad6a5e5b.jpg

For each the differential equations given, find the general solution :
$\dfrac {dy}{dx} + 2y \sin x$

$\dfrac{dy}{dx}+2y \sin x=0$

$\Rightarrow \displaystyle \int \dfrac{dy}{y} \ = \displaystyle \int -2 \sin x \ dx$

$\Rightarrow \ln\ y = 2\cos x+c$

$\Rightarrow y=e^{2\cos x+c}$

$\Rightarrow y=ke^{2 \cos x}$

${ye}^{x/y}dx=\left ( { {xe}^{x/y}+{y}^{2} }\right)\ dy$ .

$ye^{x/y}dx= (xe^{x/y}+y^{2})dy$

$\dfrac{dx}{dy}= \dfrac{xe^{x/y}+y^{2}}{ye^{x/y}}$

$=\dfrac{x}{y}e^{x/y}+y/e^{x/y}$

$x= vy$

$\dfrac{dx}{dy}= v+y\dfrac{dv}{dy}$

$v+y\dfrac{dv}{dy}= ve^{v}+y/e^{v}$

$y\dfrac{dy}{dy}= \dfrac{ve^{v}}{e^{v}}-v+\dfrac{y}{ev}$

$y\dfrac{dv}{dy}= v-v+\dfrac{y}{e^{v}}$

$\displaystyle \int e^{v}dv= \int \dfrac{y}{y}dy$

$e^{v}+e=y$

$\Rightarrow y=e^{x/y}+e$

1155595_1144281_ans_356a0bef42f74d12ae5bbf41c139f3ee.jpeg

$xy\dfrac {dy}{dx}={x}^{2}-{y}^{2}$

1155593_1144283_ans_45dfe0da607e44b8990040a7d91dee98.pdf

$y = 2\dfrac {dy}{dx} + \left (\dfrac {dy}{dx}\right )^{2} y$ .

Given,

$y=2\dfrac{dy}{dx}+\left ( \dfrac{dy}{dx} \right )^2y$

substitute

$\dfrac{dy}{dx}=p$

$y=2p+p^2y$ ......(1)

From Clairaut's equation, differentiate both sides w.r.t. p

$\dfrac{dy}{dp}=2+2py$

to find the solution, substitute

$\dfrac{dy}{dp}=0$

$0=2+2py$

$\therefore p=-\dfrac{1}{y}$

substitute the value of p in equation (1)

$y=2\left ( -\dfrac{1}{y} \right )+\left ( -\dfrac{1}{y} \right )^2y$

$y=-\dfrac{2}{y}+\dfrac{1}{y}$

$\therefore y^2-1=0$

Is the required equation.

Solve $x^{2}\dfrac {dy}{dx}=x^{3}y$

$x^{2}\dfrac{dy}{dx} = x^{3}y$

$\dfrac{dy}{dx}-xy = 0$

$I.F = C^{f-xdy} = C^{-x^{2}/2}$

$G.S : ye^{-x^{2}/2} = 0$

1240528_1145444_ans_79fb03c32ff04b0caae57ed6478248bd.jpg

Find the solution of the differential equation $\left( x\log { x } \right) \frac { dy }{ dx } +y=2x\log { x,\left( x\ge 1 \right) }$ .

$x\log x\dfrac{dy}{dx}+y=2x\log x$

$\dfrac{dy}{dx}+\displaystyle\int\dfrac{1}{x\log x}y=\dfrac{2x\log x}{x\log x}$

$I.F=\displaystyle\int\dfrac{1}{x\log x}dx$

Let

$\log =4=e^{\displaystyle\int \dfrac{1}{u}du}=e^{\log 4}=u=\log x$

$\dfrac{1}{x}dx=dx$

$y\times \log x=2\displaystyle\int\log x dx+C$

Using

$ILATE$ rule

$y\times \log x=2(x\log x-x)+C$

$y\times \log x=2x\log x-2x+C$

$y=2x-\dfrac{2x}{\log x}+\dfrac{C}{\log x}$

Solve the following rational equations
$2(y+3)-xy\dfrac {dy}{dx}=0$ given that $y(1)=-2$

Given that,

$2(y+3)-xy\dfrac{dy}{dx}=0$

$\Rightarrow 2(y+3)=xy\dfrac{dy}{dx}$

$\Rightarrow 2\dfrac{dx}{x} =\left(\dfrac{y}{(y+3)}\right)dy$

$\Rightarrow 2\dfrac{dx}{x} =\left(\dfrac{y+3-3}{(y+3)}\right)dy$

$\Rightarrow 2\dfrac{dx}{x} =\left(1-\dfrac{3}{(y+3)}\right)dy$
On integrating both sides, we get

$\displaystyle 2\dfrac{dx}{x} = \int \left(1-\dfrac{3}{(y+3)}\right)dy$

$2 \log x=y-3 \log (y+3)+C ...(i)$
When

$x=1\ and\ y=-2$ , then

$2 \log 1=-2-3 \log (-2+3)+C$

$\Rightarrow 2.0=-2-3.0+C$

$\Rightarrow C=2$
On substituting the value of

$C$ in Eq.

$(i)$ , we get

$2 \log x=y-3 \log (y+3)+2$

$\Rightarrow 2 \log x +3 \log (y+3)=y+2$

$\Rightarrow \log x^2 + \log (y+3)^3=(y+2)$

$\Rightarrow \log x^2 (y+3)^3=y+2$

$\Rightarrow x^2 (y+3)^3=e^{y+2}$

Find the differential equation of the family of all the circles.
(A) touching X-axis at the origin.
(B) touching Y-axis at the origin.

(A) Family of all the circles touching $x$ -axis at origin

We know that, equation of circle is-

${\left( x - h \right)}^{2} + {\left( y - k \right)}^{2} = {r}^{2}$

Whereas,

Centre of circle

$\equiv \left( h, k \right)$

Radius of circle

$= r$

Since the circle touches the

$x$ -axis at origin, thus centre will be on

$y$ -axis.

$\therefore h = 0$

$\therefore$ Centre of the circle

$\equiv \left( 0, k \right)$

Thus, radius of the circle

$= k$

Now, equation of the circle-

${\left( x - 0 \right)}^{2} + {\left( y - k \right)}^{2} = {k}^{2}$

${x}^{2} + {y}^{2} + {k}^{2} - 2yk = {k}^{2}$

${x}^{2} + {y}^{2} - 2yk = 0 ..... \left( 1 \right)$

As the above equation has only one variable, thus differentiating the above equation once w.r.t.

$x$ , we have

$2x + 2y \cfrac{dy}{dx} - 2k \cfrac{dy}{dx} = 0$

$x + y y' - k y' = 0$

$\Rightarrow k = \cfrac{x + yy'}{y'}$

Substituting the value of

$k$ in

${eq}^{n} \left( 1 \right)$ , we have

${x}^{2} + {y}^{2} - 2y \left( \cfrac{x + y y'}{y'} \right) = 0$

$\left( {x}^{2} + {y}^{2} \right) y' = 2y \left( x + yy' \right)$

${x}^{2} y' + {y}^{2} y' = 2xy + 2 {y}^{2} y'$

$\Rightarrow {x}^{2} y' + {y}^{2} y' - 2 {y}^{2} y' = 2xy$

$\Rightarrow y' \left( {x}^{2} - {y}^{2} \right) = 2xy$

$\Rightarrow y' = \cfrac{2xy}{\left( {x}^{2} - {y}^{2} \right)}$

(B) Family of all the circles touching $y$ -axis at origin

We know that, equation of circle is-

${\left( x - h \right)}^{2} + {\left( y - k \right)}^{2} = {r}^{2}$

Whereas,

Centre of circle

$\equiv \left( h, k \right)$

Radius of circle

$= r$

Since the circle touches the

$y$ -axis at origin, thus centre will be on

$x$ -axis.

$\therefore k = 0$

$\therefore$ Centre of the circle

$\equiv \left( h, 0 \right)$

Thus, radius of the circle

$= h$

Now, equation of the circle-

${\left( x - h \right)}^{2} + {\left( y - 0 \right)}^{2} = {k}^{2}$

${x}^{2} + {h}^{2} - 2xh + {y}^{2} = {h}^{2}$

${x}^{2} + {y}^{2} - 2hx = 0 ..... \left( 1 \right)$

As the above equation has only one variable, thus differentiating the above equation once w.r.t.

$x$ , we have

$2x + 2y \cfrac{dy}{dx} - 2h = 0$

$x + y y' - h = 0$

$\Rightarrow h = x + yy'$

Substituting the value of

$h$ in

${eq}^{n} \left( 1 \right)$ , we have

${x}^{2} + {y}^{2} - 2x \left( x + y y' \right) = 0$

$\left( {x}^{2} + {y}^{2} \right) = 2x \left( x + yy' \right)$

${x}^{2} + {y}^{2} = 2{x}^{2} + 2 xy y'$

$\Rightarrow {x}^{2} + {y}^{2} - 2 {x}^{2} = 2xy y'$

$\Rightarrow \left( {y}^{2} - {x}^{2} \right) = 2xy y'$

$\Rightarrow y' = \cfrac{\left( {y}^{2} - {x}^{2} \right)}{2xy}$

Hence family of all the circles touching

$x$ -axis at origin is

$y' = \cfrac{2xy}{\left( {x}^{2} - {y}^{2} \right)}$ while family of all the circles touching

$y$ -axis at origin is

$y' = \cfrac{\left( {y}^{2} - {x}^{2} \right)}{2xy}$ .

$e^{x}\tan y dx+(1-e^{x})\sec^{2}y dy=0$

$e^x\tan y dx+(1-e^x)\sec^2 ydy=0$

$e^x\tan y dx=-(1-e^x)\sec^2 y dy$

$\dfrac{e^x}{e^x-1}.dx=\dfrac{\sec^2 y}{\tan y}.dy$

$\displaystyle \int{\dfrac{e^x}{e^x-1}.dx}=\displaystyle \int{\dfrac{\sec^2 y}{\tan y}.dy}----(1)$

Put

$e^x-1=u$

$\tan y=v$

$e^x.dx=du$

$\sec^2y dy=dv$

$e^x=\dfrac{du}{dx}$

$\sec^2y dy=dv$

$e^x=\dfrac{du}{dx}$

$\sec^2y dy=dv$

$dx=\dfrac{du}{e^x}$

$dy=\dfrac{dv}{\sec^2y}$

Put in

$(1)$

$\displaystyle \int{\dfrac{e^x}{u}.\dfrac{du}{e^x}}=\displaystyle \int{\dfrac{\sec^2y}{v}.\dfrac{dv}{\sec^2 y}}$

$\displaystyle \int{\dfrac{du}{u}}=\displaystyle \int{\dfrac{dv}{v}}$

$\log u+c_1=\log v$

$\log (e^x-1)+c_1=\log(\tan y)$

$\log |e^x-1|+\log c=\log |\tan y|$

$\log |c(e^x-1)|=\log |\tan y|$

$c(e^x-1)=\tan y$

Find the $I.F$ of the differential equation:
$\cos^{2}x\dfrac {dy}{dx}+y=\tan x$

$\dfrac{dy}{dx} +y sec^2x =tanx .sec^2x$

It is linear differential equation,

$IF=e^{\int sec^2x dx}$

Therefore,

$IF=e^{tanx}$

Find the solution of $(\sqrt{a+x})\left(\dfrac{dy}{dx}\right)+x=0$ .

$\left( \sqrt { a+x } \right) \dfrac { dy }{ dx } +x=0$

$\Rightarrow dy=\dfrac { -x }{ \sqrt { a+x } } dx$

$\Rightarrow y=\int { \dfrac { -x }{ \sqrt { a+x } } } dx$

Let

$\sqrt { a+x } =t$

$\Rightarrow x={ t }^{ 2 }-a$

$dx=2tdt$

$\therefore$

$y=\int { -\dfrac { \left( { t }^{ 2 }-a \right) .2tdt }{ t } }$

$=2\int { \left( a-{ t }^{ 2 } \right) } dt$

$=2\left( at-\dfrac { { t }^{ 3 } }{ 3 } \right) +c$

$\Rightarrow y=2\left( a\left( \sqrt { a+x } \right) -\dfrac { { \left( \sqrt { a+x } \right) }^{ 3 } }{ 3 } \right) +c$

Solve : $\dfrac{dy}{dx} = \dfrac{y^2 - x}{2y (x + 1)}$

$\dfrac{dy}{dx}=\dfrac{y^2-x}{2y(x+1)}$

$[2y(x+1)]dy=(y^2-x)dx$

$(2xy+2y)dy=(y^2-x)dx$

$(x+1)y^2=y^2x-1$

$xy^2+y^2=xy^2-1$

$y^2+1=0$

The solution of $\dfrac{dy}{dx}=\dfrac{\sqrt{x^2-y^2}+y}{x}$ .

$\cfrac{dy}{dx}=\cfrac{\sqrt{x^2-y^2}+y}{x}$

Put

$y=x\sin\theta$

$\implies \cfrac{dy}{dx}=\sin\theta+x\cos\theta\cfrac{dx}{d\theta$$}$

$\sin\theta+x\cos\theta\cfrac{dx}{d\theta}$

$=\cfrac{\sqrt{x^2-x^2\sin^2\theta}+x\sin\theta}{x}$

$\implies \sin\theta+x\cos\theta\cfrac{dx}{d\theta}$

$=\cfrac{x(\cos\theta+\sin\theta)}{x}$

$\implies d\theta=\cfrac{dx}{x}$

$\implies \theta =\ln x+c$

$\implies \sin^{-1}\cfrac{y}{x}=\ln x+c$

Find the differential equation of all ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ , where $a$ and $b$ are constant.

$\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$

Differentiating both sides, w.r.t

$x$ we get

$\dfrac{1}{{a}^{2}}2x+\dfrac{1}{{b}^{2}}2y\dfrac{dy}{dx}=0$

$\Rightarrow\,\dfrac{1}{{b}^{2}}2y\dfrac{dy}{dx}=-\dfrac{1}{{a}^{2}}2x$

$\Rightarrow\,\dfrac{y}{x}\dfrac{dy}{dx}=-\dfrac{{b}^{2}}{{a}^{2}}$

Differentiating both sides, w.r.t

$x$ we get

$\Rightarrow\,\dfrac{y}{x}\dfrac{{d}^{2}y}{d{x}^{2}}+\dfrac{x\dfrac{dy}{dx}-y}{{x}^{2}}\dfrac{dy}{dx}=0$

$\Rightarrow\,\dfrac{y}{x}\dfrac{{d}^{2}y}{d{x}^{2}}+\dfrac{1}{x}{\left(\dfrac{dy}{dx}\right)}^{2}-\dfrac{y}{{x}^{2}}\dfrac{dy}{dx}=0$

$\Rightarrow\,\dfrac{y{x}^{2}}{x}\dfrac{{d}^{2}y}{d{x}^{2}}+\dfrac{{x}^{2}}{x}{\left(\dfrac{dy}{dx}\right)}^{2}-y\dfrac{dy}{dx}=0$

$\Rightarrow\,xy\dfrac{{d}^{2}y}{d{x}^{2}}+x{\left(\dfrac{dy}{dx}\right)}^{2}-y\dfrac{dy}{dx}=0$ is the required differential equation.

If $f\left( x \right) = x\sin \,x,\,$ find $f'\left( \pi \right)$ , using first principle.

Given,

$f(x)=x\sin x$

$f'(x)=\sin x +x\cos x$

$f'(\pi)=\sin \pi +\pi \cos \pi$

$=0+\pi(-1)$

$\therefore f'(\pi)=-\pi$

solve. $xdy - (4 + 2{x^2})dx = 0$

$xdy-(4+2x^{2})dx=0$

$\Rightarrow xdy=(4+2x^{2})dx$

$\Rightarrow dy=\left(\dfrac{4}{x}+2x\right)dx$

Integrating both sides

$y=4\log x+x^{2}+C$

Solve $ydx-xdy=x^{2}ydx$

We have,

$ydx-xdy={{x}^{2}}ydx$

$ydx-{{x}^{2}}ydx=xdy$

$y\left( 1-{{x}^{2}} \right)ydx=xdy$

$\left( \dfrac{1-{{x}^{2}}}{x} \right)dx=\dfrac{dy}{y}$

$\left( \dfrac{1}{x}-\dfrac{{{x}^{2}}}{x} \right)dx=\dfrac{dy}{y}$

$\dfrac{1}{x}dx-\dfrac{{{x}^{2}}}{x}dx=\dfrac{dy}{y}$

$\dfrac{1}{x}dx-xdx=\dfrac{dy}{y}$

on integrating we get,

$\log x-\dfrac{{{x}^{2}}}{2}=\log y+\log C$

$\log x-\dfrac{{{x}^{2}}}{2}=\log yC$

$\log yC-\log x=-\dfrac{{{x}^{2}}}{2}$

$\log \dfrac{yC}{x}=-\dfrac{{{x}^{2}}}{2}$

$\dfrac{yC}{x}={{e}^{-\dfrac{{{x}^{2}}}{2}}}$

$yC=x{{e}^{-\dfrac{{{x}^{2}}}{2}}}$

$y=\dfrac{{{e}^{-\dfrac{{{x}^{2}}}{2}}}}{C}$

$y={{C}_{1}}{{e}^{-\dfrac{{{x}^{2}}}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore {{C}_{1}}=\dfrac{1}{C}$

Hence this is the answer.

Prove that $y = \frac{A}{x} + B$ is the solution of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} + \frac{2}{x}\frac{{dy}}{{dx}} = 0$ .

Given,

$y=\dfrac{A}{x}+B$

$\dfrac{dy}{dx}=-\dfrac{A}{x^2}$

$\dfrac{d^2y}{dx^2}=\dfrac{2A}{x^3}$

Given,

$\dfrac{d^2y}{dx^2}+\dfrac{2}{x}\dfrac{dy}{dx}$

$=\dfrac{2A}{x^3}+\dfrac{2}{x}\times -\dfrac{A}{x^2}$

$=\dfrac{2A}{x^3}-\dfrac{2A}{x^3}$

$=0$

Hence proved.

Simplify: $\frac{1}{{{x^2} - 5x + 6}} - \frac{1}{{{x^2} - 3x + 2}}$

Given,

$\dfrac{1}{x^2-5x+6}-\dfrac{1}{x^2-3x+2}$

$=\dfrac{1}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}$

$=\dfrac{x-1}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}-\dfrac{x-3}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}$

$=\dfrac{x-1-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}$

$=\dfrac{2}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}$

If $y = \log \left( {1 + 2{t^2} + {t^4}} \right),\,\,x = {\tan ^{ - 1}}t,$ find $\dfrac{{{d^2}y}}{{d{x^2}}}$

$y=\log (1+2t^{2}+t^{4})$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=\dfrac{1}{(1+2t^{2}+t^{4})}(4t+4t^{3})$

$\dfrac{x=\tan^{-1}t}{\dfrac{dy}{dx}=\dfrac{1}{1+t^{2}}}$

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dr}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{4t(1+t^{2})}{(1+2t^{2}+t^{4})}}{\dfrac{1}{(1+t^{2})}}$

$=\dfrac{4t(1+t^{2})^{2}}{1+2t^{2}+t^{4}}$

$=\dfrac{4t(1+t^{4}+2t^{2}}{1+2t^{2}+t^{4}}$

$=4t$

$\dfrac{d^{2}y}{dx^{2}}=d\left(\dfrac{dy}{dx}\right).\dfrac{dt}{dx}$

$=4\times (1+t^{2})$

$\dfrac{d^{2}y}{dx^{2}}=4(1+t^{2})$

Can $y=ax+\cfrac{b}{a}$ be a solution of the following differential equation?
$y=x\cfrac{dy}{dx}+\cfrac{b}{\cfrac{dy}{dx}}$ .......( $^{\ast}$ )
If no, find the solution of the D.E ( $^{\ast}$ ).

Given,

$y = ax + \dfrac{b}{a}$

$y = x \dfrac{dy}{dx} + \dfrac{b}{\dfrac{dy}{dx}}$

R :H:S =

$x \dfrac{dy}{dx} + \dfrac{b}{\dfrac{dy}{dx}}$

$= x (a) + \dfrac{b}{a}$

$= ax + \dfrac{b}{a}$

$= y$

$L : H : S = R : H : S$

Hence,

$y = ax + \dfrac{b}{a}$ is solution of given differential equation.

If $dy = (4t^3+3t^2+4t) dt$ find the value of y.

Given that,

$dy=4{{t}^{3}}+3{{t}^{2}}+4t$

On integrating

$\int{dy}=\int{\left( 4{{t}^{3}}+3{{t}^{2}}+4t \right)dt}$

$y={{t}^{4}}+{{t}^{3}}+2{{t}^{2}}$

Hence, this is the required solution

Evaluate: $\int \frac { 1 + v } { 1 - v } d v = \int \frac { d x } { x }$

1970876_1183876_ans_e9fbfdc1b4ad45b6aa8df998a109c622.jpg

Find $\frac{{dy}}{{dx}}$ when $y = {e^{\sqrt x + \sin x}}$

$y=e^{\sqrt{x}+\sin x}$

$\dfrac{dy}{dx}=e^{\sqrt{x}+\sin x}\left(\dfrac{1}{2\sqrt{x}+\cos x}\right)$

$\dfrac{dy}{dx}=e^{\sqrt{x}+\sin x\left(\dfrac{1+2\sqrt{x}\cos x}{2\sqrt{x}}\right)}$

$\dfrac{dy}{dx}=\dfrac{e^{\sqrt{x}+\sin x}(1+2\sqrt{x}\cos x}{2\sqrt{x}}$

1086097_1170566_ans_e30dbafdee3c4495be7626a00579c3ec.png

What is the degree of the following differential equation ?
$5x{\left( {\frac{{dy}}{{dx}}} \right)^2} - \frac{{{d^2}y}}{{d{x^2}}} - 6y = \log x.$

solution -

Degree of diff.eg. is power of highest order

derivative involved.

$5x{\left( {\frac{{dy}}{{dx}}} \right)^2} - \frac{{{d^2}y}}{{d{x^2}}} - 6y = \log x.$

Degree = power of

$\frac{d^{2}y}{dx^{2}} = 1$

1100730_1188365_ans_c0775c9c943845c98410df703f7eda70.jpg

Let $y=a\cos(\log x)+b\sin(\log x)$ is a solution of the differential equation ${x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + y = 0$ .Prove.

$\begin{array}{l} y=a\cos \left( { \log x } \right) +b\sin \left( { \log x } \right) \\ \frac { { dy } }{ { dx } } =-a\sin \left( { \log x } \right) \frac { 1 }{ x } +b\cos \left( { \log x } \right) \frac { 1 }{ x } \\ \frac { { dy } }{ { dx } } =\frac { 1 }{ x } \left( { b\cos \left( { \log x } \right) -a\sin \left( { \log x } \right) } \right) \\ \frac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } =-\frac { 1 }{ { { x^{ 2 } } } } \left( { -b\cos \left( { \log x } \right) +a\sin \left( { \log x } \right) } \right) +\frac { 1 }{ x } \left( { b\sin \left( { \log x } \right) \frac { 1 }{ x } -a\cos \left( { \log x } \right) \frac { 1 }{ x } } \right) \\ \frac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } =\frac { 1 }{ { { x^{ 2 } } } } \left( { -b\cos \left( { \log x } \right) +a\sin \left( { \log x } \right) -b\sin \left( { \log x } \right) -a\cos \left( { \log x } \right) } \right) \\ { x^{ 2 } }\frac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } +x\frac { { dy } }{ { dx } } +y=-b\cos \left( { \log x } \right) +a\sin \left( { \log x } \right) -b\sin \left( { \log x } \right) -a\cos \left( { \log x } \right) \\ +b\cos \left( { \log x } \right) -a\sin \left( { \log x } \right) +a\cos \left( { \log x } \right) +b\sin \left( { \log x } \right) \\ =0 \end{array}$

solve the differential equation : $\dfrac{{dy}}{{dx}} + y {\,\sec ^2}\,x = \tan x\,{\sec ^{2\,}}y;\,y\,(0) = 1$

$\dfrac{dy}{dx} + y \sec^2x = \tan x .\sec^2 y$

$y(0) = 1$

Integrating Factor,

$e^{\int \sec^2x\, dx} = e^{\tan x}$

solution,

$y(I.F) = \int \tan x.sec^2x(I.F) dx$ ......(1)

$y.e^{\tan x} = \int \tan x. \sec^2x. e^{\tan x}.dx$

$\int \tan x. \sec^2x.e^{\tan x} dx$

Let

$\tan x = t$

$\Rightarrow \sec^2 xdx = dt$

$\int \tan x.\sec^2x.e^{\tan x}dx = \int t.e^t dt$

$= te^t -e^t$

$=(\tan x-1)e^{tan x}$

$\therefore$ solution

$\to ye^{\tan x}\Rightarrow (\tan x-1)e^{\tan x}+C$

Now when

$x=0$

$y=1$

$1 = -1 + c \Rightarrow C=2$

$\therefore ye^{\tan x} = \tan x. e^{\tan x} - e^{\tan x}+ 2$

$\dfrac{dy}{dx}=\dfrac{1-cos\:x}{1+cos\:x}$

$\dfrac{dy}{dx}=\dfrac{1-\cos{x}}{1+\cos{x}}$

We know that

$\cos{x}=2{\cos}^{2}{\dfrac{x}{2}}-1=1-{\sin}^{2}{\dfrac{x}{2}}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1-\left(1-{\sin}^{2}{\dfrac{x}{2}}\right)}{1+\left(2{\cos}^{2}{\dfrac{x}{2}}-1\right)}$

$=\dfrac{1-1+{\sin}^{2}{\dfrac{x}{2}}}{1+2{\cos}^{2}{\dfrac{x}{2}}-1}$

$=\dfrac{{\sin}^{2}{\dfrac{x}{2}}}{2{\cos}^{2}{\dfrac{x}{2}}}$

$={\tan}^{2}{\dfrac{x}{2}}$

$={\sec}^{2}{\dfrac{x}{2}}-1$

$\Rightarrow \dfrac{dy}{dx}={\sec}^{2}{\dfrac{x}{2}}-1$

$\Rightarrow dy=\left({\sec}^{2}{\dfrac{x}{2}}-1\right)dx$

$\Rightarrow \int{dy}=\int{\left({\sec}^{2}{\dfrac{x}{2}}-1\right)dx}$

$\Rightarrow \int{dy}=\int{{\sec}^{2}{\dfrac{x}{2}}dx}-\int{dx}$

$\Rightarrow y=\dfrac{\tan{\frac{x}{2}}}{\frac{1}{2}}-x+c$ where

$c$ is the constant of integration

$\therefore y=2\tan{\dfrac{x}{2}}-x+c$ where

$c$ is the constant of integration

Find the order and degree of : $\sqrt[3]{\dfrac{d^2y}{dx^2}} = \sqrt{\dfrac{dy}{dx}}$ .

$3\sqrt { \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } } =\sqrt { \cfrac { dy }{ dx } }$

${ \left( \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) }^{ 3/2 }={ \left( \cfrac { dy }{ dx } \right) }^{ 1/2 }$

squaring on both sides

${ \left( \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) }^{ 3 }-\left( \cfrac { dy }{ dx } \right) =0$

In this equation order

$=2$

degree

$=3$

Evaluate:
$( 1 + \cos x ) d y = ( 1 - \cos x ) d x$

$(1 + \cos x) dy = (1 - \cos x) dx$

$\displaystyle \int dy = \int \dfrac{1 - \cos x}{1 + \cos x } dx$

$\cos x = 1 - 2 \sin^2 (x /2)$

$\cos x=2 \cos^2 (x/2) -1$

$y = \displaystyle \int \dfrac{1 - (1 - 2 \sin^2 (x /2))}{1 + 2 \cos^2 (x /2) - 1} dx = \int \dfrac{\sin^2 (x/ 2)}{\cos^2 ( x /2)} dx$

$y = \displaystyle \int \tan^2 x \, dx = \int (\sec^2 x - 1) dx$

$[\because \sec^2 x - \tan^2 x = 1]$

$y = \tan x - x + c$ .

Solve: $\dfrac{dy}{dx} + \dfrac{4}{\left(\dfrac{dy}{dx}\right)} = 5$

Let

$\cfrac{dy}{dx}=k$

$\Rightarrow$

$k+\cfrac{4}{k}=5$

${k}^{2}-5k+4=0$

$\Rightarrow$

$k=4,1$

$\cfrac{Dy}{dx}=4\Rightarrow$

$y=4x$

$\cfrac{dy}{dx}=1\Rightarrow$

$y=x$

Evaluate: $\frac { d y } { d x } = \sin ( x + y ) + \cos ( x + y )$

$\begin{array}{l} \dfrac { { dy } }{ { dx } } =\sin \left( { x+y } \right) +\cos \left( { x+y } \right) \\ put\, \, x+y=t \\ 1+\dfrac { { dy } }{ { dx } } =\dfrac { { dt } }{ { dx } } \\ \dfrac { { dt } }{ { dx } } =\sin t+\cos t+1 \\ \int { \dfrac { 1 }{ { 1+\sin t+\cos t } } } dt=\int { dx+C } \\ =\int { \dfrac { { { { \sec }^{ 2 } }\dfrac { t }{ 2 } st } }{ { 1+{ { \tan }^{ 2 } }\dfrac { t }{ 2 } +2\tan \dfrac { t }{ 2 } +1-{ { \tan }^{ 2 } }\dfrac { t }{ 2 } } } =x+c } \\ =\int { \dfrac { { { { \sec }^{ 2 } }\dfrac { t }{ 2 } dt } }{ { 1+2\tan \dfrac { t }{ 2 } } } =x+c } \\ =\log \left\{ { 1+2\tan \left( { \dfrac { { x+y } }{ 2 } } \right) } \right\} =x+c \end{array}$

Solve: $\sqrt { 1 + x ^ { 2 } } \sqrt { 1 + y ^ { 2 } } \ d x + x y d y = 0$

$\\\sqrt{1+x^2}\sqrt{1+y^2}dx=-xydy\\\int(\frac{\sqrt{1+x^2}}{x})dx=\int(\frac{-y}{\sqrt{1+y^2}})dy\\let 1+x^2=t^2\>or\>x^2=(t^2-1)\\2xdx=2tdt\>or\>dx=(\frac{tdt}{x})\\and\>\>1+y^2=z^2\\2ydy=2zdz\\ydy=zdz\\\therefore\int(\frac{t}{x})(\frac{tdt}{x})=-\int(\frac{zdz}{z})\\\int(\frac{t^2}{t^2-1})dt=-z+C\\\int dt+\int(\frac{1}{t^2-1})dt=-z+C\\t+(\frac{1}{2})log\left | (\frac{t-1}{t+1})\right|=-z+C\\\sqrt{1+x^2}+(\frac{1}{2})log\left | (\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1})\right|=-\sqrt{1+y^2}+C$

Solve: $\dfrac{d^2 y}{dx^2} + \dfrac{dy}{dx} + e^x = 0$

$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\cfrac { dy }{ dx } +{ e }^{ x }=0\Rightarrow \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\cfrac { dy }{ dx } =-{ e }^{ x }$

Auxillary equation

${m}^{2}+m=0$

$m(m+1)=0$

$m=0,m=-1$

${ y }_{ h }(x)={ C }_{ 1 }{ e }^{ 0x }+{ C }_{ 2 }{ e }^{ -x }\Rightarrow { y }_{ h }(x)={ C }_{ 1 }+{ C }_{ 2 }{ e }^{ -x }\quad$

${ y }_{ p }(x)=A{ e }^{ x }$

$\quad { y }_{ p' }(x)=A{ e }^{ x }$

Putting the value of

${y}_{p}$ and

${Y}_{p'}$ in the differential equation we get

$A{ e }^{ x }+A{ e }^{ x }=-{ e }^{ x }$

$2A{ e }^{ x }=-{ e }^{ x }\Rightarrow 2A=-1\Rightarrow A=-\cfrac { 1 }{ 2 }$

$\therefore { y }_{ p }(x)=-\cfrac { 1 }{ 2 } { e }^{ x }$

$\therefore y(x)=\quad { y }_{ h }(x)+{ y }_{ p }(x)$

$y(x)={ C }_{ 1 }+{ C }_{ 2 }{ e }^{ -x }+\left( -\cfrac { 1 }{ 2 } \right) { e }^{ x }$

Solve the differential equation $\cfrac{dy}{dx}=\cfrac{\sqrt{1-{y}^{2}}}{\sqrt{1-{x}^{2}}}$ .

$\rightarrow \ \dfrac {dy}{dx}=\dfrac {\sqrt {1-y^2}}{\sqrt {1-x^2}}$

$\displaystyle \int \dfrac {dy}{\sqrt {1-y^2}}=\displaystyle \int \dfrac {dy}{\sqrt {1-x^2}}$

$\sin^{-1}y=\sin^{-1}x+c$

$y=\sin (\sin^{-1}x+c)$

Solve : $\left( e ^ { 3 } + 1 \right) \cos x d x + e ^ { y } \sin x d y = 0$

$(e^3+1)cosx dx=-e^ysinx dy\\(e^3+1)\int(\frac{cosx}{sinx}dx)=-\int e^ydy\\(e^3+1)log \left |sinx \right |=-e^y+c$

Show that the given differential equation is homogeneous
$({x^2} + xy)dy = ({x^2} + {y^2})dx$

$(x^2+xy)dy=(x^2+y^2)dx$

$\dfrac{dy}{dx}=\dfrac{x^2+y^2}{x^2+xy}$

Put

$\dfrac{dy}{dx}=F(x, y)$ and Find

$(\lambda x, \lambda y)$

$F(x, y)=\dfrac{x^2+y^2}{x^2+xy}$

$F(\lambda x, \lambda y)=\dfrac{\lambda^2x^2+\lambda^2y^2}{\lambda^2x^2+\lambda^2xy}$

$=\dfrac{\lambda^2(x^2+y^2)}{\lambda^2(x^2+xy)}$

$=\lambda^oF(x, y)$

Hence

$F(x, y)$ is a homogeneous function.

$\dfrac{dy}{dx}$ is a homogeneous differential equation.

Solve $\dfrac{dy}{dx}=cos\:\left(x+y\right)$

Let

$x+y=t\Rightarrow \dfrac{dx}{dx}+\dfrac{dy}{dx}=\dfrac{dt}{dx}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{dt}{dx}-1$

Given

$\dfrac{dy}{dx}=\cos{\left(x+y\right)}$

$\Rightarrow \dfrac{dt}{dx}-1=\cos{t}$

$\Rightarrow \dfrac{dt}{dx}=1+\cos{t}$

$\Rightarrow \dfrac{dt}{1+\cos{t}}=dx$

$\Rightarrow \dfrac{dt}{1+\left(2{\cos}^{2}{\dfrac{t}{2}}-1\right)}=dx$ where

$\cos{t}=2{\cos}^{2}{\dfrac{t}{2}}-1$

$\Rightarrow \dfrac{dt}{2{\cos}^{2}{\dfrac{t}{2}}}=dx$

$\Rightarrow \dfrac{1}{2}{\sec}^{2}{\dfrac{t}{2}}dt=dx$

Integrating both sides

$\Rightarrow \dfrac{1}{2}\int{{\sec}^{2}{\dfrac{t}{2}}dt}=\int{dx}$

$\Rightarrow \dfrac{1}{2}\dfrac{\tan{\dfrac{t}{2}}}{\dfrac{1}{2}}=x+c$ where

$c$ is the constant of integration.

$\Rightarrow \tan{\dfrac{x+y}{2}}=x+c$ where

$t=x+y$

$\Rightarrow \dfrac{x+y}{2}={\tan}^{-1}{\left(x+c\right)}$

$\Rightarrow x+y=2{\tan}^{-1}{\left(x+c\right)}$

$\Rightarrow y=-x+2{\tan}^{-1}{\left(x+c\right)}$

$\dfrac{dy}{dx} = (1 + x^2) (1 + y^2)$

$\begin{array}{l} \frac { { dy } }{ { dx } } =\left( { 1+{ x^{ 2 } } } \right) \left( { 1+{ y^{ 2 } } } \right) \\ \frac { { dy } }{ { 1+{ y^{ 2 } } } } =\left( { 1+{ x^{ 2 } } } \right) dx \\ Integrate\, both\, sides \\ \int _{ }^{ }{ \frac { 1 }{ { 1+{ y^{ 2 } } } } dy } =\int _{ }^{ }{ \left( { 1+{ x^{ 2 } } } \right) } dx \\ { \tan ^{ -1 } }y=\frac { { { x^{ 3 } } } }{ 3 } +x+C \end{array}$

For differential equation $\left(\dfrac{dy}{dx}\right)^{2}-x\left(\dfrac{dy}{dx}\right)+y=0$ ,find the solution

$\left( \dfrac{dy}{dx} \right)^{2}-x \left( \dfrac{dy}{dx} \right)+y=0$

Let

$\dfrac{dy}{dx}=p$ so we get

$p^{2}-xp+y=0$

Differentiating w.r.t

$'x'$

$2p \dfrac{dp}{dx}-p-x.\dfrac{dp}{dx}+ \dfrac{dy}{dx}=0$

$2p \dfrac{dp}{dx} - x . \dfrac{dp}{dx} - p + p = 0$

$2p.\dfrac{dp}{dx}-x\dfrac{dp}{dx}=0$

$2p-x=0\Rightarrow 2p=x$

$\dfrac{dy}{dx}=p$ So

$\dfrac{dy}{dx}=\dfrac{x}{2}$

Integrating both sides

$\int dy= \int \dfrac{x}{2}.dx$

$y=\dfrac{x^{2}}{4}+c$

Solve
$(1+e^{x/y})dx+e^{x/y} \left(1-\dfrac{x}{y}\right)dy=0$

$(1+e^{x/y})dx+e^{x/y}\left(1-\dfrac{x}{y}\right)dy=0$

$(1+e^{x/y})dx=-e^{x/y}\left(1-\dfrac{x}{y}\right)dy$

$\dfrac{dx}{dy}=\dfrac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$\dfrac{dx}{dy}=f(x, y)$

$f(x, y)=\dfrac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$f(\lambda x, \lambda y)=\dfrac{e^{-\lambda x/\lambda y}(1-\lambda x/\lambda y)}{1+e^{\lambda x/\lambda y}}$

$=\dfrac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$=\lambda^o f(x, y)$

Let

$x=vy$

Put

$\dfrac{dx}{dy}=v\dfrac{dy}{dy}+y\dfrac{dv}{dy}$

$\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$

$\dfrac{dx}{dy}=\dfrac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$V+y\dfrac{dv}{dy}=\dfrac{-e^{vy/y}(1-vy/y)}{1+e^{vy/y}}$

$y\dfrac{dv}{dy}=\dfrac{-e^{v}(1-v)}{1+e^{v}}-V$

$y\dfrac{dv}{dy}=\dfrac{-e^v+ve^v-v-ve^v}{1+e^v}$

$\dfrac{1+e^v}{v+e^v}dv=\dfrac{-dy}{y}$

Integrate Both sides

$\displaystyle \int{\dfrac{1+e^v}{v+e^v}dv}=-\log y+\log C$

$V+e^v=t\rightarrow (1+e^v)dv=dt$

$\displaystyle \int{\dfrac{dt}{t}=-\log y+\log C}$

now substitute

$t$

$\log t=-\log y+\log c$

$\log(v+e^v)=-\log y+\log C$

$\log(v+e^v)+\log y=\log C$

$\log y(v+e^v)=\log C$

Put

$v=\dfrac{x}{y}$

$\log y\left(\dfrac{x}{y}\right)+e^\dfrac{x}{y}$

$\log y\left(\dfrac{x}{y}+e^\dfrac{x}{y}\right)=\log c$

$y\left(\dfrac{x}{y}+e^{x/y}\right)=C$

$x+ye^{x/y}=C$ .

$x + y \dfrac { d y } { d x } = \sec \left( x ^ { 2 } + y ^ { 2 } \right)$

$\cfrac{xdx+ydy}{dx}=\sec(x^2+y^2)$

$\implies \cfrac{d(x^2+y^2)}{\sec(x^2+y^2)}=2dx\implies \cos (x^2+y^2)d(x^2+y^2)=2dx$

So, on integrating on both sides we get

$\sin(x^2+y^2)=2x+c$

Solve $\dfrac{dy}{dx}=1+x+y+xy$

$\dfrac{dy}{dx}=1+x+y+xy$

$\Rightarrow \dfrac{dy}{dx}=\left(1+x\right)+y\left(1+x\right)$ by splitting into factors

$\Rightarrow \dfrac{dy}{dx}=\left(1+x\right)\left(1+y\right)$ by grouping into factors

$\Rightarrow \dfrac{dy}{\left(1+y\right)}=\left(1+x\right)dx$

Integrating both sides we get

$\Rightarrow \int{\dfrac{dy}{\left(1+y\right)}}=\int{\left(1+x\right)dx}$

$\Rightarrow \ln\left|y+1\right|=x+\dfrac{{x}^{2}}{2}+c$ where

$c$ is the constant of integration

$\Rightarrow y+1={e}^{x+\dfrac{{x}^{2}}{2}+c}$

$\therefore y={e}^{x+\dfrac{{x}^{2}}{2}+c}-1$

Find the general solution of the equation given by $\dfrac {dy}{dx}+\dfrac {1+y^{2}}{1+x^{2}}=0$ .

1233174_1227358_ans_174075df97034f469799d142ef20528f.jpg

solve
xdy - ydx = $\sqrt{x^2 + y^2dx}$

$\rightarrow$ Let

$x = r cos\theta , y = r \sin\theta$

$\Rightarrow dx = -r\sin\theta d\theta +\cos\theta \,dr$

$dy = r\cos\theta \,d\theta +\sin\theta dr$

$\Rightarrow xdy = r^{2}\cos^{2}\theta \,d\theta +r.\sin\theta ,\cos\theta .d\theta$

$ydx = -r^{2}\sin^{2}\theta d\theta +r\sin\theta .\cos\theta d\theta$

$\Rightarrow xdy-ydx = r^{2}d\theta$

$x^{2}+y^{2} = r^{2}$

$\Rightarrow r^{2}d\theta = r(-r\sin\theta d\theta +cos\theta dr)$

$\Rightarrow (r^{2})(1+\sin\theta )d\theta = rcos\theta dr$

$\Rightarrow \int \dfrac{1+\sin\theta }{\cos\theta }d\theta = \dfrac{dr}{r}$

$\Rightarrow ln\left |\ sec\theta +tan\theta \right |+ ln \left |\ sec\theta \right | = ln r+c$

$\Rightarrow ln \left | \sec^{2}\theta +\sec\theta +\tan\theta \right | = ln\sqrt{x^{2}+y^{2}+c}$

$\Rightarrow ln \left | 1+(\dfrac{y}{x})^{2}+\dfrac{y}{x}\sqrt{1+\dfrac{y^{2}}{x^{2}}} \right | = ln \sqrt{x^{2}+y^{2}+c}$

Solve:
$\dfrac{dy}{dx}=x+e^x$ .

Given,

$\dfrac{dy}{dx}=x+e^x$

or,

$dy=(x+e^x)dx$

Now integrating we have,

$y=\dfrac{x^2}{x}+e^x+c$ . [ Where

$c$ is integrating constant]

Solve:
$\dfrac{dy}{dx}=\sqrt{4-y^{2}}(-2 < y < 2)$ .

Given,

$\dfrac{dy}{dx}=\sqrt{4-y^{2}}(-2 < y < 2)$

or,

$\dfrac{dy}{\sqrt{4-y^2}}=dx$

Now integrating we have,

$\sin^{-1}\dfrac{y}{2}=x+c$

or,

$y=\sin (x+c)$ .

Solve the differential equation $\left( 1+e^{ 2x } \right) dy+(1+y^{ 2 }){ e }^{ x }dx=0$ given that $y=1$ when $x=0$ .

1311790_1236583_ans_b512d1c993d54f5898126f960eb3068b.jpg

Solve:
$\cos ( x + y ) d y = d x$

$\\cos(x+y)dy=dx\\(\frac{dy}{dx})=(\frac{1}{cos(x+y)})\\let\>x+y=t\\then 1+(\frac{dy}{dx})=(\frac{dt}{dx})\\or\>(\frac{dy}{dx})=(\frac{dt}{dx})-1\\\therefore\>(\frac{dt}{dx})-1=(\frac{1}{cost})\\(\frac{dt}{dx})=(\frac{1}{cost})+1\\(\frac{dt}{dx})=(\frac{1+cost}{cost})\\\int(\frac{cost\>dt}{1+cost})=\int\>dx\\\int(\frac{cost(1-cost)}{1^2-cos^2t})dt=x+c\\\int(\frac{cost-cos^2t}{sin^2t})dt=x+c\\\int cot\>t\>cosect\>dt-\int cot^2t\>dt=x+c\\-cosect-\int(cosec^2t-1)dt=x+c\\-cosect+cott+t=x+c\\-cosec(x+y)+cot(x+y)+x+y=x+c\\-cosec(x+y)+cot(x+y)+y=c$

If $x = \sin t,y = \sin pt$ then prove that $(1 - {x^2}){d^2}y - xdy + {p^2}y = 0$

Solution:-

$x=\sin t$ ,

$dx/dt=\cos t=\sqrt{1-x^2}$

$y=\sin pt$ ,

$dy/dt=p\cos pt=p\sqrt{1-y^2}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{dt}/\dfrac{dx}{dt}$

$=\dfrac{p\cos p t}{\cos t}=\dfrac{p\sqrt{1-y^2}}{\sqrt{1-x^2}}$

$\Rightarrow \dfrac{d^2y}{dx^2}=-\left[\dfrac{p^2y}{1-x^2}+\dfrac{px\sqrt{1-y^2}}{(1-x^2)^{2/2}}\right]$

Substituting in

$(1-x^2)\dfrac{d^2y}{dx^2}+\dfrac{xdy}{dx}+xp^2y$

$\Rightarrow -\left[p^2y+\dfrac{px\sqrt{1-y^2}}{\sqrt{1-x^2}}\right]+\dfrac{px\sqrt{1-y^2}}{\sqrt{1-x^2}}+p^2y$

$=0$ .

Solve the differential equation : $y\log { y } \dfrac { dx }{ dy } +x-\log { y=0 }$ .

We have,

$y\log y\dfrac{dx}{dy}+x-\log y=0$

Now,

$\dfrac{dx}{dy}+\dfrac{x}{y\log y}-\dfrac{\log y}{y\log y}=0$

$\dfrac{dx}{dy}+\dfrac{x}{y\log y}=\dfrac{1}{y}$

On Comparing that,

$\dfrac{dx}{dy}+Px=Q$

$P=\dfrac{1}{y\log y},\,\,\,Q=\dfrac{1}{y}$

$I.F.$

$I.F.={{e}^{\int{Pdy}}}={{e}^{\int{\dfrac{1}{y\log y}}dy}}={{e}^{\log (\log y)}}=\log y$

Then,

$x\times I.F.=\int{Q\times I.F.+C}$

$x\log y=\int{\dfrac{1}{y}\times \log ydy}+C$

$x\log y=\dfrac{(\log y)^2}{2}+C$

Hence, this is the answer.

Solve:
$(x+y)\dfrac{dy}{dx}= 1$

$(x+y)\dfrac {dy}{dx}=1$

$=0\ (x+y)dy=dx$

$=0\ \dfrac {dx}{dy}+(-1)x=y$

I.F $=e^{\int -1dy}=e^{-y}$

$\therefore$ Required soln;

$x\times$ I.F $=\int y\times e^{-y}dy ++c$

By integration, we get $x+y+1=ce^{y}$

The general solution of the differential equation $\dfrac{dy}{dx}=(1+x^2)(1+y^2)$ .

1344720_1246541_ans_71470837130642eca5333623edde1f29.jpg

Find the general solution of the differential equation.
$(1+\tan y)(dx-dy)+2xdy=0$ .

$(1+\tan y)(dx-dy)+2xdy=0$

$\dfrac{(1+\tan y)}{2x}=\dfrac{-dy}{dx-dy}$

$\dfrac{dx-dy}{-dy}=\dfrac{2x}{1+\tan y}$

$\dfrac{-dx}{dy}+1=\dfrac{2x}{1+\tan y}$

$\dfrac{dx}{dy}+\dfrac{2x}{1+\tan y}=1$

$I.F=\displaystyle \int e^{\dfrac{2}{1+\tan y}.dy}$

$=\displaystyle \int e^{\dfrac{2\cos y}{1+\sin y+\cos y}.dy}$

$=\displaystyle \int e^{1+\dfrac{\cos y-\sin y}{\sin y+\cos y}.dx}$

$=e^{y+(n(\sin y+\cos y))}$

$=e^y.(\sin y+\cos y)$

$x.e^y.(\sin y.\cos y)=\displaystyle \int e^y(\sin y+\cos y).dy$

$x.e^y(\sin y+\cos y)=e^y.\sin y+c$

$x(\sin y+\cos y)=\sin y+ce^{-y}$

Solve the differential equation $(1+y^2)(1+log{\,}x)dx+x{\,}dy=0$ , given that when $x=1$ then $y=1$ .

$(1 + y^{2}) ( 1 + \log x) dx = x dy$

$\dfrac{dy}{1 + y^2} = \dfrac{- (1 + \log x)}{x} dx$

Integrate

$\tan^{-1} y = - \int \dfrac{1 + \log x }{x} dx$

Let

$1 + \log x = t$

$\dfrac{dx}{x} = dt$

$\tan^{-1} y = \dfrac{-1}{2} (1 + \log x)^{2} + C$

Put

$x = 1 , y = 1$

$\dfrac{\pi}{4} = \dfrac{-1}{2} + C$

$C = \dfrac{\pi}{4} + \dfrac{1}{2}$

$C = \dfrac{\pi + 2}{4}$

$\tan^{-1} y = \dfrac{-1}{2} (1 + \log x)^{2} + \dfrac{\pi + 2}{4}$

Solve $y$ $dx-(x+2y^2)dy = 0$

$ydx-(x+2y^2)dy=0$

$ydx=(x+2y^2)dy$

$\dfrac{dx}{dy}=\dfrac{x}{y}+2y$

$ydx-(x+2y^2)dy=0$

$ydx=(x+2y^2)dy$

$ydx-xdy=2y^2dy$

$\dfrac{(ydx-xdy)}{y^2}=2dy$

Integration on both sides

$\displaystyle\int d\left(\dfrac{x}{y}\right)=\displaystyle\int 2dy$

$\dfrac{x}{y}=2y+c$ .

Solve : $y^{2}dx+(x^{2}+xy)dy=0$

1190261_1266016_ans_920e4b55af3b4bc7ae11e3dfab2d9ce1.jpg

Solve the differential equation $\cos\left(\dfrac{dy}{dx}\right)=a, (a\in R)$ .

$\textbf{Hint: Solve by variable separable method}$

Solution:

$\textbf{Step 1: Simplify the given differential equation}$

$\cos\left(\dfrac{dy}{dx} \right)=a$

$\Rightarrow \dfrac{dy}{dx}=\cos^{-1}a$

$\textbf{Step 2: Solve by variable separable method}$

$\dfrac{dy}{dx}=\cos^{-1}a$

$\Rightarrow dy=\cos^{-1}a\,dx$

Integrating both the sides,

$\Rightarrow \int d y=\int \cos^{-1} a d x$

$y=x\cos^{-1}a+C$

$\textbf{ Hence, the solution of the given differential equation is }$

$y=x\cos^{-1}a+C$

Solve the differential equation $(x^{2}-y^{2})dx+2xydy=0$ .

$\left( {{x^2} - {y^2}} \right)dx + 2xydy = 0$

Now,

$\begin{array}{l} \Rightarrow { x^{ 2 } }dx={ y^{ 2 } }dx-2xydy \\ \Rightarrow -dx=\frac { { 2xydy-{ y^{ 2 } }dx } }{ { { x^{ 2 } } } } \\ \Rightarrow -dx=\frac { { 2xydy-{ y^{ 2 } }dx } }{ { { x^{ 2 } } } } \\ \Rightarrow -dx=d\left( { \frac { { { y^{ 2 } } } }{ x } } \right) \\ \Rightarrow C-x=\frac { { { y^{ 2 } } } }{ x } \\ hence,\, \, \, we\, \, get \\ \frac { { { y^{ 2 } } } }{ x } +x=C \end{array}$

Prove that $y={ ae }^{ -2x }+{ be }^{ x }$ is the solution of differential equation $\dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\dfrac { dy }{ dx } -2y=0$

We have,

$y=a{{e}^{-2x}}+b{{e}^{x}}$

On differentiating and we get,

$\dfrac{dy}{dx}=-2a{{e}^{-2x}}+b{{e}^{x}}$

Again differentiating and we get,

$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4a{{e}^{-2x}}+b{{e}^{x}}$

Now,

L.H.S.

$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{dy}{dx}-2y$

$\Rightarrow 4a{{e}^{-2x}}+b{{e}^{x}}-2a{{e}^{-2x}}+b{{e}^{x}}-2\left( a{{e}^{-2x}}+b{{e}^{x}} \right)$

$\Rightarrow 4a{{e}^{-2x}}+b{{e}^{x}}-2a{{e}^{-2x}}+b{{e}^{x}}-2a{{e}^{-2x}}-2b{{e}^{x}}$

$\Rightarrow 4a{{e}^{-2x}}-4a{{e}^{-2x}}+2b{{e}^{x}}-2b{{e}^{x}}$

$\Rightarrow 0$

Hence, this is the answer.

Construct a differential equation by eliminating the arbitrary constants $a,b$ from the equation:
$y=(ax+b)e^x$ .

Given the equation,

$y=(ax+b)e^x$ ......(1)

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=(ax+b)e^x+ae^x$

or,

$ae^x=\dfrac{dy}{dx}-y$ . [ Using (1)]......(2)

Again differentiating both sides with respect to

$x$ we get,

$\dfrac{d^2y}{dx^2}=(ax+b)e^x+2ae^x$

or,

$\dfrac{d^2y}{dx^2}=2\left(\dfrac{dy}{dx}-y\right)+y$ [ Using (1) and (2)]

or,

$\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0$ .

This is the required differential equation.

Reduce the following differential equation to the variables separable form and hence solve.
$(x-y)^2\dfrac{dy}{dx}=a^2$ .

${ (x-y) }^{ 2 }\cfrac { dy }{ dx } ={ a }^{ 2 }$

Let

$x-y=v\quad \Rightarrow 1-\cfrac { dy }{ dx } =\cfrac { dv }{ dx }$

$\cfrac { dy }{ dx } =1-\cfrac { dv }{ dx } \\ { (x-y) }^{ 2 }\cfrac { dy }{ dx } ={ a }^{ 2 }\quad \Rightarrow { v }^{ 2 }(1+\cfrac { dv }{ dx } )={ a }^{ 2 }\\ \Rightarrow 1+\cfrac { dv }{ dx } =\cfrac { { a }^{ 2 } }{ { v }^{ 2 } } \quad \Rightarrow \cfrac { dv }{ dx } =\cfrac { { a }^{ 2 } }{ { v }^{ 2 } } -1=\cfrac { { a }^{ 2 }-{ v }^{ 2 } }{ { v }^{ 2 } } \\ \Rightarrow \int { \cfrac { { v }^{ 2 } }{ { a }^{ 2 }-{ v }^{ 2 } } dv } =\int { dx } +C\\ =-[\int { \cfrac { { (a }^{ 2 }-{ v }^{ 2 })-{ a }^{ 2 } }{ { a }^{ 2 }-{ v }^{ 2 } } dv } ]=\int { dx } +C\\ \Rightarrow [\int { (1-\cfrac { { a }^{ 2 } }{ { a }^{ 2 }-{ v }^{ 2 } } )dv } ]=\int { dx } +C\\ \Rightarrow -[v-{ a }^{ 2 }\times \cfrac { 1 }{ 2a } \ln { \left| \cfrac { a+v }{ a-v } \right| } ]=x+C\\ \Rightarrow -v+\cfrac { a }{ 2 } \ln { \left| \cfrac { a+v }{ a-v } \right| } =x+C\\ v=x-y\\ \Rightarrow -(x-y)+\cfrac { a }{ 2 } \ln { \left| \cfrac { a+x-y }{ a-x+y } \right| } =x+C\\ \Rightarrow -y-x+\cfrac { a }{ 2 } \ln { \left| \cfrac { a+x-y }{ a-x+y } \right| } =x+C\\ \cfrac { a }{ 2 } \ln { \left| \cfrac { a+x-y }{ a-x+y } \right| } +y=2x+C\\ \Rightarrow a\ln { \left| \cfrac { a+x-y }{ a-x+y } \right| } +2y=4x+2C$

Solve the following differential equation.
$y-x\dfrac{dy}{dx}=0$ .

$y -x\dfrac{dy}{dx}=0$

$y =x\dfrac{dy}{dx}$

$\dfrac{dx}{x}=\frac{dx}{dy}$

Integrate both the sides

$\int \dfrac{dx}{x} =\int \dfrac{dy}{y}$

$\dfrac {-1}{x^2}+c_1=\dfrac{-1}{y^2}+c_2$

$\dfrac{1}{y^2}-\dfrac{1}{x^2}+c=0$

Solve the differential equation $ye^{ { \dfrac { x }{ y } } }dx=\left( xe^{ { \dfrac { x }{ y } } }+y^{ { 2 } } \right) dy(y\neq 0).$

1311792_1322493_ans_085cdf656a9f4ce0bff4898507357f49.jpg

$\dfrac { dx }{ dy } +\dfrac { x }{ 1+{ y }^{ 2 } } =\dfrac { { tan }^{ -1 }y }{ 1+{ y }^{ 2 } }$

1188507_1320551_ans_70b2528ca5e648d99037db2f6750ab02.jpg

The solution of differential equation $(sinx + cosx)dy$ + $(cosx - sinx)dx=0$ is

$(\sin x+\cos x)dy+(\cos x-\sin x)dx=0$

$(\sin x+\cos x)dy=-(\cos x-\sin x)dx$

$\dfrac{dy}{dx}=-\dfrac{(\cos x-\sin x)}{(\sin x+\cos x)}$

$dy=-\dfrac{(\cos x-\sin x)}{(\sin x+\cos x)}dx$

$\displaystyle\int dy=-\displaystyle\int \dfrac{\cos x-\sin x}{\sin x+\cos x}dx$

$\sin x+\cos x=t$

$(\cos x-\sin x)dx=dt$

$(\sin x-\cos x)dx=-dt$

$y=\displaystyle\int -\dfrac{dt}{t}+c$

$=-log |t|+c$

$y=-log|\sin x+\cos x|+c$ .

1226487_1316850_ans_d271eb7fc1144845831daa7c10cd7ab3.jpg

Solve:
$y - x\dfrac{{dy}}{{dx}} = a\left( {{y^2} + \dfrac{{dy}}{{dx}}} \right)$

Given the differential equation,

$y - x\dfrac{{dy}}{{dx}} = a\left( {{y^2} + \dfrac{{dy}}{{dx}}} \right)$

or,

$(x+a)\dfrac{dy}{dx}=y-ay^2$

or,

$\dfrac{dy}{y-ay^2}=\dfrac{dx}{x+a}$

or,

$\left[\dfrac{1}{y}+\dfrac{a}{1-ay}\right] dy=\dfrac{dx}{x+a}$

Now integrating both sides we get,

$\log|y|-\log|1-ay|=\log|a+x|+\log |c|$ [ Where

$c$ is integrating constant]

or,

$\dfrac{y}{1-ay}=c(a+x)$ .

Solve the differential equation: ${ x }dy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx$

Given equation can be written as

$x^2 \dfrac{dy}{dx}-xy=2 \cos^2 \left(\dfrac{y}{2x}\right),\ x \ne 0$

$\Rightarrow \dfrac{x^2 \dfrac{dy}{dx}-xy}{2 \cos^2 \left(\dfrac{y}{2x}\right)}=1 \Rightarrow \dfrac{sec^2 \left(\dfrac{y}{2x}\right)}{2} \left[x^2 \dfrac{dy}{dx}-xy \right]=1$
Dividing both sides by

$x^3$ , we get

$\dfrac{sec^2 \left(\dfrac{y}{2x}\right)}{2} \left[\dfrac{x \dfrac{dy}{dx}-y}{x^2} \right] =\dfrac{1}{x^3} \Rightarrow \dfrac{d}{dx}\left[ \tan \left(\dfrac{y}{2x} \right)\right]=\dfrac{1}{x^3}$
Integrating both sides, we get

$\tan \left(\dfrac{y}{2x} \right) =\dfrac{-1}{2x^2}+k$
Substituting

$x=1,\ y=\dfrac{\pi}{2}$ , we get

$k=\dfrac{3}{2},$ therefore,

$\tan\left(\dfrac{y}{2x} \right) =\dfrac{1}{2x^2} +\dfrac{3}{2} \tan \left(\dfrac{y}{2x} \right)=\dfrac{-1}{2x^2} +\dfrac{3}{2}$ is the required solution.

Solve :
$\dfrac{dy}{dx}=\dfrac{1}{y^2+\sin y}, y\neq 0$

$\dfrac{dy}{dx}=\dfrac{1}{y^2+\sin y}$

$\dfrac{dx}{dy}=y^2+\sin y$

$dx=(y^2+\sin y)dy$

Integrating both sides,

$\int dx = \int (y^2+\sin y)dy$

$x=\dfrac{y^3}{3}-\cos y+C$

This is the required solution.

Solve the differential equation $\dfrac{dy}{dx}=\dfrac{1+y}{x}$

$\frac{dy}{dx}=\frac{1+y}{x}$

$\frac{1}{1+y}dy=\frac{1}{x}dx$

on integerating both sides.

$\int \frac{1}{1+y}dy=\int \frac{1}{x}dx$

$log (1+y)= log x+c$

1200601_1382673_ans_f75362fa80d046b181b3e929a2d68e3c.JPG

Find the differential equation whose solution represents the family: $y = a e ^ { 3 x } + b e ^ { x }$

$\begin{array}{l} y=a{ e^{ 3x } }+b{ e^{ x } } \\ \dfrac { { dy } }{ { dx } } =3a{ e^{ 3x } }+b{ e^{ x } } \\ \dfrac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } =9a{ e^{ 3x } }+b{ e^{ x } } \\ \dfrac { { dy } }{ { dx } } -y=2a{ e^{ 3x } } \\ \frac { 1 }{ 2 } \left[ { \dfrac { { dy } }{ { dx } } -y } \right] =a{ e^{ 3x } } \\ y-\dfrac { 1 }{ 2 } \left[ { \dfrac { { dy } }{ { dx } } -y } \right] =b{ e^{ x } } \\ \dfrac { { 3y } }{ 2 } -\frac { 1 }{ 2 } \dfrac { { dy } }{ { dx } } =b{ e^{ x } } \\ \dfrac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } =\dfrac { 9 }{ 2 } \left[ { \frac { { dy } }{ { dx } } -y } \right] +\dfrac { { 3y } }{ 2 } -\dfrac { 1 }{ 2 } \frac { { dy } }{ { dx } } \\ \dfrac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } =4\dfrac { { dy } }{ { dx } } -3y \\ \dfrac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } -4\dfrac { { dy } }{ { dx } } +3y=0 \end{array}$

Find the differential equation of the family of curve $y=A{e}^{2x}+B{e}^{-2x}$ for different values of $A$ and $B$ .

Given:

$y=A{e}^{2x}+B{e}^{-2x}$

${y}^{\prime}=2A{e}^{2x}-2B{e}^{-2x}$

${y}^{\prime\prime}=4A{e}^{2x}+4B{e}^{-2x}=4y$

$\Rightarrow\,{y}^{\prime\prime}-4y=0$

Verify whether $y=\sqrt{1+x^{2}}$ is a solution of the differential equation $y'=\dfrac{x}{\sqrt{1+x^{2}}}$

$y=\sqrt{1+{x}^{2}}$

squaring both sides, we get

${y}^{2}=1+{x}^{2}$

$2y\dfrac{dy}{dx}=2x$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{y}$

Where

$y=\sqrt{1+{x}^{2}}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+{x}^{2}}}$

Solve:
$\dfrac { d y } { d x } = x+1 ;$ find $y$ when $x = 2$

$\dfrac{dy}{dx}=x+1$

$y=\displaystyle \int x+1dx$

$y=\dfrac{x^2}2+x$

$x=2$

$y=\dfrac{(2)^2}2+2$

$y=4$

Solve:
$\dfrac { d y } { d x } = 2 x ^ { 2 } + x ;$ find $y$ when $x = 0$

$\dfrac{dy}{dx}=2x^2+x$

$y=\displaystyle \int 2x^2+xdx$

$y=\dfrac {2x^3}{3}+\dfrac{x^2}2$

$y=\dfrac{2(0)}{3}+\dfrac 02$

$y=0$

Solve $x \frac { d y } { d x } + 2 x = x ^ { 2 } \log x$

$x\dfrac{dy}{dx}+2x=\log x\\\dfrac{dy}{dx}+2=\dfrac{\log x}{x}\\\dfrac{dy}{dx}=\dfrac{\log x}x-2\\y=\displaystyle \int (\dfrac{\log x}x-2 )dx\\y=\dfrac{(\log x)^2}{2}-2x+c$

Find the general solution for the following differential equation:
$x^{2}dy+y(x+y)dx=0$

$x^2dy = -y(x+y)dx$

$\implies \dfrac{dy}{dx}= -\dfrac{y(x+y)}{x^2}$

$\dfrac{dy}{dx}= -\dfrac{xy}{x^2}-\dfrac{y^2}{x^2}$

$\dfrac{dy}{dx}= -\dfrac{y}{x}-(\dfrac{y}{x})^2$

The above differential equation is homogeneous.

$\therefore$ put

$vx=y........(1)$

$\dfrac{dy}{dx}= -v-v^2.....(2)$

differentiating equation (1),

$\dfrac{dy}{dx}= v+ x\dfrac{dv}{dx}......(3)$

by equation (2) & (3),

$v+ x\dfrac{dv}{dx}= -v^2 - v$

$x\dfrac{dv}{dx}= -v^2-2v$

$\implies \dfrac{1}{v^2+2v}dv= -\dfrac{1}{x}dx$

integrating both sides,

$\displaystyle \int\dfrac{1}{v^2+2v}dv= -\int \dfrac{1}{x}dx$

$\displaystyle \int\dfrac{1}{(v^2+2v+1)-1}dv= -ln|x|+ln|c|$

$\displaystyle \int\dfrac{1}{(v+1)^2-1^2}dv=-ln|x|+ln|c|$

using

$\displaystyle\int\dfrac{1}{x^2-a^2}dx=\dfrac{1}{2a}ln|\dfrac{x-a}{x+a}|+c$

$\dfrac{1}{2}ln\left|\dfrac{v+1-1}{v+1+1}\right|=-ln|x|+ln|c|$

$\dfrac{1}{2}ln\left|\dfrac{v}{v+2}\right|=-ln|x|+ln|c|$

$ln\left|\dfrac{\sqrt{v}}{\sqrt{v+2}}\right|=-ln|x|+ln|c|$

$ln\left|\dfrac{\sqrt{v}}{\sqrt{v+2}}\right|+ln|x|=ln|c|$

$ln\left|\dfrac{x\sqrt{v}}{\sqrt{v+2}}\right|=ln|c|$

$\implies\dfrac{x\sqrt{v}}{\sqrt{v+2}}=c$

from equation (1)

$v=\dfrac{y}{x}$ ,

$\dfrac{x\sqrt{\dfrac{y}{x}}}{\sqrt{\dfrac{y}{x}+2}}=c$

squaring both sides,

$\dfrac{xy}{\left(\dfrac{y+2x}{x}\right)}=c^2$

Therefore,

general solution

$:x^2y=c^2(y+2x)$

Let $y=\sin^{-1}x$ then prove that $(1-x^2)y_2-xy_1=0$ . Where $y_1$ and $y_2$ are first and second order derivatives of $y$ with respect to $x$ respectively.

Given,

$y=\sin^{-1}x$

Now differentiating both sides we get,

$y_1=\dfrac{1}{\sqrt{1-x^2}}$

Now squaring both sides we get,

$(1-x^2)y_1^2=1$

Now again differentiating both sides with respect to

$x$ we get,

$(1-x^2)2y_1.y_2-2x.y_1^2=0$

or,

$(1-x^2)y_2-xy_1=0$ . [ Since

$y_1\ne 0$ ]

Find $\dfrac{dy}{dx}$ for $y=\sqrt{x+\sqrt{x+\sqrt{x+.....}}}$ .

Given,

$y=\sqrt{x+\sqrt{x+\sqrt{x+.....}}}$ .....(1)

Now squaring both sides we get,

$y^2=x+\sqrt{x+\sqrt{x+.......}}$

or,

$y^2=x+y$ [ Using (1)]

Now differentiating both sides with respect to

$x$ we get,

$2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx}$

or,

$(2y-1)\dfrac{dy}{dx}=1$

or,

$\dfrac{dy}{dx}=\dfrac{1}{2y-1}$ .

Write the order of the differential equation: $\left(\dfrac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}\right)=\left(\dfrac{\mathrm{dy}}{\mathrm{dx}}\right)^{3}+\mathrm{x}$

The order of the differential equation is given as order of highest order derivative term in the expansion.

Here

$2$ is the highest order derivative so ,Order is

$2$

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$y=ae^x+be^{-x}+x^2$ : $x\frac {d^2y}{dx^2}+2\frac {dy}{dx}-xy+x^2-2=0$

$Given \space equation$

$y=ae^x+be^{-x}+x^2$

$Differentiating \space wrt \space to \space x$

$\dfrac{dy}{dx}=ae^x-be^{-x}+2x...............(1)$

$Differentiating \space again \space wrt \space x$

$\dfrac{d^2y}{dx^2}=ae^x+be^{-x}+2..................(2)$

$Given \space Differential \space equation$

$x\frac {d^2y}{dx^2}+2\frac {dy}{dx}-xy+x^2-2=0$

$To \space check \space whether \space given \space equation \space satisfies \space the \space DE \space substitute \space values \space of \space y,\dfrac{dy}{dx},\dfrac{d^2y}{dx^2} \space in\space DE$

$x\frac {d^2y}{dx^2}+2\frac {dy}{dx}-xy+x^2-2$

$=x(ae^x+be^{-x}+2)+2(ae^x-be^{-x}+2x)-x(ae^x+be^{-x}+x^2)+x^2-2$

$Canceling \space terms \space we \space we \space get$

$6x+2ae^x-2be^{-x}-x^3+x^2-2 \neq 0$

$Hence \space given \space function \space not \space solution \space of \space DE$

$xdy + (y - x^{3})dx = 0$

By seperating the terms

$xdy+ydx={x}^{3}dx$

We know that

$\dfrac{d\left(xy\right)}{dx}=x\dfrac{dy}{dx}+y\dfrac{dx}{dx}$

Now,

$xdy+ydx=d\left(xy\right)$

$\Rightarrow xdy+ydx={x}^{3}dx$

$\Rightarrow d\left(xy\right)dx={x}^{3}dx$

$\Rightarrow \int{d\left(xy\right)dx}=\int{{x}^{3}dx}$

$\Rightarrow xy=\dfrac{{x}^{4}}{4}+c$ where

$c$ is the constant of integration

$\Rightarrow y=\dfrac{{x}^{3}}{4}+cx$ is the solution of the above equation

Solve: $\displaystyle y - x\frac{{dy}}{{dx}} = a\left( {{y^2}+\frac{{dy}}{{dx}}} \right)$

or,

$\ln |y|-\ln|1-ay|=\ln |(a+x)|+\ln c$ [ Where

$c$ is integrating constant]

or,

$\dfrac{y}{1-ay}=c(a+x)$ .

1211823_1428150_ans_dbf7944e5bc34215b720d28992c9bf9b.jpg

If $x=at^{ 2 },\ \ y= 2at,$ then find $y\dfrac { dy }{ dx }$

1314670_1488241_ans_4748f3ba23d84908ba0ca14d634df92d.JPG

Solve the differential equation :
$(xdy - ydx)\,y\,sin\left( {\frac{y}{x}} \right) = (ydx + xdy)X\,\cos \,\left( {\frac{y}{x}} \right)$

1203202_1420932_ans_591b9f5b74cd4dacba92a8ab9939b59d.jpg

Solve: $\left(y+3{x}^{2}\right)\dfrac{dx}{dy}=x$

$\Rightarrow \left(y+3{x}^{2}\right)=x\dfrac{dy}{dx}$

Let

$y=vx\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

$\Rightarrow \left(vx+3{x}^{2}\right)=x\left(v+x\dfrac{dv}{dx}\right)$

$\Rightarrow x\left(v+3x\right)=x\left(v+x\dfrac{dv}{dx}\right)$ by removing the common on the left hand side.

$\Rightarrow v+3x=v+x\dfrac{dv}{dx}$ by removing the common terms

$\Rightarrow 3x=x\dfrac{dv}{dx}$

$\Rightarrow 3=\dfrac{dv}{dx}$ by removing the common terms

$\Rightarrow 3dx=dv$

$\Rightarrow \int{dv}=\int{3dx}=3\int{dx}$

$\Rightarrow v=3x+c$ where

$c$ is a constant of integration

$\Rightarrow \dfrac{y}{x}=3x+c$ since

$y=vx\Rightarrow v=\dfrac{y}{x}$

$\therefore y=3{x}^{2}+cx$

Solve: $\dfrac{dy}{dx} + 1 = e^{x + y}$ .

Let

$x+y=t$

$\Rightarrow 1+\dfrac {dy} {dx} =\dfrac {dt} {dx}$

Substituting in question, we get ;

$\dfrac {dt} {dx} =e^{t}$

$\Rightarrow e^{-t} dt=dx$

Integrate on both sides

$\Rightarrow \dfrac {e^{-t}} {-1}=x+c$

$\Rightarrow x+e^{-t} +c=0$

$\Rightarrow x+e^{-(x+y)} +c=0$ ('c' is a constant)

Find the particular solution of differential equal $x\left( {{x^2} - 1} \right)\dfrac{{dy}}{{dx}} = 1;y = 0$ and $x = 2.$

$x(x^2-1)dy=dx$

$dy=\dfrac{dx}{x(x^2-1)}$

Integrating on both sides

$\displaystyle\int dy=\displaystyle\int \dfrac{dx}{x(x^2-1)}$

$y=\displaystyle\int \dfrac{dx}{x(x+1)(x-1)}$ …………..

$(1)$

We can write

$\dfrac{1}{x(x+1)(x-1)}=\dfrac{A}{x}+\dfrac{B}{(x+1)}+\dfrac{C}{(x-1)}$

By cancelling the denominators

$1=A(x-1)(x+1)+Bx(x-1)+C(x+1).x$

Putting

$x=0$

$1=A(0-1)(0+1)+B.0(0-1)+C.0(0+1)$

$1=-A$

$A=-1$

Similarly putting

$x=-1$

$1=A(-1-1)(-1+1)+B(-1)(-1-1)+C(-1)(-1+1)$

$1=A(-2)(0)+B(-1)(-2)+C:0$

$1=0+2B+0$

$2B=1$

$B1/2$

Putting

$x=1$

$1=A(1-1)(1+1)+B(1)(1-1)+C.1(1+1)$

$1=20.A+0.B+2C$

$1=2C$

$C=1/2$

Therefore

$\dfrac{1}{x(x+1)(x-1)}=\dfrac{-1}{x}+\dfrac{1/2}{(x+1)}-\dfrac{1/2}{(x-1)}$

Now, from

$(1)$

$y=\displaystyle\int \dfrac{1}{(x(x+1)(x-1)}\cdot dx$

$=\displaystyle\int \dfrac{1}{x}+dx+1/2\displaystyle\int \dfrac{dx}{x+1}+\dfrac{1}{2}\displaystyle\int \dfrac{dx}{x-1}$

$=log |x|+\dfrac{1}{2}.log |x+1|+\dfrac{1}{2}log|x-1|+c$

$=\dfrac{-2}{2}log |x|+\dfrac{1}{2} log |x+1|+\dfrac{1}{2}log|x-1|+c$

$=\dfrac{1}{2}\left[-2log |x|-2+log |(x+1)(x-1)|\right]+c$

$=\dfrac{1}{2}\left[ log x^{-2}log|(x+1)(x-1)|\right]+c$

$=\dfrac{1}{2}\left[log |x^{-2}(x^2-1)|\right]+c$

$[\because log a+log b-log ab]$

$=\dfrac{1}{2}log \left|\dfrac{x^2-1}{x^2}\right|+c$

$\therefore y=\dfrac{1}{2}log\left|\dfrac{x^2-1}{x^2}\right|+c$ …………..

$(1)$

Given that

$x=2; y=0$

Substituting values in

$(1)$ , we get

$0=\dfrac{1}{2}log\left|\dfrac{2^2-1}{2^2}\right|+c$

$0=\dfrac{1}{2}log\dfrac{3}{4}+c$

$c=-\dfrac{1}{2}log \dfrac{3}{4}$

Putting value of c in

$(1)$

$y=\dfrac{1}{2}log \left|\dfrac{x^2-1}{x^2}\right|-\dfrac{1}{2}log 3/4$ .

1225142_1462510_ans_9fda4390bd7143d4ad103c687d767406.jpg

If y = sec $\sqrt { x }$ , Then find $\dfrac { dy }{ dx }$

1324820_1492417_ans_d6d17ae8ac784c948a98e44b22d767a0.JPG

If $y={ x }^{ x }+{ x }^{ 1/x }$ , then find $\dfrac { dy }{ dx }$ .

1320128_1532858_ans_f2212dd3f7e94c86b07e8191b8abf827.jpeg

Find the differential equation representing the family of curves $y=ae^{bx+5}$ where a and b are arbitrary constants.

1487144_1605190_ans_d4d24b69f28e4e25a22da6acb4d7d0af.jpg

Find the differential equation of the family of curves, $x = A \cos nt + B \sin nt$ , where $A$ and $B$ are arbitrary constants.

1545349_1553885_ans_37f5b0f079c746dfa5433f8085720fa2.jpg

Solve the differential equation :
$y \ \log y \ dx-xdy=0$

1318175_1538906_ans_7796dfec10d243e9926bec0fa01fe8c2.jpeg

Solve the differential equation $(x -1 ) \dfrac{dy}{dx} = 2x^3y$

$(x-1) \frac{dy}{dx} = 2x^{3}y.$

$\frac{dy}{y} = \frac{2x^{3} dx}{x-1}$

$\frac{dy}{y} = \frac{2(x^{3} - 1 + 1)}{x-1} dx$

$\frac{dy}{y} = \left \{ 2 \frac{(x-1)(x^{2}+x+1)}{x-1} + \frac{2}{x-1}\right \} dx$

$\int \frac{dy}{y} = \int 2(x^{2} + x + 1) dx + \int \frac{2dx}{(x-1)}$

$\Rightarrow log y = 2 \left \{ \frac{x^{3}}{3} + \frac{x^{2}}{2} + 2 log (x-1) + c \right \}$

$\Rightarrow log y = log(x-1)^{2} + \frac{2x^{3}}{3} + x^{2} 2x+c.$

1219593_1567197_ans_b220e49c448d445b9bc40d5ecfd84446.JPG

Solve the following differential equation.
$(1+x^2)\dfrac{dy}{dx}-x=2\tan^{-1}x$

We have,

$(1+x^2)\dfrac{dy}{dx}-x=2\tan^{-1}x$

$\Rightarrow (1+x)^2\dfrac{dy}{dx}=x+2\tan^{-1}x$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{1+x^2}+2\dfrac{\tan^{-1}}{1+x^2}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\left(\dfrac{2x}{1+x^2}\right)+2\tan^{-1}x\cdot \dfrac{1}{1+x^2}$

Integrating both sides, we get

$y=\dfrac{1}{2} log(1+x^2)+2\displaystyle\int \tan^{-1}xd(\tan^{-1}x)$

or,

$y=\dfrac{1}{2} log(1+x^2)+2\dfrac{\tan^{-1}x)^2}{2}+C$

or,

$y=\dfrac{1}{2}log(1+x^2)+(\tan^{-1}x)^2+C$ .

Show that $y=e^x(A\cos x+B\sin x)$ is the solution of the differential equation $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

We have,

$y=e^x(A\cos x+B\sin x)$ ……(i)

$\Rightarrow \dfrac{dy}{dx}=e^x(A\cos x+B\cos x)+e^x(-A\sin x+B\cos x)$

$\Rightarrow \dfrac{dy}{dx}=y+e^x(-A\sin x+B\cos x)$ …….(ii)

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)+e^x(-A\cos x-B\sin x)$

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)-y$ …….(iii)

Subtracting (ii) from (iii), we get

$\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=\dfrac{dy}{dx}-2y\Rightarrow \dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

This is the given differential equation. So,

$y=e^x(A\cos x+B\sin x)$ satisfies the differential equation. Hence, it is the solution of the given differential equation.

Show that the function $y=A\cos x+B\sin x$ is a solution of the differential equation $\dfrac{d^2y}{dx^2}+y=0$ .

We have,

$y=A\cos x+B\sin x$

$\Rightarrow \dfrac{dy}{dx}=-A\sin x+B\cos x$ and

$\dfrac{d^2y}{dx^2}=-A\cos x-B\sin x$

$\Rightarrow \dfrac{d^2y}{dx^2}=-y$ or,

$\dfrac{d^2y}{dx^2}+y=0$

This is the given differential equation. So,

$y=A\cos x+B\sin x$ satisfies the given differential equation. Hence,

$y=A\cos x+B\sin x$ is the solution of the given differential equation.

Solve the following differential equation.
$\cos x\dfrac{dy}{dx}-\cos 2x=\cos 3x$

$y=\sin 2x-x+2\sin x-log |\sec x+\tan x|+C$ .

We have,

$\dfrac{dy}{dx}=\dfrac{\cos 3x+\cos 2x}{\cos x}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{4\cos^3x-3\cos x+2\cos^2x-1}{\cos x}$

$\Rightarrow \dfrac{dy}{dx}=4\cos^2x-3+2\cos x-\sec x=2+2\cos 2x-3+2\cos x-\sec x$

$\Rightarrow y=\sin 2x-x+2\sin x-log (\sec x+\tan x)+C$ .

Write the order of the differential equation of the family of circles touching $Y$ -axis at the origin.

Let

$a$ be the radius of the circle, touching the

$y-$ axis at the origin.

Then the center of the circle will be

$(a,0)$ .

Then the equation of the circle will be

$x^2+y^2-2ax=0$ .......(1).

Now differentiating both sides of (1) with respect to

$x$ we get,

$2x+2y\dfrac{dy}{dx}=2a$ ......(2).

Using value of

$2a$ from (2) in (1) we get,

$(x^2+y^2)=2x^2+2xy\dfrac{dy}{dx}$

or,

$(y^2-x^2)=2xy\dfrac{dy}{dx}$ .

So the order of the differential equation is

$1$ .

Write the order of the differential equation of all non-horizontal lines in a plane.

Any non-horizontal line in a plane can be written as

$y=mx+c$ .......(1) where

$m$ and

$c$ are constants.

Now differentiating both sides of (1) with respect to

$x$ we get,

$\dfrac{dy}{dx}=m$ .

Now again differentiating both sides with respect to

$x$ we get,

$\dfrac{d^2y}{dx^2}=0$ .

This is the differential equation related to the equation (1) and the order of the differential equation is

$2$ .

Solve the following differential equation.
$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$

$y=\dfrac{1}{2}log|x+1|+\dfrac{3}{4}log(x^2+1)-\dfrac{1}{2}\tan^{-1}x+C$ .

We have,

$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$

$\Rightarrow (x+1)(x^2+1)\dfrac{dy}{dx}=2x^2+x$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x^2+x}{(x+1)(x^2+1)}$

Integrating both sides with respect to x, we get

$y=\displaystyle\int \dfrac{2x^2+x}{(x+1)(x^2+1)}dx$

$\Rightarrow y=\displaystyle\int \dfrac{1}{2(x+1)}+\dfrac{\dfrac{3}{2}x-\dfrac{1}{2}}{x^2+1}dx$ [Using partial fractions]

$\Rightarrow y=\dfrac{1}{2}\displaystyle\int \dfrac{1}{x+1}dx+\dfrac{1}{2}\displaystyle\int\dfrac{3x-1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2}\displaystyle\int \dfrac{1}{x+1}dx+\dfrac{1}{2}\displaystyle\int \dfrac{3x}{x^2+}dx-\dfrac{1}{2}\displaystyle\int \dfrac{1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2} log|x+1|+\dfrac{3}{4}\displaystyle\int \dfrac{2x}{x^2+1}dx-\dfrac{1}{2}\displaystyle\int \dfrac{1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2} log|x+1|+\dfrac{3}{4} log|x^2+1|-\dfrac{1}{2}\tan^{-1}+C$ .

Write the order of the differential equation of the family of circles touching $X$ -axis at the origin.

Let

$a$ be the radius of the circle, touching the

$x-$ axis at the origin.

Then the center of the circle will be

$(0,a)$ .

Then the equation of the circle will be

$x^2+y^2-2ay=0$ .......(1).

Now differentiating both sides of (1) with respect to

$x$ we get,

$2x+2y\dfrac{dy}{dx}-2a\dfrac{dy}{dx}=0$

or,

$2x+2y\dfrac{dy}{dx}=2a\dfrac{dy}{dx}$ ......(2).

Using value of

$2a$ from (2) in (1) we get,

$(x^2+y^2)\dfrac{dy}{dx}=2xy+2y^2\dfrac{dy}{dx}$

or,

$(x^2-y^2)\dfrac{dy}{dx}=2xy$ .

So the order of the differential equation is

$1$ .

Solve the following differential equation.
$\dfrac{dy}{dx}=\sec^2x$ .

$\dfrac {dy}{dx}=\sec ^2 x$

Integrating on both sides

$\displaystyle \int \dfrac {dy}{dx}=\int \sec ^2 xdx$

$y=\tan x$

Solve the following differential equation.

$\dfrac{dy}{dx}=x-1$

$\dfrac{dy}{dx}=x-1$ .

Integrating on both sides

$\Rightarrow$

$\displaystyle \int \dfrac {dy}{dx}=x-1$

$\Rightarrow$

$\int dy =\int x-1\ dx$

$\Rightarrow$

$y=\dfrac{x^2}2-x +c$

Solve the following differential equation.
$\dfrac{dy}{dx}+1=\sin x$

$dy=(sinx-1)dx$

On integrating both sides, we get

$\implies \int dy =\int(sinx-1)dx$

$\implies y=-cosx-x+C$ ans.

Show that $y = e^{x} (A\cos x + B\sin x)$ is the solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} - 2 \dfrac {dy}{dx} + 2y = 0$ .

We have,

$y=e^x(A\cos x+B\sin x)$ ……(i)

differentiating with respect to x

$\Rightarrow \dfrac{dy}{dx}=e^x(A\cos x+B\cos x)+e^x(-A\sin x+B\cos x)$

$\Rightarrow \dfrac{dy}{dx}=y+e^x(-A\sin x+B\cos x)$ …….(ii)

differentiating with respect to x

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)+e^x(-A\cos x-B\sin x)$

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)-y$ …….(iii)

Subtracting (ii) from (iii), we get

$\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=\dfrac{dy}{dx}-2y$

$\Rightarrow \dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

This is the given differential equation. So,

$y=e^x(A\cos x+B\sin x)$ satisfies the differential equation. Hence, it is the solution of the given differential equation.

Write the order of the differential equation whose solution is $y=a\cos x+b\sin x+ce^{-x}$ .

Given the solution of a differential equation is

$y=a\cos x+b\sin x+ce^{-x}$ .

Since the solution contains three arbitrary constants we've the order of the differential equation associated to the solution is

$3$ .

Hence the order of the differential equation is

$3$ .

Differentiate the following function
$(x+2)(x+3)$ .

$f(x) = (x + 2) (x + 3)$

$f(x) = x^2 + 2x + 3x + 6$

$f(x) = x^2 + 5x + 6$

Now Differentiating

$f'(x) = 2x + 5$

Find the solution of the differential equation $\dfrac{dy}{dx}=\log x$ ,

Given the differential equation,

$\dfrac{dy}{dx}=\log x$ .

Now integrating both sides we get,

$y=\displaystyle\int \log x dx+c$ [ Where

$c$ is integrating constant]

$y=\displaystyle \log x\int 1.dx-\int \left(\dfrac{d(\log x)}{dx} \int 1.dx \right)dx$

$y=\displaystyle \log x\cdot x -\int \left(\dfrac{1}{x} \cdot x \right)dx$

$y=x\log x-x+c$ .

This is the required solution.

Find the area of the triangle with vertices at the points :
$(0,0),(6,0) \ and \ (4,3)$

Given Points are

$(0,0),(6,0),(4,3)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

$\therefore$ Area of triangle

$:-$

$=\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|0(0-3)+6(3-0)+4(0-0) |$

$=\dfrac 12|0+18+0|=9$

If $\sin x$ is an integrating factor of the differential equation $\dfrac{dy}{dx}+Py=Q$ , then write the value of $P$ .

Given that

$\sin{x}$ is an integrating factor of the differential equation

$\cfrac{dy}{dx} + Py = Q$

Therefore,

${e}^{\int{P \; dx}} = \sin{x}$

$\Rightarrow \int{P \; dx} = \log{\left| \sin{x} \right|}$

$\Rightarrow \int{P \; dx} = \int{\cot{x} \; dx} \quad \left( \because \int{\cot{x} \; dx} = \log{\left| \sin{x} \right|} + C \right)$

$\Rightarrow P = \cot{x}$

For the following differential equation, find the general solution.
$\dfrac{dy}{dx}+x=1$ .

$\dfrac{dy}{dx}+x=1$ .

$\dfrac {dy}{dx}=1-x$

Integrating both sides

$\displaystyle \int dy=\int (1-x) dx$

Using,

$\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+C$ , we get

$\Rightarrow y=x-\dfrac {x^2}2+c$

Form the differential equation from the following primitive where constant is arbitrary.
$xy=a^2$ .

Given

$xy=a^2$

Differentiating w.r.t. x, we get,

$y+x\dfrac{dy}{dx}=0$ .

The equations of family of primitives is

$xy=a^2$ .

Differentiating with respect to x, we get

$x\dfrac{dy}{dx}+y=0$ , which is the required differential equation.

Represent the following families of curves by forming the corresponding differential equation(a, b being parameters).
$x^2+y^2=a^2$ .

$x^2+y^2=a^2$

Differentiating w.r.t. x, we get,

$\Rightarrow 2x+2y\dfrac {dy}{dx}=0$

$\Rightarrow x+y\dfrac {dy}{dx}=0$

Solve the following differential equation.
$x dy=(-y+2x^4+x^2)dx$ .

Given,

$x dy=(-y+2x^4+x^2)dx$

or,

$(xdy+ydx)=(2x^4+x^2)dx$

or,

$d(xy)=2x^4 dx+x^2 dx$

Now integrating both sides we get,

$xy=\dfrac{2x^5}{5}+\dfrac{x^3}{3}+c$ [ where

$c$ is integrating constant]

Solve the following differential equation.
$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$ .

We have,

$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$

$\Rightarrow (x+1)(x^2+1)\dfrac{dy}{dx}=2x^2+x$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x^2+x}{(x+1)(x^2+1)}$

Integrating both sides with respect to x, we get

$y=\displaystyle\int \dfrac{2x^2+x}{(x+1)(x^2+1)}dx$

$\Rightarrow y=\displaystyle\int \dfrac{1}{2(x+1)}+\dfrac{\dfrac{3}{2}x-\dfrac{1}{2}}{x^2+1}dx$ [Using partial fractions]

$\Rightarrow y=\dfrac{1}{2}\displaystyle\int \dfrac{1}{x+1}dx+\dfrac{1}{2}\displaystyle\int\dfrac{3x-1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2}\displaystyle\int \dfrac{1}{x+1}dx+\dfrac{1}{2}\displaystyle\int \dfrac{3x}{x^2+}dx-\dfrac{1}{2}\displaystyle\int \dfrac{1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2} log|x+1|+\dfrac{3}{4}\displaystyle\int \dfrac{2x}{x^2+1}dx-\dfrac{1}{2}\displaystyle\int \dfrac{1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2} log|x+1|+\dfrac{3}{4} log|x^2+1|-\dfrac{1}{2}\tan^{-1}x+C$ .

Form the differential equation from the following primitives where constant is arbitrary.
$y=ax^2+bx+c$ .

$y=ax^2+bx+c$

Differentiating w.r.t. x, we get,

$\implies \dfrac{dy}{dx}=\dfrac d{dx}(ax^2+bx+c)\\\dfrac {dy}{dx}=2ax+b$

Differentiating w.r.t. x, we get,

$\implies \dfrac {d^2y}{dx^2}=\dfrac d{dx}(2ax+b)\\\dfrac {d^2y}{dx^2}=2a$

$\dfrac {d^3y}{dx^3}=\dfrac d{dx}(2a)$

Differentiating w.r.t. x, we get,

$\dfrac {d^3y}{dx^3}=0$

Show that $y=ax^3+bx^2+c$ is a solution of the differential equation $\dfrac{d^3y}{dx^3}=6a$ .

Consider,

$y=ax^3+bx^2+c$

Differentiating w.r.t. x, we get,

$\Rightarrow \dfrac {dy}{dx}=3ax^2+2bx$

Differentiating w.r.t. x, we get,

$\Rightarrow$

$\dfrac {d^2y}{dx^2}=6ax+2b$

Differentiating w.r.t. x, we get,

$\Rightarrow$

$\dfrac {d^3y}{dx^3}=6a$

Hence proved.

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$ is equal to:

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$

$\sin x=t\implies \cos x dx=dt$

$x\to 0\to \dfrac \pi 2$

$t\to 0\to 1$

$\implies \displaystyle \int _0^{\pi/2} t dt$

$\left.\dfrac {t^2}2\right|^1_0$

$\dfrac 12-0=\dfrac 12$

Represent the following families of curves by forming the corresponding differential equation.(a, b being parameters).
$x^2-y^2=a^2$ .

$x^2-y^2=a^2$

Differentiating w.r.t. x, we get,

$\Rightarrow2x-2y\dfrac {dy}{dx}=0\\$

$\Rightarrow x-y\dfrac{dy}{dx}=0$ .

Show that $y=e^x(A\cos x+B\sin x)$ is the solution of the differential
equation $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

We have,

$y=e^x(A\cos x+B\sin x)$ ……(i)

$\Rightarrow \dfrac{dy}{dx}=e^x(A\cos x+B\cos x)+e^x(-A\sin x+B\cos x)$

$\Rightarrow \dfrac{dy}{dx}=y+e^x(-A\sin x+B\cos x)$ …….(ii)

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)+e^x(-A\cos x-B\sin x)$

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)-y$ …….(iii)

Subtracting (ii) from (iii), we get

$\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=\dfrac{dy}{dx}-2y\Rightarrow \dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

This is the given differential equation. So,

$y=e^x(A\cos x+B\sin x)$ satisfies the differential equation. Hence, it is the solution of the given differential equation.

Solve the following differential equation.

$\dfrac{dy}{dx}+y\tan x=\cos x$ .

Given:

$\dfrac{dy}{dx}+y\,\tan{x}=\cos{x}$

This is a linear differential equation, comparing it with

$\dfrac{dy}{dx}+Py=Q$

where

$P=\tan{x},\,Q=\cos{x}$

I.F

$={e}^{\int{P\,dx}}={e}^{\int{\tan{x}\,dx}}={e}^{\log{\left|\sec{x}\right|}}=\sec{x}$

Solution of the equation is given by

$y\left(I.F\right) =\displaystyle\int{Q.\left(I.F\right)dx} + c$

$\Rightarrow\,y\sec{x}=\displaystyle\int{\cos{x}\sec{x}dx}+c$

$\Rightarrow\,\dfrac{y}{\cos{x}}=x+c$

$\therefore\,y=x\cos{x}+c\cos{x}$ is the solution of the differential equation.

Answer the following question in one word or one sentence or as per exact requirement of the question.
Write the differential equal obtained eliminating the arbitrary constant C in the equation $xy=C^2$ .

$xy=C^2$

Differentiating both sides we get,

$xy'+y=0$ which is required differential equation .

Solve the following differential equation.
$\dfrac{dy}{dx}=x^2$ .

Consider,

$\dfrac {dy}{dx}=x^2$

Integrating on both sides

$\displaystyle \int {dy}=\int x ^2 dx$

$y=\dfrac {x^3}3$

Solve the following differential equation.
$\dfrac{dy}{dx}=\sec^2x$ .

Consider,

$\dfrac {dy}{dx}=\sec ^2 x$

Integrating on both sides

$\Rightarrow$

$\displaystyle \int {dy}=\int \sec ^2 xdx$

$\Rightarrow$

$y=\tan x+c$

Show that $y = e^{x} (A \cos x + B \sin x)$ is the solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} - 2 \dfrac {dy}{dx} + 2y = 0$ .

We have,

$y=e^x(A\cos x+B\sin x)$ ……(i)

$\Rightarrow \dfrac{dy}{dx}=e^x(A\cos x+B\cos x)+e^x(-A\sin x+B\cos x)$

$\Rightarrow \dfrac{dy}{dx}=y+e^x(-A\sin x+B\cos x)$ …….(ii)

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)+e^x(-A\cos x-B\sin x)$

$\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+e^x(-A\sin x+B\cos x)-y$ …….(iii)

Subtracting (ii) from (iii), we get

$\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=\dfrac{dy}{dx}-2y\Rightarrow \dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

This is the given differential equation. So,

$y=e^x(A\cos x+B\sin x)$ satisfies the differential equation. Hence, it is the solution of the given differential equation.

Solve the following differential equation.
$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$ .

We have,

$(x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x$

$\Rightarrow (x+1)(x^2+1)\dfrac{dy}{dx}=2x^2+x$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x^2+x}{(x+1)(x^2+1)}$

Integrating both sides with respect to x, we get

$y=\displaystyle\int \dfrac{2x^2+x}{(x+1)(x^2+1)}dx$

$\Rightarrow y=\displaystyle\int \dfrac{1}{2(x+1)}+\dfrac{\dfrac{3}{2}x-\dfrac{1}{2}}{x^2+1}dx$ [Using partial fractions]

$\Rightarrow y=\dfrac{1}{2}\displaystyle\int \dfrac{1}{x+1}dx+\dfrac{1}{2}\displaystyle\int\dfrac{3x-1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2}\displaystyle\int \dfrac{1}{x+1}dx+\dfrac{1}{2}\displaystyle\int \dfrac{3x}{x^2+}dx-\dfrac{1}{2}\displaystyle\int \dfrac{1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2} log|x+1|+\dfrac{3}{4}\displaystyle\int \dfrac{2x}{x^2+1}dx-\dfrac{1}{2}\displaystyle\int \dfrac{1}{x^2+1}dx$

$\Rightarrow y=\dfrac{1}{2} log|x+1|+\dfrac{3}{4} log|x^2+1|-\dfrac{1}{2}\tan^{-1}+C$ .

Form the differential equation of the family of curves represented by $y^2=(x-c)^3$ .

The equation of the family of curves is

$y^2=(x-c)^3$ .

Differentiating with respect to x, we get

$\Rightarrow$

$2yy_1=3(x-c)^2$

$\left[\because \dfrac{dy}{dx}=y_1\right]$

$\Rightarrow (2yy_1)^3=27(x-c)^6$

$\Rightarrow 8y^3y^3_1=27(y^2)^2$

$[\because (x-c)^3=y^2]$

$\Rightarrow 8y^3_1=27y$ , which is the required differential equation.

Show that $y=\dfrac1x$ is a solution of the differential equation $\dfrac{dy}{dx}=\log x$

Consider,

$y=\dfrac 1x$

Differentiating on both sides

$\dfrac {dy}{dx}=\log x$

Hence proved.

Show that the function $y=A\cos x+B\sin x$ is a solution of the differential equation $\dfrac{d^2y}{dx^2}+y=0$ .

We have,

$y=A\cos x+B\sin x$

Differentiating w.r.t. x, twice we get,

$\Rightarrow \dfrac{dy}{dx}=-A\sin x+B\cos x$ and

$\dfrac{d^2y}{dx^2}=-A\cos x-B\sin x$

$\Rightarrow \dfrac{d^2y}{dx^2}=-y$ or,

$\dfrac{d^2y}{dx^2}+y=0$

This is the given differential equation. So,

$y=A\cos x+B\sin x$ satisfies the given differential equation. Hence,

$y=A\cos x+B\sin x$ is the solution of the given differential equation.

Solve the following differential equation.
$\dfrac{dy}{dx}=x^5+x^2-\dfrac{2}{x}, x\neq 0$ .

Given

$\dfrac{dy}{dx}=x^5+x^2-\dfrac{2}{x}$

Integrating on both sides

$\displaystyle \int {dy}=\int x^5+x^2-\dfrac 2x dx$

$\Rightarrow y=\dfrac {x^6}{6}+\dfrac {x^3}{3}-2\log x + c$ [

$\because\int x^n=\dfrac{x^{n+1}}{n+1}$ ]

Solve the following differential equation.
$\dfrac{dy}{dx}+2x=e^{3x}$ .

Given

$\dfrac{dy}{dx}+2x=e^{3x}$ .

Integrating on both sides

$\displaystyle \int dy +\int 2x dx$ =

$\int e^{3x}dx$

$\Rightarrow y+x^2=\dfrac 13 e^{3x}+c$ [

$\because\int x^n=\dfrac{x^{n+1}}{n+1}$ ]

Show that $y=\dfrac1x$ is a solution of the differential equation $\dfrac{dy}{dx}=\log x$

Given

$y=\dfrac 1x$

Differentiating on both sides w.r.t. x,

$\Rightarrow \dfrac {dy}{dx}=\log x$

$[hence \, proved]$

Solve the following differential equation.
$\dfrac{dy}{dx}={1-\cos x}$

$\Rightarrow\dfrac{dy}{dx}={1-\cos x}$

Integrating on Both sides

$\Rightarrow\displaystyle \int dy=\int (1-\cos x)dx$

$\Rightarrow y=x+\sin x+c$

Show that $y=ax^3+bx^2+c$ is a solution of the differential equation $\dfrac{d^3y}{dx^3}=6a$ .

Given

$y=ax^3+bx^2+c$

Differentiating w.r.t. x, we get,

$\Rightarrow \dfrac {dy}{dx}=3ax^2+2bx$

Differentiating w.r.t. x, we get,

$\Rightarrow \dfrac {d^2y}{dx^2}=6ax+2b$

Differentiating w.r.t. x, we get,

$\Rightarrow \dfrac {d^3y}{dx^3}=6a$ .

$[hence \, proved]$

Solve the following differential equation.
$(x^3-2y^3)dy+3x^2ydx=0$ .

Given the differential equation

$(x^3-2y^3)dy+3x^2ydx=0$ can be written as,

$x^3dy+3x^2y dx-2y^3dy=0$

or,

$d(x^3y)-2y^3 dy=0$

Now integrating both sides we get,

$x^3y-2\dfrac{y^4}{4}=c$ [ Where

$c$ is integrating constant]

or,

$x^3y-\dfrac{y^4}{2}=c$ .

Solve the following differential equation.
$\dfrac{dy}{dx}=x^2+x-\dfrac{1}{x}, x\neq 0$ .

Given

$\dfrac{dy}{dx}=x^2+x-\dfrac{1}{x}$

Integrating on both sides

$\displaystyle \int {dy}=\int x^2+x-\dfrac 1x dx$

$\Rightarrow y=\dfrac {x^3}{3}+\dfrac {x^2}{2}-\log x + c$ [

$\because\int x^n=\dfrac{x^{n+1}}{n+1}$ ]

Solve the following differential equation.
$(x^2+1)\dfrac{dy}{dx}=1$ .

$\Rightarrow(x^2+1)\dfrac{dy}{dx}=1$

$\Rightarrow\dfrac {dy}{dx}=\dfrac 1{1+x^2}$

Taking Integration on both the side

$\Rightarrow\displaystyle \int dy=\int \dfrac {1}{1+x^2} dx$

$\Rightarrow y=\tan ^{-1}x+c$ .

Solve the following differential equation.
$\dfrac{dy}{dx}=x^2+3x+7$ .

$\Rightarrow\dfrac{dy}{dx}=x^2+3x+7$ .

Integrating on Both sides

$\Rightarrow\displaystyle \int dy=\int (x^2+3x+7)dx$

$\Rightarrow y=\dfrac{x^3}3+3\dfrac {x^2}2+7x+c$

For the following differential equation, find the general solution.
$x dy=dx$ .

$\Rightarrow x dy=dx$ .

$\Rightarrow dy =\dfrac 1x dx$

Taking Integration on both the side

$\Rightarrow \displaystyle \int dy=\int \dfrac 1x dx$

$\Rightarrow y=\log x +c$

For the following differential equation, find the general solution.
$\dfrac{dy}{dx}=\sin x$ .

$\Rightarrow\dfrac{dy}{dx}=\sin x\\$ .

Taking Integration on both the side

$\Rightarrow\displaystyle \int dy=\int \sin x dx$

$\Rightarrow y=-\cos x +c$

Form the differential equation corresponding to $y^2-2ay+x^2=a^2$ by eliminating a.

The equation of the one-parameter family of curves is

$y^2-2ay+x^2=a^2$ .

Differentiating with respect to x, we get

$\Rightarrow 2y\dfrac{dy}{dx}-2a\dfrac{dy}{dx}+2x=0\Rightarrow a=\dfrac{x+y\dfrac{dy}{dx}}{\dfrac{dy}{dx}}$

$\Rightarrow y^2-2ay+x^2=a^2$

$\Rightarrow y^2-2\left(\dfrac{x+y\dfrac {dy}{dx}}{\dfrac{dy}{dx}}\right)y+x^2=\left(\dfrac{x+y\dfrac {dy}{dx}}{\dfrac {dy}{dx}}\right)^2$

$\Rightarrow \left(y\dfrac {dy}{dx}\right)^2-2xy\dfrac {dy}{dx}-2\left(y\dfrac {dy}{dx}\right)^2+\left(x\dfrac {dy}{dx}\right)^2=x^2+\left(y\dfrac {dy}{dx}\right)^2+2xy\dfrac {dy}{dx}$

$\Rightarrow x^2\left(1-\dfrac {dy}{dx}\right)+4xy\dfrac {dy}{dx}+2\left(y\dfrac {dy}{dx}\right)^2$

Differentiate the following function with respect to x.
$\dfrac{x^2\cos \dfrac{\pi}{4}}{\sin x}$ .

We have,

$f(x)=\dfrac{x^2\cos \dfrac{\pi}{4}}{\sin x}$

$=\dfrac{d}{dx}\left(\dfrac{x^2\cos \dfrac{\pi}{4}}{\sin x}\right)$

$=\dfrac{1}{\sqrt{2}}\dfrac{d}{dx}(x^2 cosec x)$

$=\dfrac{1}{\sqrt{2}}\left\{cosec x\dfrac{d}{dx}(x^2)+x^2\dfrac{d}{dx}(cosec x)\right\}$

$=\dfrac{1}{\sqrt{2}}(2 x cosec x-x^2 cosec x\cot x)$ .

Show that $y=ax^3+bx^2+c$ is a solution of the differential equation $\dfrac{d^3y}{dx^3}=6a$ .

Here,

$y=ax^3+bx^2+c$

$\implies \dfrac {dy}{dx}=3ax^2+2bx$

$\dfrac {d^2y}{dx^2}=6ax+2b$

$\dfrac {d^3y}{dx^3}=6a$

Hence proved.

Solve the following differential equation.
$\dfrac{dy}{dx}={1-\cos x}$

$\dfrac{dy}{dx}={1-\cos x}$

Integrating on Both sides

$\displaystyle \int dy=\int (1-\cos x)dx$

$y=x+\sin x+c$

For the following differential equation, find the general solution.
$\dfrac{dy}{dx}=\sin x$ .

$\dfrac{dy}{dx}=\sin x\\$ .

$\Rightarrow \displaystyle \int dy=\int \sin x dx$

$\therefore y=-\cos x +c$

For the following differential equation, find the general solution.
$x dy=dx$ .

$x dy=dx$ .

$\Rightarrow dy =\dfrac 1x dx$

$\Rightarrow\displaystyle \int dy=\int \dfrac 1x dx$

$\therefore y=\log x +c$

Verify that $y = e^{m \cos^{-1} x}$ is a solution of the differential equation $(1 - x^{2}) \dfrac {d^{2}y}{dx^{2}} - x\dfrac {dy}{dx} - m^{2}y = 0$ .

1674549_1706801_ans_cbb1318ddc3c43ef8318dc21a2a4361d.jpeg

Verify that $Ax^{2} + By^{2} = 1$ is a solution of the differential equation $x\left \{y \dfrac {d^{2} y}{dx^{2}} + \left (\dfrac {dy}{dx}\right )^{2} \right \} = y \dfrac {dy}{dx}$ .

Let

$Ax^{2} + By^{2} = 1$ be given. Then,

$2Ax + 2By \dfrac {dy}{dx} = 0\\$

$\Rightarrow Ax + By \dfrac {dy}{dx} = 0 ....(i)\\$

$\Rightarrow Ax = -By\dfrac {dy}{dx}\\$

$\Rightarrow \dfrac {A}{B} = \dfrac {-y}{x} \cdot \dfrac {dy}{dx}.... (ii)$

On differentiating (i) w.r.t.

$x$ , we get

$A + B \left [y \dfrac {d^{2}y}{dx^{2}} + \left (\dfrac {dy}{dx}\right )^{2}\right ] = 0$

$\Rightarrow y \dfrac {d^{2}y}{dx^{2}} + \left (\dfrac {dy}{dx}\right )^{2} = \dfrac {-A}{B}$

$\Rightarrow \left \{y \dfrac {d^{2}y}{dx^{2}} + \left (\dfrac {dy}{dx}\right )^{2}\right \} = \dfrac {y}{x}\cdot \dfrac {dy}{dx}$ . [from (ii)].

So,

$Ax^{2} + By^{2} = 1$ is a solution of the differential equation

$x\left \{y \dfrac {d^{2} y}{dx^{2}} + \left (\dfrac {dy}{dx}\right )^{2} \right \} = y \dfrac {dy}{dx}$ .

Find the general solution of each of the following differential equations:
$(x - 1) \dfrac {dy}{dx} = 2x^{3}y$ .

1549822_1707012_ans_e511a626dec8412dab97c6e805113dfd.jpg

Verify that $y = ae^{bx}$ is a solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} = \dfrac {1}{y} \left (\dfrac {dy}{dx}\right )^{2}$ .

1549945_1706827_ans_d2a3df7a297049b3a0877d4d39f1d59a.jpg

Find the general solution of each of the following differential equations:
$x^{4} \dfrac {dy}{dx} = -y^{4}$ .

$x^{4} \dfrac {dy}{dx} = -y^{4}$

$\Rightarrow \dfrac {-1}{y^{4}} dy = \dfrac {1}{x^{4}} dx$

$\Rightarrow \int x^{-4} dx + \int y^{-4} dy = C_{1}$ .......

$\because \displaystyle \int x^n \ dx=\dfrac {x^{n+1}}{n+1}+C$

$\Rightarrow \left [ \dfrac {x^{-4+1}}{-4+1}\right]+\left [ \dfrac {y^{-4+1}}{-4+1}\right]=C_1$

$\Rightarrow \dfrac {1}{x^{3}} + \dfrac {1}{y^{3}} = -3C_{1}$

$\Rightarrow \dfrac {1}{x^{3}} + \dfrac {1}{y^{3}} = C$

Verify that $y = \dfrac {c - x}{1 + cx}$ is a solution of the differential equation $(1 + x^{2}) \dfrac {dy}{dx} + (1 + y^{2}) = 0$ .

Given,

$\implies y = \dfrac {c - x}{1 + cx}$
Let

$c = \tan A$ and

$x = \tan B$ .

Then,

$y = \dfrac {\tan A - \tan B}{1 + \tan A \tan B} = \tan (A - B)$

$\Rightarrow \tan^{-1} y = A - B$

$\Rightarrow \tan^{-1} y = \tan^{-1} (c) - \tan^{-1} x$

Differentiate with respect to

$x$

$\Rightarrow \dfrac {1}{(1 + y^{2})} \dfrac {dy}{dx} = 0 - \dfrac {1}{(1 + x^{2})}$

$\Rightarrow (1 + x^{2}) \dfrac {dy}{dx} + (1 + y^{2}) = 0$ .

$\implies y = \dfrac {c - x}{1 + cx}$ is a solution of the differential equation

$(1 + x^{2}) \dfrac {dy}{dx} + (1 + y^{2}) = 0$ .

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = (1 + x^{2})(1 + y^{2})$ .

1549793_1707007_ans_fd3408f71ad64f63b9ee612bebdc335c.jpg

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = 1 - x + y - xy$ .

1549823_1707011_ans_d2f9e559548140a19f6ab9c65c32fc7b.jpg

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = e^{x + y}$ .

$\dfrac {dy}{dx} = e^{x} \cdot e^{y}$

$\Rightarrow \dfrac {1}{e^{y}} dy = e^{x} dx$

$\Rightarrow \int e^{-y} dy = \int e^{x} dx$

$\Rightarrow -e^{-y}=e^x-C$

$\Rightarrow e^x+e^{-y}=C$

Find the general solution of the following differential equations:
$\dfrac {dy}{dx} = \dfrac {1 + x}{1 + y^{2}}$ .

1551876_1707353_ans_41963982064a4b93b3d19b2dc3919644.jpg

Find the general solution of each of the following differential equations:
$(e^{x} + e^{-x}) dy - (e^{x} - e^{-x}) dx = 0$ .

$(e^{x} + e^{-x}) dy - (e^{x} - e^{-x}) dx = 0$

dividing both sides by

$(e^x+e^{-x})$ , we get

$\Rightarrow dy - \dfrac {(e^{x} - e^{-x})}{(e^{x} + e^{-x})} dx = 0$

$\Rightarrow \int dy - \displaystyle \int \dfrac {e^{x} - e^{-x}}{e^{x} + e^{-x}} dx = C$

$\Rightarrow y - \log (e^{x} + e^{-x}) = C$ .

Find the general solution of each of the following differential equations:
$y\log y\ dx - x\ dy = 0$ .

$y\log y\ dx - x\ dy = 0$

Applying integration on both sides,

$\int \dfrac {1}{x} dx = \int \dfrac {1}{y\log y} dy \Rightarrow \int \dfrac {1}{x} dx = \int \dfrac {dt}{t}$ , where

$\log y = t$ .

$\therefore \log |x| = \log |t| + \log |C_{1}|\Rightarrow \log |x| - \log |\log y| = \log |C_{1}|$

$\Rightarrow \log \left |\dfrac {x}{\log y}\right | = \log C_{1} \Rightarrow \dfrac {x}{\log y} = \pm C_{1} = C$ .

Find the general solution of each of the following differential equations:
$(1 + x^{2}) \dfrac {dy}{dx} = xy$ .

$(1 + x^{2}) \dfrac {dy}{dx} = xy$

$\displaystyle \int \dfrac {1}{y} dy = \dfrac {1}{2} \int \dfrac {2x}{(1 + x^{2})} dx$

$\Rightarrow \log |y| -\ \dfrac {1}{2} \log |1 + x^{2}| = \log |C|$

$\Rightarrow \log \left |\dfrac {y}{\sqrt {1 + x^{2}}}\right | = \log |C| \Rightarrow \dfrac {y}{\sqrt {1 + x^{2}}} = \pm C = C_{1}$ .

Find the general solution of each of the following differential equations:
$\cos x (1 + \cos y)dx - \sin y(1 + \sin x)dy = 0$ .

$\cos x (1 + \cos y)dx - \sin y(1 + \sin x)dy = 0$

$\cos x (1 + \cos y)dx = \sin y(1 + \sin x)dy$

Integrating both sides,

$\displaystyle \int \dfrac {\cos x}{(1 + \sin x)} dx + \int \dfrac {-\sin y}{(1 + \cos y)} dy = \log |C_{1}|$

$\Rightarrow \log |1 + \sin x| + \log |1 + \cos y| = \log |C_{1}|$

$\Rightarrow \log |(1 + \sin x)(1 + \cos y)| = \log |C_{1}|$

$\Rightarrow |(1 + \sin x)(1 + \cos y)| = C_{1} \Rightarrow (1 + \sin x)(1 + \cos y) = \pm C_{1} = C$ .

Find the general solution of the following differential equations:
$\dfrac {dy}{dx} = \dfrac {x}{x^{2} + 1}$

1551779_1707352_ans_b7ee7c73892a46789596dcd22d0dd776.jpg

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = e^{x - y} + x^{2} e^{-y}$ .

$\dfrac {dy}{dx} = e^{x} \cdot e^{-y} + x^{2} e^{-y} = (e^{x} + x^{2}) e^{-y}$

$\Rightarrow \int e^{y} dy = \int (e^{x} + x^{2}) dx$

$\Rightarrow e^y=e^x+ \left [\dfrac{x^3}{3}\right ]+C$

$\Rightarrow e^y-e^x- \left (\dfrac{x^3}{3}\right )=C$

Find the general solution of each of the following differential equations:
$\sec^{2} x \tan y \ dx + \sec^{2} y\tan x\ dy = 0$ .

$\displaystyle \int \dfrac {\sec^{2}x}{\tan x}dx + \int \dfrac {\sec^{2}y}{\tan y}dy = \log |C_{1}|$

$\Rightarrow \log |\tan x| + \log |\tan y| = \log |C_{1}|$

$\therefore \log |\tan x \tan y| = \log |C_{1}|$

$\Rightarrow |\tan x \tan y| = |C_{1}|$ .

$\therefore \tan x \tan y = \pm C_{1} = C$ .

Find the general solution of each of the following differential equations:
$x\dfrac {dy}{dx} + y = y^{2}$ .

1551875_1707368_ans_0ee3fc2c11574bf4bae95b495881c55d.jpg

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = e^{x + y} + e^{x - y}$ .

1567128_1707419_ans_0c4e9a27645546058c0de883c2f8df01.jpg

Find the general solution of each of the following differential equations:
$(1 - x^{2}) dy + xy (1 - y)dx = 0$ .

Given

$(1 - x^{2}) dy + xy (1 - y)dx = 0$

$\displaystyle \int \dfrac {1}{y(1 - y)}dy + \int \dfrac {x}{(1 - x^{2})} dx = \log |C_{1}|$

$\displaystyle \Rightarrow \left \{\dfrac {1}{y} + \dfrac {1}{(1 - y)}\right \} dy - \dfrac {1}{2} \int \dfrac {-2x}{(1 - x^{2})} dx = \log |C_{1}|$

$\Rightarrow \log |y| - \log |1 - y| - \dfrac {1}{2} \log |1 - x^{2}| = \log |C_{1}|$

$\Rightarrow \log \left |\dfrac {y}{(1 - y)\sqrt {1 - x^{2}}}\right | = \log |C_{1}|$

$\Rightarrow \dfrac {y}{(1 - y)\sqrt {1 - x^{2}}} = \pm C_{1} = C$ is the required general solution.

Find the general solution of each of the following differential equations:
$(e^{y} + 1)\cos x\ dx + e^{y}\sin x\ dy = 0$ .

1567129_1707421_ans_8a2428248ada4d3185ae0b9096d39c8d.jpg

Find the general solution of each of the following differential equations:
$x(x^{2} - x^{2}y^{2}) dy + y(y^{2} + x^{2}y^{2}) dx = 0$ .

1551993_1707374_ans_d5ae0e3fbc2945baba2747a7a3dd709b.jpg

Find the general solution of the following differential equations:
$(x^{2} - xy^{2}) dy + (y^{2} + xy^{2}) dx = 0$ .

1551986_1707407_ans_4a4f604eff6b4410b28769ca70f63cf3.jpg

Find the general solution of the following differential equations:
$3e^{x} \tan y\ dx + (1 - e^{x})\sec^{2} y\ dy = 0$ .

Given,

$3e^{x} \tan y\ dx + (1 - e^{x})\sec^{2} y\ dy = 0$ .

$\displaystyle \int \dfrac {3e^{x}}{(1 - e^{x})} dx + \int \dfrac {\sec^{2}y}{\tan y} dy = \log |C_{1}|$

$\displaystyle \Rightarrow -3\int \dfrac {-e^{x}}{(1 - e^{x})} dx + \int \dfrac {\sec^{2}y}{\tan y}dy = \log |C_{1}|$

$\Rightarrow -3\log |1 - e^{-x}| + \log |\tan y| = \log |C_{1}|$

$\displaystyle \Rightarrow \log |(1 - e^{-x})^{-3} \tan y| = \log |C_{1}|$

$\Rightarrow \dfrac {\tan y}{(1 - e^{-x})^{3}} = \pm C_{1} = C$ .

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = \dfrac {3e^{2x} + 3e^{4x}}{e^{x} + e^{-x}}$ .

1567126_1707416_ans_497b3e499d4a4f418a6fd34c394f14d2.jpg

Find the general solution of each of the following differential equations:
$(x^{2}y - x^{2}) dx + (xy^{2} - y^{2}) dy = 0$ .

1552129_1707408_ans_de4f664cfd4441f1b64ed24be6311dab.jpg

Find the general solution of each of the following differential equations:
$e^{y}(1 + x^{2}) dy - \dfrac {x}{y}dx = 0$ .

1567127_1707418_ans_fd23ffb9ea9a4c7ba102512cf17eb940.jpg

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = e^{x + y} + x^{2}e^{y}$ .

1567125_1707412_ans_a6ef2beb508a450ca2925a72f9dc41fc.jpg

Find the general solution of each of the following differential equations:
$cosec x\log y \dfrac {dy}{dx} + x^{2} y = 0$ .

1567131_1707438_ans_17188de213bd4d9d8b74b738717ca66f.jpg

Find the general solution of each of the following differential equations:
$\dfrac {1}{x} \cdot \dfrac {dy}{dx} = \tan^{-1} x$ .

1567134_1707441_ans_761496e03d4444f2ba4935b3f598bb9a.jpg

Find the general solution of the following differential equations:
$e^{x} \sqrt {1 - y^{2}} dx + \dfrac {y}{x} dy = 0$ .

1567135_1707443_ans_0f1bdc60875d40dfbd63e60762820e01.jpg

Find the general solution of each of the following differential equations:
$\sqrt {1 - x^{4}} dy = xdx$ .

1557369_1707436_ans_ffe40587733f4353a71996da020a1763.jpg

Find the general solution of each of the following differential equations:
$(\cos x) \dfrac {dy}{dx} + \cos 2x = \cos 3x$ .

Given,

$(\cos x) \dfrac {dy}{dx} + \cos 2x = \cos 3x$ .

$\dfrac {dy}{dx} = \dfrac {\cos 3x - \cos 2x}{\cos x} = \dfrac {(4\cos^{3}x - 3\cos x) -(2\cos^{2}x - 1)}{\cos x}$

$\Rightarrow \dfrac {dy}{dx} = 4\cos^{2}x - 2\cos x - 3 + \sec x$

$\Rightarrow \dfrac {dy}{dx} = \dfrac {4(1 + \cos 2x)}{2} - 2\cos x - 3 + \sec x$

$\Rightarrow \dfrac {dy}{dx} = 2\cos 2x - 2\cos x - 1 + \sec x$ .

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} + \dfrac {\cos x \sin y}{\cos y} = 0$ .

1558162_1707454_ans_975504764d8949889773bc605404ed67.jpg

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = \dfrac {1 - \cos x}{1 + \cos x}$ .

1558156_1707446_ans_d5833074828a44d5af4a7278c8822c02.jpg

Find the general solution of each of the following differential equations:
$ydx + (1 + x^{2}) \tan^{-1} x\ dy = 0$ .

1567133_1707439_ans_8b16cc24df89479eb26302eaf9046439.jpg

Find the general solution of the following differential equations:
$\cos x (1 + \cos y) dx - \sin y(1 + \sin x)dy = 0$ .

Given,

$\cos x (1 + \cos y) dx - \sin y(1 + \sin x)dy = 0$ .

$\displaystyle \int \dfrac {\cos x}{(1 + \sin x)}dx - \int \dfrac {\sin y}{(1 + \cos y)}dy = \log |C_{1}|\\$

$\Rightarrow \log |1 + \sin x| + \log |1 + \cos y| = \log |C_{1}|\\$

$\Rightarrow \log |(1 + \sin x)(1 + \cos y)| = \log |C_{1}| \\$

$\Rightarrow (1 + \sin x)(1 + \cos y) = \pm C_{1} = C\\$ .

$\Rightarrow (1 + \sin x)(1 + \cos y) = C$ is the required general solution.

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} + \dfrac {(1 + \cos 2y)}{(1 - \cos 2x)} = 0$ .

1563532_1707452_ans_652334bd412a4565bb2e72ca991a5dc1.jpg

Find the general solution of each of the following differential equations:
$\dfrac {1}{x}\cos^{2}y dy + \dfrac {1}{y}\cos^{2} x dx = 0$ .

1567158_1707465_ans_1bbf0827e19048fc96f1a03e1d114ffc.jpg

Find the general solution of each of the following differential equations:
$\sin^{3} x dx - \sin y\ dy = 0$ .

1567137_1707462_ans_d6ab622e23f24bfbb92640a9d05bd0d1.jpg

Solve $\dfrac {dy}{dx} = y^{2}\tan 2x$ , given that $y = 2$ when $x = 0$ .

1558458_1707533_ans_a52a4b33ee944b6db0008b49af799983.jpg

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} + \sin (x + y) = \sin (x - y)$ .

Given,

$\dfrac{dy}{dx}+\sin \left( x+y \right)=\sin \left( x-y \right)$

$\Rightarrow \dfrac{dy}{dx}=\sin \left( x-y \right)-\sin \left( x+y \right)$

$\Rightarrow \dfrac{dy}{dx}=2\cos \left( \frac{x-y+x+y}{2} \right)\sin \left( \frac{x-y-x-y}{2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \sin C-\sin D=2\cos \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right) \right]$

$\Rightarrow \dfrac{dy}{dx}=2\cos \left( \frac{x+x}{2} \right)\sin \left( \frac{-y-y}{2} \right)$

$\Rightarrow \dfrac{dy}{dx}=2\cos \left( x \right)\sin \left( -y \right)$

$\Rightarrow \dfrac{dy}{dx}=-2\cos \left( x \right)\sin \left( y \right)$

$\Rightarrow \dfrac{dy}{\sin x}=-2\cos xdx$

$\operatorname{int}egrating\,both\,sides$

$\int{\dfrac{dy}{\sin x}}=-2\int{\cos x}dx$

$\Rightarrow \ln \sin \left( \frac{x}{2} \right)=-2\sin x+c$

Find the general solution of each of the following differential equations:
$\dfrac {dy}{dx} = \sin^{3}x\cos^{2} x + xe^{x}$ .

1567142_1707466_ans_a09b239091004b29bab029dafbf478c8.jpg

Solve the differential equation $\dfrac {dy}{dx} = y\sin 2x$ , given that $y(0) = 1$ .

1558452_1707507_ans_12eb4d41af3e4e79bc3a5c4508801808.jpg

Solve the differential equation $(x + 1)\dfrac {dy}{dx} = 2xy$ , given that $y(2) = 3$ .

1567154_1707509_ans_cfa7c649ba914a2182c8dc5d99fa68a5.jpg

Solve $\dfrac {dy}{dx} = y\tan x$ , given that $y = 1$ when $x = 0$ .

1567144_1707531_ans_4fc0cdb7b7b44d409d056d258ffc1f75.jpg

Solve $\dfrac {dy}{dx} = y\cot 2x$ , given that $y = 2$ when $x = \dfrac {\pi}{4}$ .

$\dfrac {dy}{dx} = y\cot 2x$

$\int \dfrac {dy}{y} = \int \dfrac {\cos 2x}{\sin 2x} dx$

$\Rightarrow \log |y| = \dfrac {1}{2} \log |\sin 2x| + \log |C_{1}|$

$\therefore \left |\dfrac {y}{\sqrt {\sin 2x}}\right | = C_{1}$

$\Rightarrow \dfrac {y}{\sqrt {\sin 2x}} = \pm C_{1} = C (say) .....(i)$

Putting

$x = \dfrac {\pi}{4}$ and

$y = 2$ in (i), we get

$C = 2$ .

Hence,

$y = 2\sqrt {\sin 2x}$ .

If $y=\left(1 / 2^{n-1}\right) \cos \left(n \cos ^{-1} x\right)$ , then prove that $y$ satisfies the differential equation $\left(1-x^{2}\right) \dfrac{d^{2} y}{d x^{2}}-x \dfrac{d y}{d x}+n^{2} y=0$

$y=\left(1 / 2^{n-1}\right) \cos \left(n \cos ^{-1} x\right)$

$\therefore \dfrac{d y}{d x}=-\dfrac{1}{2^{n-1}} \sin \left(n \cos ^{-1} x\right)\left[\dfrac{-n}{\sqrt{\left(1-x^{2}\right)}}\right]$

$\Rightarrow\left(1-x^{2}\right)\left(\dfrac{d y}{d x}\right)^{2} =\dfrac{n^{2}}{2^{2 n-2}} \sin ^{2}\left(n \cos ^{-1} x\right)$

$=\dfrac{n^{2}}{2^{2 n-2}}\left[1-\cos ^{2}\left(n \cos ^{-1} x\right)\right]$

$=n^{2}\left[\dfrac{1}{2^{2 n-2}}-y^{2}\right]$

Differentiating both sides w.r.t.x

$\left(1-x^{2}\right) 2 \dfrac{d y}{d x} \dfrac{d^{2} y}{d x^{2}}-2 x\left(\dfrac{d y}{d x}\right)^{2}=-2 n^{2} y \dfrac{d y}{d x}$

$\Rightarrow\left(1-x^{2}\right) \dfrac{d^{2} y}{d x^{2}}-x \dfrac{d y}{d x}+n^{2} y=0$

The integrating factor of $\dfrac {dy}{dx}+y=\dfrac {1+y}{x}$ is _______ .

Given differential equation is

$\dfrac {dy}{dx}+y=\dfrac {1+y}{x}$

$\dfrac {dy}{dx}+y=\dfrac {1}{x}+\dfrac {y}{x}$

$\Rightarrow \ \dfrac {dy}{dx}+y\left(1-\dfrac {1}{x}\right)=\dfrac {1}{x}$

$\therefore IF=e^{\displaystyle \int \left(1-\dfrac {1}{x}\right)dx}$

$=e^{(x-\log x)}$

$=e^x e^{-\log x}=\dfrac {e^x}{x}$

If the dependent variable y is changed to $'z'$ by the substitution $y = \tan z$ and the differential equation $\dfrac{d^2 y}{dx^2} = 1 + \dfrac{2 (1 + y)}{1 + y^2} \left(\dfrac{dy}{dx}\right)^2$ is changed to $\dfrac{d^2z}{dx^2} = \cos^2 z + k \left(\dfrac{dz}{dx}\right)^2$ , then the value of k equals.

Given

$y = \tan z$

$\dfrac{dy}{dx} = \sec^2 z \dfrac{dz}{dx}$ ___(1)
Now

$\dfrac{d^2 y}{dx^2} = \sec^2 z. \dfrac{d^2 z}{dx^2} + \dfrac{dz}{dx} . \dfrac{d}{dx} (\sec^2 z)$ [using product rule]

$= \sec^2 z. \dfrac{d^2 z}{dx^2} + \dfrac{dz}{dx} . \dfrac{d}{dz} (\sec^2 z) \dfrac{dz}{dx}$

$\dfrac{d^2 y}{dx^2} = \sec^2 z. \dfrac{d^2z}{dx^2} + \left(\dfrac{dz}{dx} \right)^2 . 2 \sec^2 z. \tan z$ ___(2)
Now

$1 + \dfrac{2 (1 + y)}{1 + y^2} \left(\dfrac{dy}{dx}\right)^2$

$= 1 + \dfrac{2 (1 + \tan z)}{\sec^2 z} . \sec^4 z . \left(\dfrac{dz}{dx} \right)^2$

$= 1 + 2 (1 + \tan z) . \sec^2 z. \left(\dfrac{dz}{dx} \right)^2$

$= 1 + 2 \sec^2 z \left(\dfrac{dz}{dx} \right)^2 + 2 \tan z. \sec^2 z \left(\dfrac{dz}{dx} \right)^2$ ____(2)
from (2) and (3), we have RHS of (2) = RHS of (3)

$\sec^2 z. \dfrac{d^2z}{dx^2} = 1 + 2 \sec^2 z \left(\dfrac{dz}{dx} \right)^2$

$\Rightarrow \dfrac{d^2 z}{dx^2} = \cos^2 z + 2 \left(\dfrac{dz}{dx} \right)^2$

$\Rightarrow k = 2$

General solution of the differential equation of the type
$\dfrac {dy}{dx}+P_1x=Q_1$ is given by _______ .

Given differential equation is

$\dfrac {dy}{dx}+P_1x =Q_1\ \dfrac {dy}{dx}+P_1x=Q_1$
The general solution

$xIF=\displaystyle \int Q(IF)dy +C$ i.e.,

$xe\displaystyle \int^{Pdy}=\displaystyle \int Q\left\{e^{\displaystyle \int pdy}\right\} dy +C$

$x.e^{\displaystyle \int P_1 dy}=\displaystyle \int Q_1\left\{e^{\displaystyle \int P_1 dy}\right\}dy+C$

If the independent variable x is changed to y, then the differential equation $x \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^3 - \dfrac{dy}{dx} = 0$ is changed to $x \dfrac{d^2 x}{dy^2} + \left(\dfrac{dx}{dy}\right)^2 = k$ where k equals

$\dfrac{dy}{dx} = \dfrac{1}{dx/dy} ; \dfrac{d^2y }{dx^2} = \dfrac{d}{dy} \left(\dfrac{1}{dx/dy} \right) . \dfrac{dy}{dx} = - \dfrac{1}{(dx/dy)^3} \dfrac{d^2 x}{dy^2}$
hence

$x \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx} \right)^3 - \dfrac{dy}{dx} = 0$
becomes

$-x. \dfrac{1}{(dx/dy)^3} \dfrac{d^2 x}{dy^2} + \dfrac{1}{(dx/dy)^3} - \dfrac{1}{(dx/dy)} = 0$
or

$x \dfrac{d^2 x}{dy^2} - 1 + \left(\dfrac{dx}{dy} \right)^2 = 0 \Rightarrow x \dfrac{d^2x }{dy^2} + \left(\dfrac{dx}{dy}\right)^2 = 1$

$\therefore k = 1$

Integrating factor of the differential equation $x\dfrac {dy}{dx}-y=\sin x$ is _________.

$\dfrac 1x$ ; given differential equation can be written as

$\dfrac{dy}{dx}-\dfrac {y}{x}=\dfrac {\sin x}{x}$ and therefore

$I.F.= e^{\displaystyle\int{-\dfrac 1x dx}}=e^{-\log x}=\dfrac 1x$ .

The number pf arbitrary constants in a particular solution of the differential equation $\tan xdx +\tan y dy=0$ is __________.

Zero; any particular solution of a differential equation has no arbitrary constant.

$y = \cos x + C : y’+ \sin x = 0$

Given

$y = \cos x + C$

$\Rightarrow(y^{\prime}=\frac{d}{d x})(\cos x + C) = - \sin x$

Given,

$y’ + sinx = 0$

$-sinx + sinx = 0$

LHS = RHS

Hence

$y = cos x + C$ , is solution of given differential equation

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$y=e^x(a\, \cos\,x+b\, sin\, x)$ : $\cfrac{d^2y}{dx^2}-2\cfrac{dy}{dx}+2y=0$

1910867_1859622_ans_f3c215eddc5340b588f416d24061f3f4.jpg

$y= x^2 + 2x + C : y' - 2x - 2 = 0$

Given

$y= x^2 + 2x + C$

$\Rightarrow y^{\prime}=\frac{d}{d x}(x^2 + 2x + C)= 2x + 2$

$y’- 2x -2 = 0$

Hence,

$y = x^2 + 2x + C$ , is solution of given differential equation.

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$(\begin{aligned}y=\& \sqrt{a^{2}-x^{2}}, &\,x \in(-a, a): x+y \frac{d y}{d x}=0(y \neq 0)\end{aligned})$

1910807_1858798_ans_c28d393fca194a9684793e2b38744e7e.jpg

$y = e^x + 1 : y’’- y’ = 0$

Given

$y = e^x + 1\Rightarrow y'=\dfrac{dy}{dx}=e^x$

$y=y''=\dfrac{d}{dx}(e^x)=e^x$

Given,

$y’’- y’= e^x – e^x =0$

∴ Hence given function is a solution of the corresponding differential equation.

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y = Ax : xy = y (x \neq 0)$

Given,

$y = Ax$ (1)

Differentiate w.r.t.

$x$ , we get,

${ y }^{ \prime }=\frac { d }{ dx } \left( Ax \right)$

$\therefore { y }^{ \prime }=A$ (2)

Given,

$x y’ = y$

$\therefore { y }^{ \prime }=\frac { y }{ x }$

From equation (1), we can write,

${ y }^{ \prime }=A$ (3)

From equation (2) and (3), we can say that

$y = Ax$ is solution of given differential equation.

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$y = x \,\sin 3x$ : $\cfrac{d^2y}{dx^2}+9y-6\,\cos\,3x=0$

$y=x\sin 3x ; \cfrac{d^2y}{dx^2}+9y-6\,\cos\,3x=0$

$\dfrac{dy}{dx} = \sin 3x+3x\cos 3x$

$\dfrac{d^2y}{dx^2}=3\cos 3x+3\cos 3x-9x\sin3x$

or

$\dfrac{d^2y}{dx} = 6 \cos 3x-9x\sin 3x$

where

$x\sin 3x=y$

$\dfrac{d^2y}{dx^2}=9y-6\cos 3x=0$

This is the proof that

$y=x\sin 3x$ is the solution of the given differential equation.

Match the statements/expressions in List 1 with the open intervals in List 2.

$({ x }-3)^{ 2 }{ y }'+{ y }=0\Rightarrow ({ x }-3)^{ 2 }\cfrac { dy }{ dx } +{ y }=0\Rightarrow \int \cfrac { dx }{ { \left( x-3 \right) }^{ 2 } } =-\int \cfrac { dy }{ y } \Rightarrow \cfrac { 1 }{ x-3 } =\ln { | } { y }|+{ C }$
Hence domain is

${ R }-\{ 3\}$

B)

$\int _{ 1 }^{ 5 } ({ x }-1)({ x }-2)({ x }-3)({ x }-4)({ x }-5){ d }{ x }$
Substituting

${ x }={ t }+3$ , we get

$\int _{ -2 }^{ 2 } ({ t }+2)({ t }+1){ t }({ t }-1)({ t }-2)dt=\int _{ -2 }^{ 2 }{ t } ({ t }^{ 2 }-1)({ t }^{ 2 }-4)dt=0$

C)

$f\left( x \right) =\cos ^{ 2 }{ x } +sinx=\cfrac { 5 }{ 4 } -\left( \sin { x } -\cfrac { 1 }{ 2 } \right) ^{ 2 }$
So

$f\left( x \right)$ is maximum for

$sinx=\cfrac { 1 }{ 2 } \Rightarrow x=\cfrac { \pi }{ 6 }$

D)

$f\left( x \right) =\tan ^{ -1 }{ \left( \sin { x } +\cos { x } \right) }$ is increasing when

$\Rightarrow { f' }({ x })>0\Rightarrow cosx>sinx$

List 1 contains different functions and List 2 contains differential equations. Match options in List 1 with corresponding differential equation from List 2

(A)

$e^{\sqrt{x}}+e^{-\sqrt{x}}=y$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}e^{\sqrt{x}}-\dfrac{1}{2\sqrt{x}}e^{-\sqrt{x}}$

$\Rightarrow 2\sqrt{x}\dfrac{dy}{dx}=e^{\sqrt{x}}-e^{-\sqrt{x}}$

$\displaystyle \Rightarrow 2(\sqrt{\mathrm{x}}\dfrac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\dfrac{1}{2\sqrt{\mathrm{x}}}\dfrac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}})=\dfrac{1}{2\sqrt{\mathrm{x}}}.y$

$\Rightarrow 4x\dfrac{d^{2}y}{dx^{2}}+2\dfrac{dy}{dx}-y=0$

(B)

$4x^{2}+9y^{2}=36\Rightarrow 8x+18y\dfrac{dy}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{4x}{9y}=0$

(C)

$y^{2}=x^{2}-cx\Rightarrow 2y\dfrac{dy}{dx}=2x-c$

$\therefore$ substituting the value of c in the given equation

$y^{2}-\left ( 2x-2y\dfrac{dy}{dx} \right )x=x^{2}$

i.e.,

$2xy\dfrac{dy}{dx}=3x^{2}-y^{2}$

(D)

$y=A\, cos\, 2x+Bsin\, 2x$

$\therefore \dfrac{dy}{dx}-2A\,sin\, 2x+2B\, cos\, 2x$

$\Rightarrow \dfrac{d^{2}y}{dx^{2}}=-4A\, cos2x-4B\, sin\: 2x=-4y$

$\Rightarrow \dfrac{d^{2}y}{dx^{2}}+4y=0$

For each of the differential equations, find the general solution:
$\cfrac{dy}{dx}+2y=\sin\,x$
2. $\cfrac{dy}{dx}+3y=e^{-2x}$
$\cfrac{dy}{dx}+\cfrac{y}{x}=x^{2}$
4. $\cfrac{dy}{dx}+(\sec\,x)y=\tan\,x$ ... $\left(0\le x < \cfrac{\pi}{2}\right)$
5. $\cos^2x\cfrac{dy}{dx}+y=\tan\,x$ ... $\left(0\le x < \cfrac{\pi}{2}\right)$
$x\cfrac{dy}{dx}+2y=x^2\log\, x$
$x\, \log\, x\cfrac{dy}{dx}+y = \cfrac{2}{x} \log x$
$(1+x^2)dy + 2xy dx = \cot x dx$ ... $(x \neq 0)$
$x\cfrac{dy}{dx}+y-x+xy\, \cot\, x=0$ ... $(x\neq 0)$
$(x+y)\cfrac{dx}{dy}=1$
$y\, dx+(x-y^2)dy=0$
12. $(x+3y^2)\cfrac{dy}{dx}=y$ ... $(y > 0)$

(1) $\dfrac { dy }{ dx } +2y=\sin \theta \\ \dfrac { dy }{ dx } +Py=Q\\ I.F={ e }^{ \int { Pdx } }={ e }^{ 2x }\\ G.S=y(I.F)=\int { Q(I.F)dx } \\ \Rightarrow _{ }^{ ax }{ \sin bx }=\dfrac { { a }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } x\\ \Rightarrow (a\sin bx-b\cos bx)+C\\ y({ e }^{ 2x })=\int { \sin x{ e }^{ 2x }dx } \\ y(e^{ 2x })=\dfrac { { e }^{ 2x } }{ 5 } (2\sin x-\cos x)+C\\ 2.\dfrac { dy }{ dx } +3y={ e }^{ -2x }\\ \dfrac { dy }{ dx } +Py=Q\\ I.F={ e }^{ \int { Pdx } }={ e }^{ 2x }={ e }^{ 3x }\\ G.S=y(I.F)=\int { Q(I.F)dx } \\ =y({ e }^{ 3x })=\int { { e }^{ -2x }{ e }^{ 3x }dx } \\ =y({ e }^{ 3x })=\int { { e }^{ x }dx } \\ =y({ e }^{ 3x })={ e }^{ x }+C\\ 3.\dfrac { dy }{ dx } +\dfrac { y }{ x } ={ x }^{ 2 }\\ \dfrac { dy }{ dx } +Py=Q\\ I.F={ e }^{ \int { Pdx } }={ e }^{ \int { \dfrac { 1 }{ x } dx } }=e^{ \log x }=x\\ G.S=y(I.F)=\int { Q(I.F)dx } \\ =y(x)=\int { { x }^{ 2 } } (x)dx\\ =xy=\dfrac { { x }^{ 3 } }{ 3 } +C\\ ={ x }^{ 3 }-3xy+k=0\\ 4.\dfrac { dy }{ dx } +(\sec x)y=\tan x\\ \dfrac { dy }{ dx } +Py=Q\\ I.F={ e }^{ \int { \sec xdx } }={ e }^{ \log \left| \sec x+\tan x \right| }=\sec x+\tan x\\ G.S=y(I.F)=\int { Q(I.F)dx } \\ \Rightarrow y(\sec x+\tan x)=\int { \sec x\tan xdx+\int { { \tan }^{ 2 }dx } } \\ =\sec x+\int { ({ \sec }^{ 2 }x-1)dx } \\ \Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C\\ 5.\cos ^{ 2 }\dfrac { dy }{ dx } +y=\tan x\\ \dfrac { dy }{ dx } +\sec ^{ 2 }x(y)=\dfrac { \tan x }{ \cos ^{ 2 }x } =\dfrac { \sin x }{ \cos ^{ 3 }x } \\ I.F={ e }^{ \int { pdx } }={ e }^{ \int { \sec ^{ 2 }x } dx }={ e }^{ \tan x }\\ G.S=y(I.F)=\int { Q(I.F)dx } \\ =y({ e }^{ \tan x })=\int { \tan x } e^{ \tan x }{ \sec }^{ 2 }dx\\ =y({ e }^{ \tan x })=\int { \tan x\sec ^{ 2 }xe^{ \tan x }dx } +C\\ let\quad \tan x\quad =\quad t\\ \sec ^{ 2 }xdx=dt\\ y{ e }^{ t }=\int { t{ e }^{ t }dt } +C\\ \Rightarrow { y }e^{ \tan x }=(\tan x-1)e^{ \tan x }+C$

$6.x\dfrac { dy }{ dx } +2y={ x }^{ 2 }\log x\\ \dfrac { dy }{ dx } +\dfrac { 2 }{ x } (y)=x\log x\\ \dfrac { dy }{ dx } +Py=Q\\ I.F={ e }^{ \int { Pdx } }={ e }^{ \int { \dfrac { 2 }{ x } dx } }={ e }^{ 2\log x }={ x }^{ 2 }\\ G.S\quad y(I.F)=\int { Q(I.F) } dx\\ =y(x^{ 2 })=\int { x^{ 3 }\log xdx } \\ y(x^{ 2 })=\log x(\dfrac { { x }^{ 4 } }{ 4 } )-\int { \dfrac { 1 }{ x } } (\dfrac { { x }^{ 4 } }{ 4 } )dx\\ { x }^{ 2 }y=\dfrac { { x }^{ 4 }\log x }{ 4 } -\dfrac { 1 }{ 16 } { x }^{ 4 }+C\\ 7.x\log x+y=\dfrac { 2 }{ x } \log x\\ \dfrac { dy }{ dx } +\dfrac { 1 }{ x\log x } (y)=\dfrac { 2 }{ { x }^{ 2 } } \\ I.F={ e }^{ \int { \dfrac { 1 }{ x\log x } dx } }=\log x\\ G.S\quad y(I.F)=\int { Q(I.F) } dx\\ =y(\log x)=\int { \dfrac { 2 }{ x^{ 2 } } \log xdx } \\ =\int { \dfrac { 1 }{ { x }^{ 2 } } \log xdx=\log x(-\dfrac { 1 }{ x } )-\int { \dfrac { 1 }{ x } (-\dfrac { 1 }{ x } )dx } } \\ =-\log x(\dfrac { 1 }{ x } )+\int { \dfrac { 1 }{ x^{ 2 } } dx } \\ =-\dfrac { 1 }{ x } \log x-\dfrac { 1 }{ x } C\\ solution\\ \Rightarrow y(\log x)=-\dfrac { 1 }{ x } \log x-\dfrac { 1 }{ x } +C\\ 8.(1+{ x }^{ 2 })dy+2xydx=\cot xdx\\ (1+{ x }^{ 2 })\dfrac { dy }{ dx } =\cot x-2xy\\ \dfrac { dy }{ dx } +(\dfrac { 2x }{ 1+{ x }^{ 2 } } )y=\dfrac { \cot x }{ 1+{ x }^{ 2 } } \\ I.F={ e }^{ \int { Pdx } }={ e }^{ \int { \dfrac { 2x }{ 1+{ x }^{ 2 } } dx } }=1+{ x }^{ 2 }\\ G.S\quad y(I.F)=\int { Q(I.F) } dx\\ y(1+{ x }^{ 2 })=\int { \dfrac { \cot x }{ 1+{ x }^{ 2 } } \times 1+{ x }^{ 2 } } dx\\ y(1+{ x }^{ 2 })=\log \left| \sin x \right| +C\\ 9.x\dfrac { dy }{ dx } +y-x+xy\cot x=0\\ \dfrac { dy }{ dx } +\dfrac { 1 }{ x } (y)-1+y\cot x=0\\ \dfrac { dy }{ dx } +[\dfrac { 1 }{ x } +\cot x]y=1\\ I.F={ e }^{ \int { Pdx } }={ e }^{ \int { \dfrac { 1 }{ x } +\cot xdx } }={ e }^{ \log x+\log \left\lfloor \sin x \right\rfloor }\\ G.S\quad y(I.F)=\int { Q(I.F) } dx\\ =y(x\sin x)=\int { x\sin x } dx\\ =y(x\sin x)=-x\cos x+\sin x+C\\ 10.(x+y)\dfrac { dy }{ dx } =1\\ \dfrac { dy }{ dx } -x=y\\ \dfrac { dy }{ dx } +Px=Q\\ I.F={ e }^{ \int { Pdy } }={ e }^{ \int { -1dy } }=e^{ -y }\\ G.S=x({ e }^{ -y })=\int { ye^{ -y } } dy\\ x({ e }^{ -y })=-y{ e }^{ -y }-e^{ -y }+C$

$11.ydx+(x-{ y }^{ 2 })dy=0\\ ydx+({ y }^{ 2 }-x)dy\\ \dfrac { dx }{ dy } =y-\dfrac { x }{ y } \\ \dfrac { dx }{ dy } +\dfrac { x }{ y } =y\\ I.F={ e }^{ \int { Pdy } }=e^{ \int { \dfrac { 1 }{ y } dy } }=y\\ G.S=x(y)=\int { y(y)dy } \\ =x(y)=\dfrac { { y }^{ 3 } }{ 3 } +C\\ 12.(x+{ 3y }^{ 2 })\dfrac { dy }{ dx } =y\\ \dfrac { dx }{ dy } =\dfrac { x }{ y } +3y\\ \dfrac { dx }{ dy } -\dfrac { x }{ y } =3y\\ I.F={ e }^{ \int { Pdy } }=e^{ \int { -\dfrac { 1 }{ y } dy } }=\dfrac { 1 }{ y } \\ G.S=x(\dfrac { 1 }{ y } )=\int { 3y(\dfrac { 1 }{ y } )dy } \\ \dfrac { x }{ y } =3y+C$

For each of the differential equations, find the general solution:
$\cfrac{dy}{dx} = \cfrac{1-\cos\, x}{1+\cos\, x}$
$\cfrac{dy}{dx} = \sqrt{4-y^2}(-2 < y < 2)$
$\cfrac{dy}{dx} = 1(y\neq 1)$
$\sec^2x\, \tan \,y\, dx+\sec^2\, y\, \tan \, x\, dy = 0$
5. $(e^x + e^{-x}) dy- (e^x -e^{-x}) dx = 0$
$\cfrac{dy}{dx} = (1+x^2)(1+y^2)$
$y\, \log\, y\, dx- x \,dy = 0$
$x^4\cfrac{dy}{dx} = -y^5$
$\cfrac{dy}{dx}=\sin^{-1}x$
$e^x \,\tan \,y \,dx + (1 -e^x) \sec^2\, y\, dy = 0$

1) $\dfrac{dy}{dx} = \dfrac{1-\cos x}{1+\cos x}$

$\int dy = \displaystyle \int \left[\dfrac {(1-\cos x)}{(1+\cos x)}\right)]dx$

$\int dy =\displaystyle \int \dfrac{2\sin^2(\frac{x}{2})}{2\cos^2 (\frac{x}{2})}dx$

$\int dy = \displaystyle \int \tan^2(\dfrac{x}{2})dx=\int (\sec^2 (\frac{x}{2}-1)dx$

$=\displaystyle \int \sec^2(\frac{x}{2} -1)dx$

$=2\tan \left (\dfrac{x}{2}-1\right)+c$

$y=2\tan \left (\dfrac{x}{2}-1\right)+c$

2) $\dfrac{dy}{dx} = \sqrt{4-y^2} (-2 < y < 2)$

$\Rightarrow \dfrac{dy}{\sqrt{4-y^2}} = dx$

$\Rightarrow \dfrac{dy}{\sqrt{1-(\frac{y}{2})^2}}=dx$

Let $\dfrac{y}{2}=t$

Thus $dy=2dt$

$\Rightarrow \dfrac{2dt}{\sqrt{1-t^2}}$

$\Rightarrow \int \dfrac{2dt}{\sqrt{1-t^2}} = \int dx$

$=2\sin ^{-1}(f) = x+c$

$=2\sin y (\frac{y}{2} = x+c$

3) $\dfrac{dy}{dx} = 1 (y \neq 1)$

$\int dy = \int dx$

$y=x \neq c$

4) $\sec^2x \tan y dx+\sec^2y \tan x dy = 0$

$\dfrac{dy}{dx} = \dfrac{\sec^2x \tan y}{\sec^2y \tan x}$

$\dfrac{\sec^2y}{\tan t}dy = -\dfrac{\sec^2x}{\tan x}dx$

$\tan y= u, \tan x = u$

$\sec^2ydy = du$ , $\sec^2xdx=dQ$

$\Rightarrow \int \dfrac{du}{u} = \dfrac{du}{u}$

$\log (u) =n-\log (Q) + \log C$

$\log (u) = \log \left(\frac{c}{Q}\right)$

$\log (uQ) = \log (C)$

$uQ = C$

$\tan y \tan x = C$

5) $(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$

$(e^x+e^{-x})dy=(e^x-e^{-x})dx$

$\dfrac{dy}{dx} = \dfrac{e^x-e^{-x}}{e^x+e^{-x}} dx$

$dy=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$

Integrating both sides.

$\int dx = \int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$ ...(1)

$y=\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

Let $t=e^x+e^{-x}$

$\dfrac {dt}{dx} = (e^x-e^{-x})dx$

$dx=\dfrac{dt}{e^x-e^{-x}}$

Substituting values in (1), we get

$\int dy = \int \dfrac{e^x-e^{-x}}{t}\dfrac{dt}{e^x-e^{-x}}$

$\int dy = \int \dfrac{dt}{t}$

$y=\log |t|+C$

Putting back $t=e^x-e^{-x}$

$y=\log |e^x-e^{-x}|+C$

$y=\log (e^x-e^{-x})+C$ $(\therefore e^x-e^{-x} > 0)$

6) $\dfrac{dy}{dx} = (1+x^2)(1+y^2) \Rightarrow dy$

$\Rightarrow \int \frac{1}{1+y^2}dy = \int (1+x^2)dx$

$\Rightarrow \tan^{-1}(y) = x+\dfrac{x^3}{3} + C$

7) $y \log y \,dx-x \,dy = 0$

$y \log y \,dx = x\, dy$

$\dfrac{dx}{x} = \dfrac{dy}{y\log y}$

Intergrating both sides

$\int \dfrac{dy}{y\log y} = \int \dfrac{dx}{x}$

Put $t =\log y$

$dt =\dfrac{1}{y}dy$

$dy=y\,dt$

Hence, our equation becomes

$\int \dfrac{y\,dt}{y.t} = \int \dfrac{dx}{x}$

$\int \dfrac{dt}{t} = \int\dfrac{dx}{x}$

$\log |t| = \log |x| + \log c$

Putting $t=\log y$

$\log (\log y)=\log x+\log c$

$\log (\log y)=\log cx$ (Using $\log ab=\log a + \log b$ )

$\log y=cx$

$y=e^{cx}$

8) $x^2 \dfrac{dy}{dx} = -y^5$

$\int \dfrac{dy}{dx} = \int \dfrac{dx}{xy} \Rightarrow \int y^{-5} dy = -\int dx$

$\Rightarrow \dfrac{y^{-4}}{-4} = \dfrac{x^{-3}}{-3} +c$

$\Rightarrow \dfrac{1}{3x^3}+\dfrac{1}{4y^4}+C = 0$

9) $\dfrac{dy}{dx} = \sin^{-1}(x) \Rightarrow \int dy =\int \sin^{-1}(x)dx$

So $\therefore \int \sin^{-1}(x)dx = x\sin^{-1}(x)+\sqrt{1-x^2}$

$y=x\sin^{-1}(x) + \sqrt{1-x^2} + c$

10) $e^x\tan y \,dx +(1-e^x) \sec^2y dy = 0$

$e^x\tan y dx = (e^x-1)\sec^2y dy$

$\int \dfrac{\sec^2 y}{\tan y} dy = \int\dfrac{e^x}{e^{x}-1} dx$

$\tan y = u, e^x-1 = Q$

$\sec^2y\, dy = du$ $e^xdx=du$

$\int \dfrac{du}{u} = \int \dfrac{du}{u}$

$\log(u) = \log(Q) + \log (C)$

$\tan y = (e^x-1)=c$

Find the particular solution of the differential equation $\dfrac {dy}{dx} = \dfrac {x(2\log x + 1)}{\sin y + y\cos y}$ given that $y = \dfrac {\pi}{2}$ when $x = 1$ .

$\dfrac{dy}{dx}=\dfrac{x(2 \log x+1)}{\sin y+y \cos y}$

$\Rightarrow (\sin y+ y \cos y)dy=x(2 \log x+1)dx$

$\Rightarrow \displaystyle\int (\sin y+y\cos y)dy=\int x(2 \log x+1)dx$

$\Rightarrow \displaystyle\int \sin y dy+\int y \cos y dy = \int 2x \log x dx + \int x dx$

$\Rightarrow -\cos y+y\sin y+\cos y=2\left ( \dfrac{1}{2}x^2\ln(x)-\dfrac{x^2}{4} \right )+\dfrac{x^2}{2}+C$

$\Rightarrow y\sin y=\left ( x^2\ln(x)-\dfrac{x^2}{2} \right )+\dfrac{x^2}{2}+C$

$\Rightarrow y\sin y=x^2\ln (x)+C$

The required solution is

$x^2\ln (x)-y\sin y+C=0.$

When

$(x,y) \equiv \left(1,\dfrac\pi2\right)$ ,

We get

$C = \dfrac\pi2$

Hence, the particular solution is

$x^2\ln (x)-y\sin y+\dfrac\pi2=0.$

Find the particular solution of the differential equation $\dfrac {dy}{dx} = 1 + x + y + xy$ , given that $y = 0$ , when $x = 1$ .

$\dfrac {dy}{dx} = (1 +x)(1 + y)\Rightarrow \int \dfrac {dy}{(1 + y)} = \int (1 + x)dx \Rightarrow \log|1 + y| = x + \dfrac {x^{2}}{2} + C .... (i)$
Putting

$x = 1$ and

$y = 0$ in (i), we get

$C = \dfrac {-3}{2}$ .

$\therefore \log |1 + y| = x + \dfrac {x^{2}}{2} - \dfrac {3}{2}$ .

Show that the general solution of the differential equation $\dfrac{dy}{dx}+\dfrac{y^2+y+1}{x^2+x+1}=0$ is given by $(x+y+1)=A(1-x-y-2xy)$ , where $A$ is parameter.

$x+y+1 = A(1-x-y-2xy)$

Differentiate both sides w.r.t.

$x$

$\Rightarrow 1+y' = A(-1-y'-2y-2xy')$

$\Rightarrow (1+A+2Ax)y' = -1+ A(-1-2y)$

$\Rightarrow y' = -\cfrac{1+A(1+2y)}{1+A(1+2x)}$ .....

$(1)$

Now, from the given equation,

$A = \cfrac{x+y+1}{1-x-y-2xy}$ .

Substitute this value of

$x$ in equation

$(1)$

$\Rightarrow y'$

$= -\cfrac{1+\dfrac{(x+y+1)(1+2y)}{1-x-y-2xy}}{1+\dfrac{(x+y+1)(1+2x)}{1-x-y-2xy}}\\=-\cfrac{1-x-y-2xy+x+y+1+2xy+2y^2+2y}{1-x-y-2xy+x+y+1+2x^2+2xy+2x}\\=-\cfrac{2y^2+2y+2}{2x^2+2x+2}$

$\Rightarrow \dfrac{dy}{dx} + \dfrac{y^2+y+1}{x^2+x+1} = 0$

Hence,

$x+y+1=A(1-x-y-2xy)$ is the general solution of the given differential equation as it satisfies the equation.

Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i) $y=e^x(a\, \cos\,x+b\, sin\, x)$   :    $\cfrac{d^2y}{dx^2}-2\cfrac{dy}{dx}+2y=0$
(ii) $y = x \,\sin 3x$ :    $\cfrac{d^2y}{dx^2}+9y-6\,\cos\,3x=0$
(iii) $x^2 = 2y^2 \,\log\, y$ :    $(x^2+ y^2 ) \cfrac{dy}{dx}-xy= 0$

(i)

$y=e^x(a\cos x+b \sin x)$

Differenttiating Both sides w.r.t.x

$\dfrac{dy}{dx} = \dfrac{d}{dx}\left[e^x(a\cos x+b\sin x)\right]$

$y' = \dfrac{d(e^x)}{dx}.[a\cos x+b\sin x]+e^x\dfrac{d}{dx}[a\cos x+b\sin x]$

$y'=e^x[a\cos x+b\sin x]=e^x[-a\sin x+b\cos x]$

$y'=y+e^x[-a\sin x+b\cos x]$ ...(1)

Again Differentiating both sides w.r.t.x

$y''=y'+ \dfrac{d(e^x)}{dx}[-a\sin x+b\cos x]+e^x\dfrac{d}{dx}[-a\sin x+b\cos x]$

$y''=y'+e^x[-a\sin x+b \cos x]=e^x[-a\cos x+b(-\sin x)]$

$y''=y'+(y'-y)+e^x[-a\cos x-b\sin x]$ (From (1)]

$y''=2y'-y-e^x[a\cos x+b\sin x]$

$y''=2y'-y-y$ (Using

$y=e^x(a\cos x+b\sin x)]$

$y''=2y'-2y$

$y''-2y'+2y=0$

Which is the required differential equation

(ii)

$y=x\sin 3x ; \cfrac{d^2y}{dx^2}+9y-6\,\cos\,3x=0$

$\dfrac{dy}{dx} = \sin 3x+3x\cos 3x$

$\dfrac{d^2y}{dx^2}=3\cos 3x+3\cos 3x-9x\sin3x$

$\dfrac{d^2y}{dx} = 6 \cos 3x-9x\sin 3x$

where

$x\sin 3x=y$

$\dfrac{d^2y}{dx^2}=9y-6\cos 3x=0$

This is the proof that

$y=x\sin 3x$ is the solution of the given differential equation.

(iii)

$x^2=2y^2\log y; (x^2+ y^2 ) \cfrac{dy}{dx}-xy= 0$

Differentiating both the sides we get;

$2x=4y\log y\dfrac{dy}{dx} + \dfrac{2y^2}{y} \dfrac{dy}{dx}$

$2x=\dfrac{dy}{dx} (4y\log y + 2y)$

$\therefore x = \dfrac{dy}{dx} (2y\log y+y)$

Multiplying both the sides by y, we get,

$xy=\dfrac{dy}{dx}$

$(2y^2\log y+y^2)$

Therefore

$(x^2+y^2)\dfrac{dy}{dx} - xy=0$

Hence proved.

Find the particular solution of the differential equation $\dfrac {dy}{dx} + 2y\cot x = 4x\ cosec x (x\neq 0)$ , given that $y = 0$ , when $x = \dfrac {\pi}{2}$

$\cfrac{dy}{dx} + Py = Q$ has the integrating factor as

$\displaystyle e^{\int P dx}$ .

Here,

$P = 2\cot x, Q = 4x\ cosec x$

Integrating factor

$= e^{2 \int \cot x dx} = e^{2\log (\sin x)} = \sin^2{x}$

$\Rightarrow \cfrac{d(y\sin^2x)}{dx} = 4x \times cosec x \times \sin^2 x$

$\Rightarrow d(y\sin^2 x) = 4x\sin x dx$

Integrating both sides, we have

$y\sin^2 x = 4x(-\cos x) + 4\sin x + c$

Substituting

$y = 0$ when

$x = \cfrac{\pi}{2}$ , we get

$0 = 0 + 4 + c$

$\Rightarrow c = -4$

The solution thus becomes

$y\sin^2 x = 4 (-x\cos x + \sin x - 1)$

Solve the differential equation $2xy + y^2 - 2x^2 \dfrac{dy}{dx} = 0$ .

$\dfrac{dy}{dx}-\dfrac{y^{2}}{2x^{2}}=\dfrac{y}{x}$
Let

$y=vx$ then

$\dfrac{dy}{dx}=v+\dfrac{xdv}{dx}$
hence

$v+\dfrac{xdv}{dx}-\dfrac{v^{2}}{2}=v$

$\dfrac{xdv}{dx}=\dfrac{v^{2}}{2}$
Hence

$v^{-2}dv=0.5x^{-1}dx$
Integrating both sides give us

$-v^{-1}=0.5\ln x+c$
Or

$\dfrac{-x}{y}=0.5\ln x+c$

Solve the differential equation:
$(x + 1)dy - 2 xy\ dx = 0$

$(x + 1)dy - 2xy\ dx = 0$

$\dfrac {dy}{dx} = \dfrac {2xy}{x + 1}$

$\dfrac {1}{y}dy = \left (\dfrac {2x}{x + 1}\right ) dx$

$\dfrac {1}{y}dy = \left [\dfrac {2(x + 1)}{(x + 1)} = \dfrac {2}{(x + 1)}\right ]dx$

$\displaystyle \int \dfrac {1}{y}dy = \displaystyle \int 2dx - \displaystyle \int \dfrac {2}{(x + 1)} dx$

$\log |y| = 2x - 2 \log |x + 1| + \log c$

$\log |y| + 2\log |x + 1| - \log c = 2x$

$\dfrac {\log y (x + 1)^{2}}{c} = 2x$

$\dfrac {y(x + 1)^{2}}{c} = c^{2x}$

$y = \dfrac {ce^{2x}}{(x + 1)^{2}}$

Solve: $(x^2-yx^2)dy+(y^2+xy^2)dx=0$

Solution

Given , Differential Equation is

$\left ( x^{2} - yx^{2} \right )dy + \left (y^{2} + xy^{2} \right )dx = 0$

This can be Simplified as

$\left ( yx^{2} - x^{2} \right )dy = \left (y^{2} + xy^{2} \right )dx$

$x^{2}\left ( y - 1 \right )dy = y^{2}\left ( 1 + x\right )dx$

$\dfrac{dy(y -1)}{y^{2}} = \dfrac{dx(1+x)}{x^{2}}$

Now On Integrating both side , we get

$\int \dfrac{dy(y -1)}{y^{2}} = \int \dfrac{dx(1+x)}{x^{2}}$

$\int\left [ \dfrac{1}{y} - \dfrac{1}{y^{2}} \right ]dy = \int \left [ \dfrac{1}{x} + \dfrac{1}{x^{2}} \right ]dx$

$\int \left [ \dfrac{dy}{y} - \dfrac{dy}{y^{2}} \right ] = \int \left [ \dfrac{dx}{x} + \dfrac{dx}{x^{2}} \right ]$

$ln\left | y \right | - \dfrac{1}{y} = ln\left | x \right | -\dfrac{1}{x} + lnC$

$ln\left | y \right | - ln\left | x \right | - lnC = \dfrac{1}{y} -\dfrac{1}{x}$

$ln \left [ \dfrac{\left | y \right |}{c\left | x \right |} \right ] = \dfrac{x -y}{xy}$

Verify that $y^{2} = 4a (x + a)$ is a solution of the differential equation $y \left \{1 - \left (\dfrac {dy}{dx} \right )^{2} \right \} = 2x \dfrac {dy}{dx}$ .

Given,

${ y }^{ 2 }=4a\left( x+a \right)$

Differentiating both side w.r.t. x, we get,

$\dfrac{\mathrm{d} }{\mathrm{d} x}{ y }^{ 2 }=\dfrac{\mathrm{d} }{\mathrm{d} x}4a\left( x+a \right)$

$2y\dfrac { { dy } }{ { d }x } =\dfrac { { d } }{ { d }x } 4ax+\dfrac { { d } }{ { d }x } 4{ a }^{ 2 }$

$2y\dfrac { { dy } }{ { d }x } =4a+0$

$\dfrac { { dy } }{ { d }x } =\dfrac { 2a }{ y }$

Now,

$y\left\{ 1-\left( \dfrac { { dy } }{ { d }x } \right) ^{ 2 } \right\} =2x\dfrac { { dy } }{ { d }x } \\$

$L.H.S = y\left\{ 1-\left( \dfrac { { dy } }{ { d }x } \right) ^{ 2 } \right\}$

Put the value of

$\dfrac { { dy } }{ { d }x } =\dfrac { 2a }{ y }$

$= y\left\{ 1-\left( \dfrac { 2a }{ y } \right) ^{ 2 } \right\}$

$= y\left\{ 1-\left( \dfrac { 4{ a }^{ 2 } }{ { y }^{ 2 } } \right) \right\}$

$= y\left( \dfrac { { y }^{ 2 }-4{ a }^{ 2 } }{ { y }^{ 2 } } \right)$

Simplifying

$= \left( \dfrac { { y }^{ 2 }-4{ a }^{ 2 } }{ { y } } \right)$

Given,

${ y }^{ 2 }=4a\left( x+a \right)$

$= \left( \dfrac { 4a\left( x+a \right) -4{ a }^{ 2 } }{ \sqrt { 4a\left( x+a \right) } } \right)$

$= \left( \dfrac { 4ax+4{ a }^{ 2 }-4{ a }^{ 2 } }{ \sqrt { 4a\left( x+a \right) } } \right)$

$= \left( \dfrac { 4ax }{ \sqrt { 4a\left( x+a \right) } } \right)$

$= 2x\left( \dfrac { 2a }{ \sqrt { 4a\left( x+a \right) } } \right)$

$= 2x\left( \dfrac { 2a }{ y } \right)$

$= 2x\dfrac { { dy } }{ { d }x }$

$= R.H.S$

$\therefore { y }^{ 2 }=4a\left( x+a \right)$ is a solution for the differential equation

$y\left\{ 1-\left( \dfrac { { dy } }{ { d }x } \right) ^{ 2 } \right\} =2x\dfrac { { dy } }{ { d }x } \\$

Solve: $\left( { D }^{ 2 }-4D+1 \right) y={x}^{2}$

We have to solve

$(D^2-4D+1)y=x^2$

The characteristic equation is

$p^2-4p+1=0$

$\Rightarrow p=\dfrac{4\pm \sqrt{16-4}}{2}=\dfrac{4\pm 2\sqrt{3}}{2}=2 \pm \sqrt{3}$

Thus Complementary function

$C.F.=Ae^{(2+\sqrt{3})x}+Be^{(2-\sqrt{3})x}$

Particular integral

$P.I.=\dfrac{1}{D^2-4D+1}(x^2)$

$=[1-(4D-D^2)]^{-1}(x^2)$

$=[1+(4D-D^2)+(4D-D^2)^2+...](x^2)$

$=[1+4D+15D^2+...](x^2)$

$\therefore P.I.=x^2+8x+30$

Hence the general solution is

$y=C.F.+P.I.$

$y.=Ae^{(2+\sqrt{3})x}+Be^{(2-\sqrt{3})x}+(x^2+8x+30)$

The solution of $x^{2}y^{2}_{1} + xy\ y_{1} - 6y^{2} = 0$ are (here $y_{1} = dy/ dx)$ ?

$x^2\left(\dfrac{dy}{dx}\right)^2+xy\left(\dfrac{dy}{dx}\right)-by^2=0$

Let

$x\left(\dfrac{dy}{dx}\right)=v$

Re-writing the differentiate equation,

$\Rightarrow (v)^2+vy-by^2=0\Rightarrow v^2+vy-by^2=0$

factorizing the quadratic,

$(v+3y)(v-2y)=0$

$v=-3y$ or

$v=2y$

$\Rightarrow x\left(\dfrac{dy}{dx}\right)=-3y$ or

$x\left(\dfrac{dy}{dx}\right)=2y$

$\Rightarrow \dfrac{dy}{y}=-3\dfrac{dx}{x}$ or

$\dfrac{dy}{y}=2\dfrac{dx}{x}$

$\Rightarrow \displaystyle \int \dfrac{dy}{y}=-3\int \dfrac{dx}{x}$ or

$\displaystyle \int \dfrac{dy}{y}=2\int\dfrac{dx}{x}$

$\Rightarrow y=\dfrac{c_1}{x^3}$ or

$y=c_2x^2$

$\therefore$ We have two solutions

$y=cx^2$ or

$x^3y=c$

Let $f : \mathbb {R} \rightarrow \mathbb {R}$ be a differentiable function with $f(0) = 0$ . If $y = f(x)$ satisfies the differential equation $\dfrac {dy}{dx} = (2 + 5y)(5y - 2)$ , then the value of $\displaystyle \lim_{x\rightarrow -\infty} f(x)$ is $k$ . Then what is the value of $5k\,?$

$\dfrac {dy}{dx} = 25y^{2} - 4$
So,

$\dfrac {dy}{25y^{2} - 4} = dx$
Integrating,

$\dfrac {1}{25} \times \dfrac {1}{2\times \dfrac {2}{5}} ln \left |\dfrac {y - \dfrac {2}{5}}{y + \dfrac {2}{5}}\right | = x + c$

$\Rightarrow ln \left |\dfrac {5y - 2}{5y + 2}\right | = 20(x + c)$

Now,

$c = 0$ as

$f(0) = 0$

Hence,

$\left |\dfrac {5y - 2}{5y + 2}\right | = e^{(20x)}$

$\therefore \displaystyle \lim_{x\rightarrow -\infty} \left |\dfrac {5f(x) - 2}{5f(x) + 2}\right | = \displaystyle \lim_{x\rightarrow -\infty} e^{(20x)}$

Now,

$RHS = 0\Rightarrow \displaystyle \lim_{x\rightarrow -\infty} (5f(x) - 2) = 0$

$\Rightarrow \displaystyle \lim_{x\rightarrow -\infty} f(x) = \dfrac {2}{5} = 0.40$ .

Solve
$\int { \frac { dy }{ y } } =\int { \frac { dx }{ x } }$

$\int\cfrac{dy}{y}=\int\cfrac{dx}{x}$

$\log y=\log x+\log C$

$\cfrac{y}{x}=C$

Show that $y = ae^{2x} + be^{-x}$ is a solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} - \dfrac {dy}{dx} - 2y = 0$ .

given differential equation

$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -\cfrac{dy}{dx}-2y=0....(1)$

$y=a{e}^{2x}+b{e}^{-x}....(2)$

differentiating on both sides

$2a{e}^{2x}-b{e}^{-x}$

Again differentiating on both sides

$4a{e}^{2x}+b{e}^{-x}$

substituting in (1)

$(4a{e}^{2x}+b{e}^{-x})-(2a{e}^{2x}-b{e}^{-x})-2(a{e}^{2x}+b{e}^{-x})=0$

So, (2) satisfies eqation (1)

$y=a{e}^{2x}+b{e}^{-x}$ is a solution of (1)

$\sec^{2} x\tan{y}\ dy+\sec^{2} y\tan x\ dx=0$

Consider the following Equation.

$se{{c}^{2}}xtanydy+{{\sec }^{2}}ytanxdx=0$

$\int\limits_{{}}^{{}}{\dfrac{\tan x}{{{\sec }^{2}}x}dx=-\int\limits_{{}}^{{}}{\dfrac{\tan y}{{{\sec }^{2}}y}}}dy$

$u=\dfrac{1}{{{\sec }^{2}}x}$

$dx=\dfrac{{{\sec }^{2}}x}{2\tan x}du$

Similarly,

$v=\dfrac{1}{{{\sec }^{2}}y}$

$dy=\dfrac{{{\sec }^{2}}y}{2\tan y}dv$

Therefore,

$\int\limits_{{}}^{{}}{\dfrac{\tan x}{{{\sec }^{2}}x}}\dfrac{{{\sec }^{2}}x}{2\tan x}du=\int\limits_{{}}^{{}}{\dfrac{\tan y}{{{\sec }^{2}}y}}\dfrac{{{\sec }^{2}}y}{2\tan y}dv$

$\dfrac{1}{2}\int\limits_{{}}^{{}}{1}du=-\dfrac{1}{2}\int\limits_{{}}^{{}}{1}dv$

$\dfrac{u}{2}=-\dfrac{v}{2}+C$

$\dfrac{1}{{{2\sec }^{2}}x}=-\dfrac{1}{{{2\sec }^{2}}y} +C$

${{\cos }^{2}}x+{{\cos }^{2}}y=2C$

Hence, this is the correct answer.

Solve:
$(x-y)^{2}\dfrac {dy}{dx}=a^{2}$

Given that:

${{(x-y)}^{2}}\dfrac{dy}{dx}={{a}^{2}}$

$Let\ x-y=v$

$1-\dfrac{dy}{dx}=\dfrac{dv}{dx}$

$1-\dfrac{dv}{dx}=\dfrac{dy}{dx}$

$Then$

${{v}^{2}}\left( 1-\dfrac{dv}{dx} \right)={{a}^{2}}$

$\left( {{v}^{2}}-{{v}^{2}}\dfrac{dv}{dx} \right)={{a}^{2}}$

$\dfrac{dv}{dx}=\dfrac{{{v}^{2}}-{{a}^{2}}}{{{v}^{2}}}$

$\dfrac{dx}{dv}=\dfrac{{{v}^{2}}}{{{v}^{2}}-{{a}^{2}}}$

$dx=\left( \dfrac{{{v}^{2}}-{{a}^{2}}}{{{v}^{2}}-{{a}^{2}}}+\dfrac{{{a}^{2}}}{{{v}^{2}}-{{a}^{2}}} \right)dv$

$dx=\left( 1+\dfrac{{{a}^{2}}}{{{v}^{2}}-{{a}^{2}}} \right)dv$

$On\ \operatorname{int}egrating\ both\ side,\ we\ get$

$x=v+{{a}^{2}}\times \dfrac{1}{2a}\log \left( \dfrac{v-a}{v+a} \right)+c$

$x=x-y+{{a}^{2}}\times \dfrac{1}{2a}\log \left( \dfrac{x-y-a}{x-y+a} \right)+c$

$y=\dfrac{a}{2}\log \left( \dfrac{x-y-a}{x-y+a} \right)+c.$

This is the required solution.

Solve the following differential equation:
$\left( { x }^{ 2 }+1 \right) \frac { dy }{ dx } +2xy=\sqrt { { x }^{ 2 }+4 }.$

1116231_1139909_ans_42784c0670c84c0482f0d18ffd324847.jpeg

Solve : $(x^2 D^2 + 3xD + 1) y = \dfrac{1}{(1 - x)^2}$

Given,

$(x^2D^2+3xD+1)y=\dfrac{1}{(1-x)^2}$ ......(1)

Let,

$x=e^z,z=\ln x, x>0, \dfrac{d}{dz}\equiv \theta$

$\Rightarrow xD\equiv \theta ,x^2D^2\equiv \theta (\theta -1)$

substituting these in equation (1)

$(\theta (\theta -1)+3\theta+1)y=\dfrac{1}{(1-e^z)^2}$

$\Rightarrow (\theta ^2+2\theta +1)y=\dfrac{1}{(1-e^z)^2}$ ....(2)

The general solution is given by,

$y=y_c+y_p$

$y_c=c_1e^{-z}+c_2ze^{-z}$

$y_p=\dfrac{1}{\theta ^2+2\theta +1}\dfrac{1}{(1-e^z)^2}$

$=\dfrac{1}{\theta +1}\left \{ \dfrac{1}{\theta +1}\dfrac{1}{(1-e^z)^2} \right \}$

$=\dfrac{1}{\theta +1}\left \{ e^{-z}\int \dfrac{1}{(1-e^z)^2}e^zdz \right \}$

$=\dfrac{1}{\theta +1}\left \{ \dfrac{e^{-z}}{1-e^z} \right \}$

$=e^{-z}\int \dfrac{e^{-z}}{1-e^z}e^z dz$

$=-e^{-z}\ln |e^{-z}-1|$

$y_p=-e^{-z}\ln |e^{-z}-1|$

$\therefore y=c_1e^{-z}+c_2ze^{-z}-e^{-z}\ln |e^{-z}-1|$

$\Rightarrow y=c_1x^{-1}+c_2(\ln x)x^{-1}-x^{-1}\ln |x^{-1}-1|$

Solve $(D^2+4D+3)y=e^{-x}\sin x+x e^{3x}$ .

$(D+1)(D+3) y = e^{-x} \sin x + xe^{3x}$

C.F. =

$C_1 e^{-x} + C_2 e^{-3x}$

For P.I. , we get

$\cfrac{e^{-x} \sin x + xe^{3x}}{D^2 + 4D+3}$

$e^{-x} \cfrac{\sin x}{D(D+2)} + e^{3x} \cfrac{x}{(D+4)(D+6)}$

$e^{-x} \cfrac{(D-2) \sin x}{D(D^2 - 4)} + e^{3x} \cfrac{x}{24\cfrac{(D^2+10D)}{24} + 1}$

$=e^{-x} \cfrac{(\sin x + 2\cos x)}{-3} + e^{3x} \cfrac{5}{288}$

Solve the differential equation $(\tan^{-1} x-y)dx = (1+x^2)dy$ .

$(tan^{-1}x - y)dx = (1 + x^2)dy$

$\therefore \dfrac{dy}{dx} = \dfrac{\tan^{-1}x-y}{1+x^2}$

$\dfrac{dy}{dx} + \dfrac{1}{1+x^2}y = \dfrac{\tan^{-1}x}{1+x^2}$

Above equation is linear

$\dfrac{dy}{dx} + Py = Q$

$P = \dfrac{1}{1+x^2} Q = \dfrac{\tan^{-1}x}{1+x^2}$

IF = e^{\int Pdx}$$

$=e^{\int(1/1+x^2)dx}$

$=e^{\tan^{-1}x}$

solution is given as

$ye^{-\tan^{-1}x} = \int \left( \dfrac{\tan^{1}x}{1+x^2} e^{\tan^{-1}x}\right)dx + c$ ...(1)

$\tan^{-1}x = t$

$\therefore \dfrac{1}{1+x^2}dx = dt$

$\int te^{t}dt$

$=t\times \int e^t dt - \int \left(\dfrac{d}{dt}(t) \times \int e^tdt\right]at$

$=te^t - \int e^tdt$

$=te^t - e^t$

$=\tan^{-1}xe^{\tan^{-1}x} - e^{\tan^{-1}x}$

From (1), we get

$ye^{\tan^{-1}x} = \tan^{-1}xe^{\tan^{-1}x} - e^{\tan^{-1}x}+C=y = \tan^{-1}x-1+ce^{\tan^{-1}x}$

Consider the differential equation $y^2dx+\left(x-\dfrac{1}{y}\right)dy=0$ . It $y(1)=1$ , then x is given by?

$\begin{matrix} { y^{ 2 } }dx+\left( { x-\frac { 1 }{ y } } \right) dy=0,\, \, at\, \, y(1)=1,\, \, then\, \, is\, \, given\, \, by \\ \frac { { dy } }{ { dx } } =\frac { { { y^{ 2 } } } }{ { \left( { x-\frac { 1 }{ y } } \right) } } \\ \Rightarrow \frac { { dy } }{ { dx } } =\frac { { { y^{ 3 } } } }{ { \left( { xy-1 } \right) } } \\ \int { \left( { xy-1 } \right) dy=\int { { y^{ 3 } }dx } } \\ x\int { ydy-\int { dy=\int { { y^{ 3 } } } dx } } \\ \left[ { \frac { { x{ y^{ 2 } } } }{ 2 } -{ y^{ 2 } } } \right] =\left[ { { y^{ 3 } }.x+c } \right] \\ \left[ { \frac { 1 }{ 2 } -1 } \right] =\left[ { 1\times 1+c } \right] \\ \frac { { -1 } }{ 2 } -1=c \\ \therefore c=\frac { { -3 } }{ 2 } \\ Put\, \, c=\frac { { -3 } }{ 2 } \\ \frac { { x{ y^{ 2 } } } }{ 2 } -y={ y^{ 3 } }x\frac { { -3 } }{ 2 } \\ \frac { { x{ y^{ 2 } } } }{ 2 } -y-{ y^{ 3 } }x=\frac { { -3 } }{ 2 } \\ y\left( { \frac { { xy } }{ 2 } -{ y^{ 2 } }x-1\int { \frac { { -3 } }{ 2 } } } \right) \, \, Ans. \\ \end{matrix}$

Prove that ${x}^{2}-{y}^{2}=c\left({x}^{2}+{y}^{2}\right)^{2}$ is the general solution of differential equation $\left({x}^{3}-3x{y}^{2}\right)dx=\left({y}^{3}-{3x}^{2}y\right)dy$ , where $c$ is a parameter.

$\dfrac{dy}{dx}= \dfrac{x^{3}-3xy^{2}}{y^{3}-3x^{2}y}$

Let

$\dfrac{y}{x}= v\Rightarrow y= vx$

$y^{1}= v+xv^{1}$

$v+xv^{1}= \dfrac{1-3v^{2}}{v^{3}-3v}$

$xv^{1}= \dfrac{1-3v^{2}-v}{v^{3}-3v}\Rightarrow xv^{1}=\dfrac{1-v^{4}}{v^{3}-3v}$

$\displaystyle = \int \dfrac{v^{3}-3v}{1-v^{4}}dv= \int \dfrac{1}{x}dx$

$\displaystyle \dfrac{-1}{4}\int \dfrac{v^{3}}{1-v^{4}}dv-\int \dfrac{3v}{1-v^{4}}dv= \int \dfrac{1}{x}dx$

$\displaystyle [\dfrac{-1}{4}log(1-v^{4})-\dfrac{1}{4}log(\dfrac{1+v^{2}}{1-v^{2}})]= logx+ log C$

$\displaystyle \dfrac{-1}{4}log[\dfrac{(1+v^{2})^{4}}{(1-v^{2})^{2}}]= log\: Cx$

$log[\dfrac{(1-v^{2})^{2}}{(1+v^{2})^{4}}]^{1/4}= log\: Cx$

$Cx= [\dfrac{(1-v^{2})^{2}}{(1+v^{2})^{4}}]^{1/4}= \dfrac{(1-v^{2})^{1/2}}{(1+v^{2})}$

$xC(1+v^{2})= (1-v^{2})^{1/2}$

$C(\dfrac{x^{2}+y^{2}}{x^{2}})x= (\dfrac{x^{2}-y^{2}}{x^{2}})^{1/2}$

$C(\dfrac{x^{2}+y^{2}}{x})=\dfrac{\sqrt{x^{2}-y^{2}}}{x}$

squaring both sides, we get,

$(x^{2}-y^{2})= C(x^{2}+y^{2})^{2}$

Which is the required solution.

1146767_1144150_ans_6f7ed475047a4ff5ad12b05e8c366b91.jpeg

$y''=5y'+5(y'-5y)$ .

$y^{4}=5y^{1}+5y^{1}-25y$

$\Rightarrow y^{4}-10y^{1}+25y=0$

A.E

$m^{2}-10m+25=0$

$m=5, 5$

$\therefore y=(c_{1}x+c_{2})e^{5x}$

Evaluate : $( x + y ) ( d x - d y ) = d x + d y$

$\textbf{Step - 1: Solve } \mathbf{x, y} \textbf{ with } \mathbf{z}$

$(x+y)(dx - dy) = dx +dy$

$\Rightarrow (x+y)\left(1- \dfrac{{dy}}{{dx}} \right) = 1 + \dfrac{{dy}}{{dx}}.........(1)$

$\text{Put } x+y=z$

$\Rightarrow 1+ \dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}}$

$\textbf{Step - 2: Substitute in equation } \mathbf{(1)}$

$z\left( 1- \dfrac{{dz}}{{dx}} +1 \right) = \dfrac{{dz}}{{dx}}$

$\Rightarrow 2z - (z+1) \dfrac{{dz}}{{dx}} = 0$

$\Rightarrow \dfrac{{dz}}{{dx}} = \dfrac{{2z}}{{z+1}}$

$\Rightarrow \dfrac{{z+1}}{{z}} dz = 2dx$

$\Rightarrow \int \left(1+ \dfrac{{1}}{{z}} \right) dz = 2 \int dx$

$\Rightarrow z +\log{z} = 2z - \log{C_1}$

$\Rightarrow 2x -z = \log{z} + \log{C_1}$

$\Rightarrow x-y = \log{C_1}$

$\Rightarrow e^{x-y} = C_1 (x +y)$

$x+y = \dfrac{{1}}{{C_1}} e^{x-y}$

$\text{Put } C = \dfrac{{1}}{{C_1}}$

$x+y = C e^{x-y}$

$\textbf{Hence , } \mathbf{x+y = C e^{x-y}}$

Solve:
$\dfrac { d y } { d x } + \dfrac { \sqrt { \left( x ^ { 2 } - 1 \right) \left( y ^ { 2 } - 1 \right) } } { x y } = 0$

Given

$\dfrac{dy}{dx}+ \dfrac{\sqrt{(x^{2}-1)(y^{2}-1)}}{xy} =0$

$\dfrac{dy}{dx}+ \dfrac{\sqrt{(x^{2}-1)}}{x} \dfrac{\sqrt{y^{2}-1}}{y} =0$

$\dfrac{dy}{dx}= \dfrac{-\sqrt{(x^{2}-1) \sqrt{(y^{2}-1)}}}{xy} \Rightarrow \dfrac{y}{\sqrt{y^{2}-1}} dy= \dfrac{-\sqrt{x^{2}-1}}{x} dx$

$\int \dfrac{y}{\sqrt{y^{2}-1}} dy= -\int \sqrt{1- \dfrac{1}{x^{2}}} dx$ .

Let

$y^{2}-1 = t \Rightarrow 2ydy = dt \ ydy=dt/2$ .

and

$\int \dfrac{\sqrt{x^{2}-a^{2}}}{x}dx =\sqrt{x^{2}-a^{2}} - a \sec^{-1} \left(\dfrac{|x|}{|a|} \right)+c$

So ,

$\because \int \dfrac{1}{t^{1/2} } \dfrac{dt}{2} = - \int \sqrt{x^{2}-1}+1. \sec^{-1} |x|+c.$

$=\dfrac{1}{2} \times \left( \dfrac{2}{3} \right) t^{1/2} = -\sqrt{x^{2}-1} + \sec^{-1} |x|+c$

$\Rightarrow \sqrt{y^{2}-1} = - \sqrt{x^{2}-1} + \sec^{-1} |x|+c$

$\Rightarrow \sqrt{y^{2}-1} + \sqrt{x^{2}-1} = \sec^{-1} |x| +c$

Solve: $\left( 1+\tan { y } \right) \left( dx-dy \right) +2xdy=0$

Given:

$\left(1+\tan{y}\right)\left(dx-dy\right)+2xdy=0$

$\Rightarrow dx+\tan{y}dx-dy-\tan{y}dy+2xdy=0$

$\Rightarrow dy\left(1+\tan{y}-2x\right)=\left(1+\tan{y}\right)dx$

$\dfrac{dy}{dx}=\dfrac{1+\tan{y}}{1-2x+\tan{y}}$

$\dfrac{dx}{dy}=\dfrac{1-2x+\tan{y}}{1+\tan{y}}$

$\dfrac{dx}{dy}=\dfrac{1+\tan{y}}{1+\tan{y}}-\dfrac{2x}{1+\tan{y}}$

I.F

$=\sin{y}+\cos{y}{e}^{y}$

$x\times$ I.F

$=\int{Q}\times$ I.F

$dy+c$

$x\sin{y}+\cos{y}{e}^{y}=\int{1\times \sin{y}+\cos{y}{e}^{y}dy}+c$

$x{e}^{y}\left(\sin{y}+\cos{y}\right)=\int{{e}^{y}\sin{y}dy}+\int{{e}^{y}\cos{y}dy}+c$

Consider

$\int{{e}^{y}\sin{y}dy}$ is of the form

$\int{{e}^{ax}\sin{bx}dy}=\dfrac{{e}^{ax}}{{a}^{2}+{b}^{2}}\left(a\sin{bx}-b\cos{bx}\right)$

Here

$a=1$ and

$b=1$

Hence

$\int{{e}^{y}\cos{y}dy}=\dfrac{{e}^{y}}{2}\left(\cos{y}+\sin{y}\right)$

Now combining and substituting in the required solution we get

$x{e}^{y}\left(\sin{y}+\cos{y}\right)=\dfrac{{e}^{y}}{2}\left(\sin{y}-\cos{y}\right)+ \dfrac{{e}^{y}}{2}\left(\sin{y}+\cos{y}\right)+c$

$x{e}^{y}\left(\sin{y}+\cos{y}\right)={e}^{y}\sin{y}+c$

Dividing throughout by

${e}^{y}$ we get

$x\left(\sin{y}+\cos{y}\right)=\sin{y}+c{e}^{-y}$

Solve the differential equation : $(y-x\frac{dy}{dx}-a(y^{2}+\frac{dy}{dx})$ ) =0

$\begin{matrix} y-x\frac { { dy } }{ { dx } } -a\left( { { y^{ 2 } }-\frac { { dy } }{ { dx } } } \right) =0 \\ y-a{ y^{ 2 } }=\left( { x+a } \right) \frac { { dy } }{ { dx } } \\ \Rightarrow \frac { { dx } }{ { x+a } } =\frac { { dy } }{ { \left( { y-a{ y^{ 2 } } } \right) } } \\ \Rightarrow { \log _{ e } }\left( { x+a } \right) =2{ \log _{ e } }\left| { \frac { { 2ay } }{ { \sqrt { 4ay-4a{ y^{ 2 } } } } } } \right| \\ \Rightarrow x+a=\frac { { { a^{ 2 } }{ y^{ 2 } } } }{ { ay-{ a^{ 2 } }{ y^{ 2 } } } } \\ \Rightarrow x+a=\frac { { ay } }{ { 1-ay } } \, \, \, \, \, \, \, \, Ans. \\ \end{matrix}$

Solve that
$x \left( 1 + y ^ { 2 } \right) d x + y \left( 1 + x ^ { 2 } \right) d y = 0$

$x(1+y^2)dx+y(1+x^2)dy=0$

$xdx(1+y^2)=-ydy(1+x^2)$

$\dfrac{x}{1+x^2}dx=\dfrac{-y}{1+y^2}dy$

$\dfrac{2x}{1+x^2}dx=-\dfrac{2y}{1+y^2}dy$

Integrating,

$\displaystyle\int \dfrac{2x}{1+x^2}dx=-\displaystyle\int \dfrac{2y}{1+y^2}dy$

$log(1+x^2)=-log(1+y^2)+log c$

$log[(1+x^2)(1+y^2)]=log c$

$(1+x^2)(1+y^2)=c$ .

For each of the differential equations in Exercises from $11$ to $15$ , find the particular solutions satisfying the given condition:
$(x+y)dy+(x-y)dx=0; y=1$ when $x=1$

$(x+y)dy+(x-y)dx=0$

$\dfrac{dy}{dx}=\dfrac{-(x-y)}{x+y}$

Let

$f(x,y)=\dfrac{dy}{dx}=\dfrac{-(x-y)}{x+y}$

$f(\lambda x, \lambda y)= \dfrac{-(\lambda x- \lambda y)}{\lambda x+ \lambda y}- \dfrac{-(x-y)}{x+y}$

$=\lambda^{o} f(x, y)$

$\therefore f(x,y)$ is a homogeneous function

Put

$y=vx$

Diff w.r.t.

$'x'$

$\dfrac{dy}{dx}=x.\dfrac{dv}{dx} +v$

$\dfrac{dy}{dx}= -(x-y)/x+y$

$v+\dfrac{xdv}{dx}= \dfrac{-(x-vx)}{x+vx}$

$v+\dfrac{xdv}{dx}=\dfrac{v-1}{1+v}$

$\dfrac{xdv}{dx}=\dfrac{v-1}{1+v}-v=\dfrac{v-1-v-v^{2}}{1+v}$

$\dfrac{xdv}{dx}=\dfrac{-(1+v^{2})}{1+v}$

$\dfrac{1+v^{2}}{1+v^{2}}dv=\dfrac{-dx}{x}$

$\int \dfrac{1+v}{1+v^{2}}dv=-\int \dfrac{dx}{x}$

$\int \dfrac{1}{v^{2}+1}.dx+ \int\dfrac{v}{1+v^{2}}=-\log x+c$ .

$\int \dfrac{v}{v^{2}+1} dv + \tan^{-1} v =-\log x+c$ .

$v^{2}+1=t.$

$2vdv=dt$

$vdv=dt/2$

$\tan^{-1} v+ \dfrac{1}{2} \log t=-\log x+c.$

$\tan^{-1}v+ \dfrac{1}{2} \log (v^{2}+1)=-\log x+c.$

$\tan^{-1} \dfrac{y}{x}+\dfrac{1}{2} \log \left( \dfrac{y^{2}}{x^{2}}+1 \right)=-\log x+c.$

$\tan^{-1} \dfrac{y}{x}+ \log \sqrt{\dfrac{x^{2}+y^{2}}{x^{2}}}+ \log x=c$

$\tan^{-1} \dfrac{y}{x}+ \log \dfrac{\sqrt{x^{2}+y^{2}}}{x}+ \log x=c.$

$\tan^{-1} \dfrac{y}{x}+ \log \sqrt{x^{2}+y^{2}}=c$

Put

$x=1$ &

$y=1$

$\tan^{-1} \left( \dfrac{1}{1} \right)+\log \sqrt{1+1}=c$ .

$\tan^{-1} 1+ \log \sqrt{2} =c$ .

$\dfrac{\pi}{4}+ \dfrac{1}{2} \log 2=c$ .

Put value of

$C$ in

$(1)$

$\tan^{-1} \dfrac{y}{x}+ \log \sqrt{x^{2}+y^{2}}=\dfrac{\pi}{2} \log 2$

$\tan^{-1}\dfrac{y}{x}+\dfrac{1}{2}\log(x^2+y^2)=\dfrac{\pi}{4}+\dfrac{1}{2}\log 2$

Multiplying by

$2$ on

$B.S.$

$\log (x^{2}+y^{2})+2 \tan^{-1} \dfrac{y}{x}=\dfrac{\pi}{2}+ \log 2.$

Find the general solution of the differential equation $x\dfrac{dy}{dx}+y-x+xy$ $\cot$ $x=0,$ $(x\neq 0)$

$x\dfrac{dy}{dx}+y-x+xy\, \cot\, x=0$

Divide by x

$\dfrac{dy}{dx}+\dfrac{y}{x}-1+y\, \cot\, x=0$

i.e; in the form

$\dfrac{dy}{dx}+Py=Q$

$\dfrac{dy}{dx}+y(\dfrac{1}{x}+\cot\, x)-1=0$

$\dfrac{dy}{dx}+(\dfrac{1}{x}+\cot\,x)y=1$ ___ (1)

From (1)

$P=\dfrac{1}{x}+\cot\,x$

$Q=1$

$I.F= e^{\int p}dx$

$=e^{\int (\dfrac{1}{x}+\cot\,x)dx}$

$= e.e^{\int \dfrac{1}{x}.dx+\int \cot x\, dx}$ .

$=e^{\log\,x+\log\, \sin\, x}$

$=e^{\log,x}sin\, x$

$[\because log\, a+ log\, b= log\, ab]$

$I.F= x. \sin\, x$

$Y\times I.F=\displaystyle\int Q\times IF\, dx+C$

$Y(x\, \sin\, x)= \displaystyle \int x\, \sin\, x. dx$ ___ (2)

$I= \int x. \sin\,x.dx$

$I= x \displaystyle\int \sin x\, dx\, - \int [1.\int \sin\,x \, dx]dx$

$=x(-\cos x)- \displaystyle\int 1.(- \cos\,x)dx$

$=-x\, \cos\, x+ \displaystyle \int \cos\,x \, dx$

$=-x\, \cos\, x+ \sin x$

Put value of 1 in (2)

$Y.x.sin\,x=-x\,\cos x+\sin x+C$

$Y=\dfrac{-x\, \cos\, x+\sin x+C}{x\, \sin\, x}$

$Y= \dfrac{-x\, \cos \, x}{x\, \sin \,x}+\dfrac{\sin\,x}{x\,\sin\,x}+\dfrac{C}{x\, \sin\, x}$

$Y=-\cot\, x+\dfrac{1}{x}+\dfrac{C}{x\, \sin\,x}$

Solve the differential equation $y-x\dfrac{dy}{dx}=a\{ { y }^{ 2 }+\dfrac { dy }{ dx } \}$

2039389_1278273_ans_7b18b09e60cc40cca28522c4d321513a.jpg

Solve the following differential equation :
$(\cot^{-1}y+x)dy=(1+y^{2})dx$

For solve the solving the given equation

$\left( {1 + {y^2}} \right)\dfrac{{dy}}{{dx}} = {\cot ^{ - 1}}y + x$

$\Rightarrow \dfrac{{dy}}{{dx}} + x\left( {\dfrac{{ - 1}}{{1 + {y^2}}}} \right) = \dfrac{{{{\cot }^{ - 1}}y}}{{1 + {y^2}}}$

$\Rightarrow x{e^{\int { - \frac{{dy}}{{1 + {y^2}}}} }} = \int {{e^{\int { - \frac{{dy}}{{1 + {y^2}}}} }}} \dfrac{{{{\cot }^{ - 1}}y}}{{1 + {y^2}}}dy$

$\Rightarrow x{e^{ - {{\tan }^{ - 1}}y}} = {e^{\frac{{ - \pi }}{2}}}\int {{e^{\frac{\pi }{2}}}} {e^{ - {{\tan }^{ - 1}}y}}\dfrac{{{{\cot }^{ - 1}}y}}{{1 + {y^2}}}dy$

$\Rightarrow x{e^{ - {{\tan }^{ - 1}}y}} = {e^{\frac{{ - \pi }}{2}}}\int {{e^{{{\cot }^{ - 1}}y}}} \dfrac{{{{\cot }^{ - 1}}y}}{{1 + {y^2}}}dy$

$\Rightarrow x{e^{ - {{\tan }^{ - 1}}y}} = - {e^{\frac{{ - \pi }}{2}}}\int {{e^t}tdt}$

$\Rightarrow x{e^{ - {{\tan }^{ - 1}}y}} = - {e^{\frac{{ - \pi }}{2}}}\left[ {t{e^t} - {e^t}} \right]$

$\Rightarrow x{e^{ - {{\tan }^{ - 1}}y}} = {e^{\frac{{ - \pi }}{2}}}\left[ {{{\cot }^{ - 1}}y - 1} \right]{e^{{{\cot }^{ - 1}}y}}$

$\Rightarrow x{e^{ - {{\tan }^{ - 1}}y}} = {e^{\frac{{ - \pi }}{2}}} + C$

$x = 1 + C{e^{{{\tan }^{ - 1}}y}}$

Hence, which is the required answer.

The general solution of the differential equation $\dfrac{dy}{dx}+sin\dfrac{x+y}{2}= sin\dfrac{x-y}{2}$ is

2039514_1294956_ans_cca2edc9507f402b8beeb3706ee8b7cd.JPG

Find the general solution of the $\left( { y }^{ 2 }-{ x }^{ 2 } \right) dy=3xy\ dx$ .
If the general solution is $y^2(\alpha x^2-y^2)^3=c$ then find the value of $\alpha$ .

2033647_1306301_ans_bb7d0752489b4a53a6517bad5a11e801.jfif

the solution of the differential equation ${ sec }^{ 2 }xtan\quad y\quad dx+{ sec }^{ 2 }ytan\quad x\quad dy=0\quad is$

2040840_1355217_ans_9a06a2e0306148ae80791161a20917fd.jpg

Solve the differential equation $\dfrac{dy}{dx} = x^5tan^{-1}(x^3)$ .

$\begin{array}{l} \frac { { dy } }{ { dx } } ={ x^{ 5 } }{ \tan ^{ -1 } }{ x^{ 3 } } \\ dy=\left( { { x^{ 5 } }{ { \tan }^{ -1 } }{ x^{ 3 } } } \right) dx \\ Taking\, { { int } }egration\, on\, both\, sides \\ \int { dy } =\int { \left( { { x^{ 5 } }{ { \tan }^{ -1 } }{ x^{ 3 } } } \right) dx } \\ Let\, { x^{ 3 } }=t \\ 3{ x^{ 2 } }dx=dt \\ y=\frac { 1 }{ 3 } \int { t\, { { \tan }^{ -1 } }tdt } +C \\ =\frac { 1 }{ 3 } { \tan ^{ -1 } }t\int { tdt } -\frac { 1 }{ 3 } \int { \left[ { \frac { { d\left( { { { \tan }^{ -1 } }t } \right) } }{ { dx } } \int tdt } \right] } dt+C \\ =\frac { { { t^{ 2 } } } }{ 6 } { \tan ^{ -1 } }t-\frac { 1 }{ 6 } \int { \left[ { \frac { { { t^{ 2 } }+1-1 } }{ { 1+{ t^{ 2 } } } } } \right] } dt+C \\ =\frac { { { t^{ 2 } } } }{ 2 } { \tan ^{ -1 } }t-\frac { 1 }{ 6 } \int { \left[ { 1-\frac { 1 }{ { 1+{ t^{ 2 } } } } } \right] } dt+C \\ =\frac { { { t^{ 2 } } } }{ 2 } { \tan ^{ -1 } }t-\frac { 1 }{ 6 } \left( { t-{ { \tan }^{ -1 } }t } \right) +C \\ y=\frac { { { x^{ 6 } } } }{ 6 } { \tan ^{ -1 } }{ x^{ 3 } }-\frac { { { x^{ 3 } } } }{ 6 } +\frac { 1 }{ 6 } { \tan ^{ -1 } }\left( { { x^{ 3 } } } \right) +C. \end{array}$

Find the general solution of the differential equation $\dfrac{dy}{dx} +y.cot x = 2x+x^cot x.$

$\dfrac{dy}{dx}+y\cot{x}=2x+{x}^{2}\cot{x}$

is of the form

$\dfrac{dy}{dx}+Py=Q$ where

$P=\cot{x}$ and

$Q=2x+{x}^{2}\cot{x}$

I.F

$={e}^{\int{p.dx}}={e}^{\int{\cot{x}dx}}={e}^{\log{\sin{x}}}=\sin{x}$ since

${e}^{\log{x}}=x$

Solution is

$y\times I.F=\int{Q\times I.F dx}+c$

$\Rightarrow y\sin{x}=\int{\left(2x\sin{x}+{x}^{2}\cot{x}\sin{x}\right)dx}$

$\Rightarrow y\sin{x}=2\int{x\sin{x}dx}+\int{{x}^{2}\cos{x}dx}$

Consider

$\int{{x}^{2}\cos{x}dx}$

Take

$u={x}^{2}\Rightarrow du=2xdx$ and

$dv=\cos{x}dx\Rightarrow v=\sin{x}$

$\int{{x}^{2}\cos{x}dx}={x}^{2}\sin{x}-2\int{x\sin{x}dx}$

$\therefore y\sin{x}=2\int{x\sin{x}dx}+{x}^{2}\sin{x}-2\int{x\sin{x}dx}$

$\therefore y\sin{x}={x}^{2}\sin{x}+c$ where

$c$ is the constant of integration.

Verify that $x^{2} = 2y^{2} \log {y}$ is a solution of the differential equation $\left (x^{2} + y^{2} \right ) \dfrac {dy}{dx} - xy = 0$ .

1545280_1441334_ans_05d6b0ae0f264f01839d50ae6f0c3bd4.jpg

The general solution of differential equation $\frac{d y}{d x}=\frac{x+y}{x-y}$ is

2040883_1453171_ans_d7d12f4dd52b4567ad4b2147ebaa5d7a.jpg

Verify that $y = e^{x} \cos bx$ is a solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} - 2 \dfrac {dy}{dx} + 2y = 0$ .

1545899_1447689_ans_bb3b703522cf48429e96b41d53342cb8.jpg

If $y=x\sin y$ , show that $\dfrac{dy}{dx}=\dfrac{y}{x(1-x\cos y)}$ .

1314622_1579130_ans_203ccee238f1448dbc2fd69f8568f604.jpg

Find the general solution of the differential equation $\dfrac{dy}{dx}-y=\cos x$ .

$\dfrac{dy}{dx}-y=\cos x$

or,

$\dfrac{dy}{dx}=\cos x+y$

or,

$dy=\cos xdx+ydx$

Integrating both sides,

$\displaystyle\int dy=\displaystyle\int \cos xdx+y\displaystyle\int dx+c$

where

$c=$ constant of integration.

or,

$y=\sin x+y x+c$

or,

$y(1-x)=\sin x+c$ .

Verify that $y = \dfrac {a}{x} + b$ is a solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} + \dfrac {2}{x}\left (\dfrac {dy}{dx}\right ) = 0$ .

1549946_1706830_ans_f0082a8329264e26b6835f8b7f794458.jpg

Find the general solution of the following differential equations:
$e^{2x - 3y}dx + e^{2y - 3x}dy = 0$ .

1551845_1707313_ans_12e9677771374399badc078754c7b98a.jpg

Find the general solution of each of the following differential equations:
$x^{2} (y + 1) dx + y^{2} (x - 1) dy = 0$ .

1649620_1707370_ans_fdd419e26afe48c1a1c22b8a25aa9429.jpeg

Find the general solution of the following differential equations:
$\dfrac {dy}{dx} = \dfrac {x - 1}{y + 2}$

1551778_1707350_ans_c03e1a8dda0044dab6bcfc7bdc2d2454.jpg

Verify that $y = (a + bx)e^{2x}$ is the general solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} - 4\dfrac {dy}{dx} + 4y = 0$ .

1549933_1706803_ans_a201f7586792463f8466f61d3ab8694d.jpg

Find the general solution of each of the following differential equations:
$y(1 - x)^{2} \dfrac {dy}{dx} = x(1 + y^{2})$ .

1551995_1707371_ans_9fcf7dae5ced4798ae84351422a03807.jpg

Verify whether that $y = \log (x + \sqrt {x^{2} + a^{2}})$ satisfies the differential equation $\dfrac {d^{2}y}{dx^{2}} + x\dfrac {dy}{dx} = 0$ .

Given,

$y = \log (x + \sqrt {x^{2} + a^{2}})$

Differentiate with respect to

$x$

$\Rightarrow \dfrac {dy}{dx} = \dfrac {1}{(x + \sqrt {x^{2} + a^{2}})} \cdot \left \{1 + \dfrac {2x}{2\sqrt {x^{2} + a^{2}}} \right \}$

$\Rightarrow y_{1} = \dfrac {1}{\sqrt {x^{2} + a^{2}}}$

$\Rightarrow y_{1}^{2} (x^{2} + a^{2}) =1$

Again, differentiate with respect to

$x$

$\Rightarrow y_{1}^{2} (2x) + (x^{2} + a^{2}) 2y_{1}y_{2} = 0$

$\Rightarrow (x^{2} + a^{2}) y_{2} + xy_{1} = 0$ .

$\Rightarrow (x^{2} + a^{2}) \dfrac {d^{2}y}{dx^{2}} + x\dfrac {dy}{dx} = 0$ .

$\implies y = \log (x + \sqrt {x^{2} + a^{2}})$ does not satisfies the differential equation

$\dfrac {d^{2}y}{dx^{2}} + x\dfrac {dy}{dx} = 0$ .

Verify that $y = e^{-3x}$ is a solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} + \dfrac {dy}{dx} - 6y = 0$ .

1549639_1706866_ans_8d832b0a099247d5bdf385c4898e5745.jpg

Solve: $(x+y)(dx-dy)=dxdy$ .[Hint: Substitute $x+y=z$ after separating $dx$ and $dy$ ]

Given, differential equation is

$(x+y)(dx-dy)=dx+dy$

$\Rightarrow (x+y) \left(1-\dfrac{dy}{dx}\right)=1+\dfrac{dy}{dx} ...(i)$

Put

$x+y=z$

$\Rightarrow 1+\dfrac{dy}{dx}= \dfrac{dz}{dx}$

On substituting these values in Eq.

$(i)$ , we get

$z \left(1-\dfrac{dz}{dx}+1\right)=\dfrac{dz}{dx}$

$\Rightarrow z \left(2-\dfrac{dz}{dx}\right)=\dfrac{dz}{dx}$

$\Rightarrow 2z-z \dfrac{dz}{dx}-\dfrac{dz}{dx}=0$

$\Rightarrow 2z-(z+1)\dfrac{dz}{dx}=0$

$\Rightarrow \dfrac{dz}{dx}=\dfrac{2z}{z+1}$

$\Rightarrow \left(\dfrac{z+1}{z}\right)dz=2dx$

On integrating both sides, we get

$\displaystyle \int {\left(1+\dfrac{1}{z}\right)dz} = 2 \int dx$

$\Rightarrow z+ \log z=2x- \log C$

$\Rightarrow (x+y) +\log (x+y)=2x- \log C [\because z=x+y]$

$\Rightarrow 2x-x-y=\log C+ \log (x+y)$

$\Rightarrow x-y=\log |C (x+y)|$

$\Rightarrow e^{x-y}=C(x+y)$

$\Rightarrow (x+y)=\dfrac{1}{C}e^{x-y}$

$\Rightarrow x+y=Ke^{x-y}[\because K=\dfrac{1}{C}]$

Find the general solution of each of the following differential equations:
$(1 - x^{2}) (1 - y) dx = xy (1 + y)dy$ .

1551989_1707405_ans_27d5b547cdae48388f45ce0d29d7f9a0.jpg

Find the general solution of the following differential equations:
$\dfrac {dy}{dx} + \dfrac {xy + y}{xy + x} = 0$ .

1552126_1707423_ans_ce20d159937045bdaeaa6c0df3ae816e.jpg

Find the general solution of the following differential equations:
$(y + xy) dx + (x - xy^{2}) dy = 0$ .

1551987_1707406_ans_2b433b9882394e319ee6cc83b010e414.jpg

Find the general solution of each of the following differential equations:
$x\sqrt {1 + y^{2}} dx + y \sqrt {1 + x^{2}} dy = 0$ .

1552125_1707410_ans_eb777275ffd443e791e8b24d8707a3a9.jpg

Find the integrating factor of differential equation
$\dfrac{dy}{dx} + \dfrac{1}{\sin x} y = e^x$

1960852_1875480_ans_a4c28e1e8aa047b7bc41de39b8721a9e.jpg

Find the integrating factor of differential equation $\dfrac{dy}{dx} + y \tan x = \sin x$

1960851_1874849_ans_c8eab2aab70d4c178610b87ad344f0e7.jpg

Differential Equations - Class 12 Commerce Maths - Extra Questions

Find the order of the differential equations of all circles in the plane XOY which have their centres on x-axis and have give radius.

Order of differential equation of y=mx+cy=mx+c

Order of differential equation of y=mxy=mx

Solve the differential equation: (1+x^2)dy=(1+y^2)dx(1+x^2)dy=(1+y^2)dx.

Solve : \dfrac{dy}{dx} = 1 + x + y + xy\dfrac{dy}{dx} = 1 + x + y + xy

The integrating factor of the differential equation \dfrac { dy }{ dx } -y\tan { x } =\cos { x } \dfrac { dy }{ dx } -y\tan { x } =\cos { x } is :

Find the particular solution of the differential equation:y(1 + \log x) \dfrac {dx}{dy} - x\log x = 0y(1 + \log x) \dfrac {dx}{dy} - x\log x = 0when y = e^{2}y = e^{2} and x = ex = e.

Distinguish which one is initial valued ordinary differential equation and boundary valued ordinary differential equation:i) y''+2y=e^x, y(\pi)=1; y'(\pi)= 2y''+2y=e^x, y(\pi)=1; y'(\pi)= 2i) y''+2y=e^x, y(0)=1; y'(1)= 1y''+2y=e^x, y(0)=1; y'(1)= 1

Solve the following differential equation:\dfrac{dy}{dx} = e^{x-y}+ x^2 e^{-y}\dfrac{dy}{dx} = e^{x-y}+ x^2 e^{-y}

A particle is moving along the X-axis. Its acceleration at time tt is proportional to its velocity at that time. Find the differential equation of the motion of the particle.

Solve:\cos y\dfrac{dy}{dx} = \sin x\cos y\dfrac{dy}{dx} = \sin x.

Solve: (x+y)(dx-dy)=dx+dy(x+y)(dx-dy)=dx+dy

Solve the following differential equation:\log \left(\dfrac{dy}{dx}\right)= ax + by\log \left(\dfrac{dy}{dx}\right)= ax + by

Verify y=cx+\dfrac{1}{x}y=cx+\dfrac{1}{x} is a solution of the differential equations y\dfrac{dy}{dx}=x\left(\dfrac{dy}{dx}\right)^2+1y\dfrac{dy}{dx}=x\left(\dfrac{dy}{dx}\right)^2+1, where c is arbitrary constant.

Solve:-\sin x\frac{{dy}}{{dx}} + 3y = \cos x\sin x\frac{{dy}}{{dx}} + 3y = \cos x

Find the value of dy/dxdy/dx,{y}^{2}=4ax{y}^{2}=4ax

Solve the following Differential equation : \frac{dy}{dx}=(y-4x^2)^2\frac{dy}{dx}=(y-4x^2)^2

find the solution of (e^{x}+e^{-x})\dfrac {dy}{dx}=e^{x}-e^{-x}(e^{x}+e^{-x})\dfrac {dy}{dx}=e^{x}-e^{-x}

Prove that y = 2e^x + 3e^2xy = 2e^x + 3e^2x is not g.s. of differential equation y^2 - 3y + 2y = 0y^2 - 3y + 2y = 0

If e^x+e^y=e^{x+y}e^x+e^y=e^{x+y}, prove that \dfrac{dy}{dx}+e^{y-x}=0\dfrac{dy}{dx}+e^{y-x}=0.

The soultion of 3e^x \cos^2 ydx + (1- e^x)\cot ydy = 03e^x \cos^2 ydx + (1- e^x)\cot ydy = 0 is

\cos{x}\dfrac{dy}{dx}-\cos{2x}=\cos{3x}\cos{x}\dfrac{dy}{dx}-\cos{2x}=\cos{3x}

Find the particular sol of D.E. \dfrac {dy}{dx} = y\tan x;\ y = 1\ x = 0D.E. \dfrac {dy}{dx} = y\tan x;\ y = 1\ x = 0.

Solve \dfrac{dy}{dx}=x+y\dfrac{dy}{dx}=x+y.

(1+y^{2}y^{2} ) dx=2xy dyBy simplifying get the below expression:\int \dfrac{2y}{1+y^{2}}dy=\int \dfrac{1}{x}dx\int \dfrac{2y}{1+y^{2}}dy=\int \dfrac{1}{x}dx

Solve the following differentiate:x (x - y)dy + y^{2}dx = 0x (x - y)dy + y^{2}dx = 0.

Solve: \dfrac { d y } { d x } = 1 +\dfrac {1}{ y ^ { 2 }} \dfrac { d y } { d x } = 1 +\dfrac {1}{ y ^ { 2 }}

F(x, y) = x^3 + y^3 + 3x^2y + 3 xy^2F(x, y) = x^3 + y^3 + 3x^2y + 3 xy^2. Is this homogeneous function.

y-x\dfrac{dy}{dx}=0y-x\dfrac{dy}{dx}=0

\dfrac{dy}{dx} + y = x^{-2}\dfrac{dy}{dx} + y = x^{-2}

Verify that y=2\sin{3x}y=2\sin{3x} is the solution of the D.E \cfrac{{d}^{2}y}{d{x}^{2}}+9y=0\cfrac{{d}^{2}y}{d{x}^{2}}+9y=0

Solve\cos x \cdot \cos y\ dy-\sin x \cdot \sin y\ dx=0\cos x \cdot \cos y\ dy-\sin x \cdot \sin y\ dx=0

Find the Particular solution of the differential equations{ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0;y=1{ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0;y=1 when ,x=1x=1

Define particular solution of a differential equation

Solve the following differential equations.y - x \dfrac{dy}{dx} = 0y - x \dfrac{dy}{dx} = 0

Find \dfrac {dy}{dx}\dfrac {dy}{dx} \left( x ^ { 3 } - 2 y ^ { 3 } \right) d x + 3 x y ^ { 2 } d y = 0\left( x ^ { 3 } - 2 y ^ { 3 } \right) d x + 3 x y ^ { 2 } d y = 0

Consider the differential equation,\dfrac{dy}{dx}+\sin y=\cos x\dfrac{dy}{dx}+\sin y=\cos x.Find the order of the differential equation.

Find solution of e ^ { x } \tan y dx+ \left( 1 - e ^ { x } \right) \sec ^ { 2 } y d y = 0e ^ { x } \tan y dx+ \left( 1 - e ^ { x } \right) \sec ^ { 2 } y d y = 0.

Find general solution of \dfrac {dy}{dx} = 1 + x + y + xy\dfrac {dy}{dx} = 1 + x + y + xy

Solve the differential equation by variable separable method: \left( { e }^{ y }+1 \right) \cos { x } dx+{ e }^{ y }\sin { x } dy=0. \left( { e }^{ y }+1 \right) \cos { x } dx+{ e }^{ y }\sin { x } dy=0.

Solve:\dfrac{dy}{dx}=1+y^2\dfrac{dy}{dx}=1+y^2.

Find an expression for y given \dfrac{dy}{dx}=7x^{5}\dfrac{dy}{dx}=7x^{5}

Solve :\dfrac{dy}{dx}=(1+x^2)(1+y^2)\dfrac{dy}{dx}=(1+x^2)(1+y^2)

If ( 2 + \sin x ) \dfrac { d y } { d x } + ( y + 1 ) \cos x =0( 2 + \sin x ) \dfrac { d y } { d x } + ( y + 1 ) \cos x =0 and y ( 0 ) = 1,y ( 0 ) = 1, then y \left( \dfrac { \pi } { 2 } \right)y \left( \dfrac { \pi } { 2 } \right) is equal to :

Find the general solution of the following differential equation.\dfrac{{dy}}{{dx}} = 1 + x + y + xy\dfrac{{dy}}{{dx}} = 1 + x + y + xy

Solve\dfrac { dy }{ dx } ={ x }^{ 2 }y+y\dfrac { dy }{ dx } ={ x }^{ 2 }y+y

Solve :(x+y)^2\dfrac{dy}{dx}=a^2(x+y)^2\dfrac{dy}{dx}=a^2

Find the particular solution of the differential equation \left( 1 + x ^ { 2 } \right) \dfrac { d y } { d x } + 2 x y = \dfrac { 1 } { 1 + x ^ { 2 } }\left( 1 + x ^ { 2 } \right) \dfrac { d y } { d x } + 2 x y = \dfrac { 1 } { 1 + x ^ { 2 } } given that at x = 1 , y = 0x = 1 , y = 0

Solve :\sin^{-1}\left (\dfrac{dy}{dx}\right )=x+y\sin^{-1}\left (\dfrac{dy}{dx}\right )=x+y

Find general solution of D.E\dfrac{{dy}}{{dx}} = {x^2}y + y\dfrac{{dy}}{{dx}} = {x^2}y + y

Solve the differential equation:\dfrac{d^2y}{dx^2}-y=0\dfrac{d^2y}{dx^2}-y=0.

Solve :\dfrac{dy}{dx}=(1+x^2)(1+y^2)\dfrac{dy}{dx}=(1+x^2)(1+y^2)

Find the general solution of \dfrac{dy}{dx} = e^{x + y}\dfrac{dy}{dx} = e^{x + y}.

Define a ordinary differential equation.

Solve the differential equation :y \log y dx-xdy=0y \log y dx-xdy=0

Solve :\dfrac{dy}{dx}=e^{x+y}\dfrac{dy}{dx}=e^{x+y}

Solve the following differential equation.\dfrac{dy}{dx}=x-1\dfrac{dy}{dx}=x-1.

Solve:\dfrac{{dy}}{{dx}} = \sin x - x\dfrac{{dy}}{{dx}} = \sin x - x

Find the solution of \dfrac{dy}{dx}=2^{y-x}\dfrac{dy}{dx}=2^{y-x}.

Define a differential equation.

Solve \dfrac{dy}{dx}=e^{x+y}\dfrac{dy}{dx}=e^{x+y}

Solve the following differential equation:y^2dy-x^2dx=0y^2dy-x^2dx=0.

For the following differential equation, find the general solution.\dfrac{dy}{dx}=\sin x\dfrac{dy}{dx}=\sin x

Solve the following differential equation.\dfrac{dy}{dx}=e^x\dfrac{dy}{dx}=e^x

For the following differential equation, find the general solution.x dy=dx x dy=dx .

Solve the differential equation :\dfrac{dy}{dx}=3x\dfrac{dy}{dx}=3x

Solve the following differential equation.\dfrac{dy}{dx}+2x=e^{3x}\dfrac{dy}{dx}+2x=e^{3x}.

Solve the following differential equation.\dfrac{dy}{dx}+1=\sin x\dfrac{dy}{dx}+1=\sin x.

Show that y=e^x(A\cos x+B\sin x)y=e^x(A\cos x+B\sin x) is the solution of the differential equation \dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0.

Answer the following question in one word or one sentence or as per exact requirement of the question.How many arbitrary constants are there in the general solution of the differential equation of order 33.

Solve the following differential eqauton.y^2dy-x^2dx=0y^2dy-x^2dx=0.

Find the area of the triangle with vertices at the points :(-1,-8),(-2,-3) \ and \ (3,2)(-1,-8),(-2,-3) \ and \ (3,2)

Write the order of the differential equation associated with the primitive y=C_1+C_2e^x+C_3e^{-2x+C_4}y=C_1+C_2e^x+C_3e^{-2x+C_4}, where C_1, C_2, C_3, C_4C_1, C_2, C_3, C_4 are arbitrary constants.

Show that the function y=A\cos x+B\sin xy=A\cos x+B\sin x is a solution of the differential equation \dfrac{d^2y}{dx^2}+y=0\dfrac{d^2y}{dx^2}+y=0.

Solve the following systems of linear equations x+y= 5 x+y= 5 y+z = 3y+z = 3 x+z = 4 x+z = 4

Solve \dfrac{\ln \left ( \sec x+\tan x \right )}{\cos x}dx=\dfrac{\ln \left ( \sec y+\tan y \right )}{\cos y}dy\dfrac{\ln \left ( \sec x+\tan x \right )}{\cos x}dx=\dfrac{\ln \left ( \sec y+\tan y \right )}{\cos y}dy

Determine whether or not the following function is homogeneous: f(x, y) = \sqrt{x^{2} + 2xy + 3y^{2}}f(x, y) = \sqrt{x^{2} + 2xy + 3y^{2}}If homogeneous enter 1 else enter 0.

Find a particular solution of the differential equation (x-y)(dx+dy)=dx-dy(x-y)(dx+dy)=dx-dy, given that y=-1y=-1 when x=0x=0.

Find the equation of a curve passing through the point (0,-2) given that at any point (x, y) on the curve, the product of the slope of its tangent and yy coordinate of the point is equal to the xx coordinate of the point.

Order of differential equation of $y=mx+c$

Order of differential equation of $y=mx$

Solve the differential equation: $(1+x^2)dy=(1+y^2)dx$ .

Solve : $\dfrac{dy}{dx} = 1 + x + y + xy$

The integrating factor of the differential equation $\dfrac { dy }{ dx } -y\tan { x } =\cos { x }$ is :

Find the particular solution of the differential equation:
$y(1 + \log x) \dfrac {dx}{dy} - x\log x = 0$
when $y = e^{2}$ and $x = e$ .

Distinguish which one is initial valued ordinary differential equation and boundary valued ordinary differential equation:
i) $y''+2y=e^x, y(\pi)=1; y'(\pi)= 2$
i) $y''+2y=e^x, y(0)=1; y'(1)= 1$

Solve the following differential equation:
$\dfrac{dy}{dx} = e^{x-y}+ x^2 e^{-y}$

A particle is moving along the X-axis. Its acceleration at time $t$ is proportional to its velocity at that time. Find the differential equation of the motion of the particle.

Solve:
$\cos y\dfrac{dy}{dx} = \sin x$ .

Solve: $(x+y)(dx-dy)=dx+dy$

Solve the following differential equation:
$\log \left(\dfrac{dy}{dx}\right)= ax + by$

Verify $y=cx+\dfrac{1}{x}$ is a solution of the differential equations $y\dfrac{dy}{dx}=x\left(\dfrac{dy}{dx}\right)^2+1$ , where c is arbitrary constant.

Solve:- $\sin x\frac{{dy}}{{dx}} + 3y = \cos x$

Find the value of $dy/dx$ ,
${y}^{2}=4ax$

Solve the following Differential equation :
$\frac{dy}{dx}=(y-4x^2)^2$

find the solution of
$(e^{x}+e^{-x})\dfrac {dy}{dx}=e^{x}-e^{-x}$

Prove that
$y = 2e^x + 3e^2x$ is not g.s. of differential equation $y^2 - 3y + 2y = 0$

If $e^x+e^y=e^{x+y}$ , prove that $\dfrac{dy}{dx}+e^{y-x}=0$ .

The soultion of $3e^x \cos^2 ydx + (1- e^x)\cot ydy = 0$ is

$\cos{x}\dfrac{dy}{dx}-\cos{2x}=\cos{3x}$

Find the particular sol of $D.E. \dfrac {dy}{dx} = y\tan x;\ y = 1\ x = 0$ .

Solve $\dfrac{dy}{dx}=x+y$ .

(1+ $y^{2}$ ) dx=2xy dy
By simplifying get the below expression:
$\int \dfrac{2y}{1+y^{2}}dy=\int \dfrac{1}{x}dx$

Solve the following differentiate:
$x (x - y)dy + y^{2}dx = 0$ .

Solve: $\dfrac { d y } { d x } = 1 +\dfrac {1}{ y ^ { 2 }}$

$F(x, y) = x^3 + y^3 + 3x^2y + 3 xy^2$ . Is this homogeneous function.

$y-x\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx} + y = x^{-2}$

Verify that $y=2\sin{3x}$ is the solution of the D.E $\cfrac{{d}^{2}y}{d{x}^{2}}+9y=0$

Solve
$\cos x \cdot \cos y\ dy-\sin x \cdot \sin y\ dx=0$

Find the Particular solution of the differential equations
${ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0;y=1$ when , $x=1$

Solve the following differential equations.
$y - x \dfrac{dy}{dx} = 0$

Find $\dfrac {dy}{dx}$
$\left( x ^ { 3 } - 2 y ^ { 3 } \right) d x + 3 x y ^ { 2 } d y = 0$

Consider the differential equation,
$\dfrac{dy}{dx}+\sin y=\cos x$ .
Find the order of the differential equation.

Find solution of $e ^ { x } \tan y dx+ \left( 1 - e ^ { x } \right) \sec ^ { 2 } y d y = 0$ .

Find general solution of
$\dfrac {dy}{dx} = 1 + x + y + xy$

Solve the differential equation by variable separable method: $\left( { e }^{ y }+1 \right) \cos { x } dx+{ e }^{ y }\sin { x } dy=0.$

Solve:
$\dfrac{dy}{dx}=1+y^2$ .

Find an expression for y given
$\dfrac{dy}{dx}=7x^{5}$

Solve :
$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$

If $( 2 + \sin x ) \dfrac { d y } { d x } + ( y + 1 ) \cos x =0$ and $y ( 0 ) = 1,$ then $y \left( \dfrac { \pi } { 2 } \right)$ is equal to :

Find the general solution of the following differential equation.
$\dfrac{{dy}}{{dx}} = 1 + x + y + xy$

Solve
$\dfrac { dy }{ dx } ={ x }^{ 2 }y+y$

Solve :
$(x+y)^2\dfrac{dy}{dx}=a^2$

Find the particular solution of the differential equation $\left( 1 + x ^ { 2 } \right) \dfrac { d y } { d x } + 2 x y = \dfrac { 1 } { 1 + x ^ { 2 } }$ given that at $x = 1 , y = 0$

Solve :
$\sin^{-1}\left (\dfrac{dy}{dx}\right )=x+y$

Find general solution of D.E
$\dfrac{{dy}}{{dx}} = {x^2}y + y$

Solve the differential equation:
$\dfrac{d^2y}{dx^2}-y=0$ .

Solve :
$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$

Find the general solution of $\dfrac{dy}{dx} = e^{x + y}$ .

Solve the differential equation :
$y \log y dx-xdy=0$

Solve :
$\dfrac{dy}{dx}=e^{x+y}$

Solve the following differential equation.
$\dfrac{dy}{dx}=x-1$ .

Solve:
$\dfrac{{dy}}{{dx}} = \sin x - x$

Find the solution of $\dfrac{dy}{dx}=2^{y-x}$ .

Solve $\dfrac{dy}{dx}=e^{x+y}$

Solve the following differential equation:
$y^2dy-x^2dx=0$ .

For the following differential equation, find the general solution.
$\dfrac{dy}{dx}=\sin x$

Solve the following differential equation.
$\dfrac{dy}{dx}=e^x$

For the following differential equation, find the general solution.
$x dy=dx$ .

Solve the differential equation :
$\dfrac{dy}{dx}=3x$

Solve the following differential equation.
$\dfrac{dy}{dx}+2x=e^{3x}$ .

Solve the following differential equation.
$\dfrac{dy}{dx}+1=\sin x$ .

Show that $y=e^x(A\cos x+B\sin x)$ is the solution of the differential equation $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$ .

Answer the following question in one word or one sentence or as per exact requirement of the question.
How many arbitrary constants are there in the general solution of the differential equation of order $3$ .

Solve the following differential eqauton.
$y^2dy-x^2dx=0$ .

Find the area of the triangle with vertices at the points :
$(-1,-8),(-2,-3) \ and \ (3,2)$

Write the order of the differential equation associated with the primitive $y=C_1+C_2e^x+C_3e^{-2x+C_4}$ , where $C_1, C_2, C_3, C_4$ are arbitrary constants.

Show that the function $y=A\cos x+B\sin x$ is a solution of the differential equation $\dfrac{d^2y}{dx^2}+y=0$ .

Solve the following systems of linear equations
$x+y= 5$
$y+z = 3$
$x+z = 4$

Solve $\dfrac{\ln \left ( \sec x+\tan x \right )}{\cos x}dx=\dfrac{\ln \left ( \sec y+\tan y \right )}{\cos y}dy$

$f(x, y) = \sqrt{x^{2} + 2xy + 3y^{2}}$
If homogeneous enter 1 else enter 0.

Find a particular solution of the differential equation $(x-y)(dx+dy)=dx-dy$ , given that $y=-1$ when $x=0$ .

Find the equation of a curve passing through the point (0,-2) given that at any point (x, y) on the curve, the product of the slope of its tangent and $y$ coordinate of the point is equal to the $x$ coordinate of the point.

Find the particular solution of the differential equation
$(1+e^{2x})dy+(1+y^2)e^xdx=0$ , given that $y=1$ when $x=0$ .

The general solution of the differential equation $\dfrac{dy}{dx}=e^{x+y}$ is

Find the general solution of the differential equation $\dfrac {dy}{dx} + \sqrt {\dfrac {1 - y^{2}}{1 - x^{2}}} = 0$ .

Find a particular solution of the differential equation $(x+1)\cfrac{dy}{dx}=2\,e^{-y}-1$ , given that $y=0$ when $x=0$ .

Solve the differential equation:
$(\tan^{-1}y-x)dy = (1+y^2)dx$ .