Class 12 Engineering Maths - Differential Equations

Find the order and degree of the differential equations.
a) $y ^ { \prime \prime } + 3 y ^ { \prime ^ { 2 } } + y ^ { 3 } = 0$
b) $\left( \dfrac { d y } { d x } \right) ^ { 2 } + \dfrac { 1 } { d y d x } = 2$
c) $\dfrac { d ^ { 2 } y } { d x ^ { 2 } } + 4 y = 0$

Order is basically the highest order of differentiation used & Degree is the power raised on the highest order derivative term when expressions are written in the form of polynomial.

So (a) Order: 2 Degree: 1

(b) Order: 1 Degree: 2

$\cos \left( \frac { d y } { d x } \right) = a ( a \in \mathbf { R } ) ; y = 1$ when $x = 0$ .Find $y$ ?

Given

$\cos \bigg(\dfrac{d y}{d x}\bigg)=a$

$\implies \dfrac{d y}{d x}=\text{cos}^{-1} a$

$\implies d{y}=\text{cos}^{-1}a d x$

Integrating on both sides

$y=x\text{cos}^{-1} a+C$

Given

$y=1$ when

$x=0$

$\implies 1=0+C\implies C=1$

$y=x\text{cos}^{-1} a+1$

Show that $y=cx+\dfrac{a}{c}$ is a solution of the differential equation $y=x\dfrac{dy}{dx}+\dfrac{a}{\dfrac{dy}{dx}}$ .

$y=cx+\dfrac{a}{c}$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=c$

Therefore,

$x \dfrac{dy}{dx}+\dfrac{a}{\dfrac{dy}{dx}}=cx+\dfrac{a}{c}$

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

This shows that

$y=cx+\dfrac{a}{c}$ is a solution of the given differential equation.

Write the integrating factor of the differential equation:
$\cos x\dfrac{dy}{dx}+y=\sin x; 0\le x <\dfrac{\pi}{2}$ .

Suppose we have the first order differential equation

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

where

$P$ and

$Q$ are functions involving

$x$ only.

We multiply both sides of the differential equation by the integrating factor

$I$ which is defined as

${ e }^{ \int { Pdx } }$

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

$\Longrightarrow \dfrac { dy }{ dx } +y\sec x=\tan x$

Here

$P=\sec x$

Hence, the integrating factor

$I$ becomes

$\displaystyle{ e }^{ \int { \sec xdx } }={ e }^{\ln { |\sec x+\tan x| } }= { |\sec x+\tan x| }=\sec x+\tan x$ , because both

$secx,tanx$ are positive in first quadrant.

Find the integrating factor of $\cfrac { dy }{ dx } +\cfrac { y }{ x } ={ x }^{ 2 }$ .

Suppose we have the first order differential equation

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

where

$P$ and

$Q$ are functions involving x only.

We multiply both sides of the differential equation by the integrating factor

$I$ which is defined as

${ e }^{ \int { Pdx } }$

Here

$P=\frac { 1 }{ x }$

Hence, the integrating factor

$I$

$={ e }^{ \int { \frac { 1 }{ x } dx } }={ e }^{ \ln { x } }=x$

Find the differential coefficient of $\tan ^{ -1 }{ x }$ w.r to $x$ .

Let

$tan^{-1}x = y$

$\therefore\ x=tany$

Now, differentiate wrt

$x$

$\therefore\ 1=\dfrac{d(tany)}{dx}$ (differentiation of

$tant$ is

$sec^2t$ )

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

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

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

$\because sec^2t=1+tan^2t$ )

But,

$tany = x$

$\therefore\ \dfrac{d(tan^{-1}x)}{dx}=\dfrac{1}{1+x^2}$

Hence the answer is

$\dfrac{1}{1+x^2}$

Consider the following equation, $\dfrac {dy}{dx} + P(x)y = Q(x)$
(i) If two particular solutions of given equation $u(x)$ and $v(x)$ are known, find the general solution of the same equation in terms of $u(x)$ and $v(x)$ .
(ii) If $\alpha$ and $\beta$ are constants such that the linear combinations $\alpha \cdot u(x) + \beta \cdot v(x)$ is a solution of the given equation, find the relation between $\alpha$ and $\beta$ .
(iii) If $w(x)$ is the third particular solution different from $u(x)$ and $v(x)$ then find the ratio $\dfrac {v(x) - u(x)}{w(x) - u(x)}$ .

The following is a linear differential equation, whose solution is given by,

$I . y = \int I . Q(x) dx + c$ , where

$I = e^{\int P(x)dx}$ is the integrating factor and '

$c$ ' is the constant of integration.

Since

$u(x)$ is a solution of the equation, we can write:

$u'(x) + P(x)u(x) = Q(x)$ ...(1) ....(

$u'(x)$ is the first derivative of

$u(x)$ wrt

$x$ ).

Similarly, we can write:

$v'(x) + P(x)v(x) = Q(x)$ ...(2)

i) Using equation (1) and (2), we write

$P(x)$ and

$Q(x)$ in terms of

$u(x)$ and

$v(x)$ . Subtracting (2) from (1), we get

$P(x) = \dfrac{v'(x)-u'(x)}{u(x)-v(x)}$ .

Similarly,

$Q(x) = \dfrac{v'(x)u(x) - u'(x)v(x)}{u(x)-v(x)}$

Therefore,

$I = e^{\displaystyle 1234\int \dfrac{-(u'(x)-v'(x))}{u(x)-v(x)}dx} = e^{ln(u(x)-v(x))^{-1}} = \dfrac{1}{u(x)-v(x)}$

Hence, the general solution is,

$\big(u(x) - v(x)\big) . \displaystyle \int \dfrac{-\big( u'(x)v(x) - v'(x)u(x) \big)}{(u(x)-v(x))^2}dx + c \,. (u(x) - v(x))$

ii.) Given

$\alpha u(x) + \beta v(x)$ is also a solution.

Hence, we write,

$(\alpha u'(x) + \beta v'(x)) + P(x).(\alpha u(x) + \beta v(x)) = Q(x)$ .

Rearranging the terms, we get:

$\alpha(u'(x) + P(x)u(x)) + \beta(v'(x) + P(x)v(x)) = Q(x)$

Using equation (1) and (2), we get:

$\alpha Q(x) + \beta Q(x) = Q(x)$

We can say:

$\alpha + \beta = 1$

iii.) Let

$c_1, \, c_2$ and

$c_3$ be the constants of integration for

$u(x), v(x)$ and

$w(x)$ respectively.

$v(x) - u(x) =$

$\dfrac{c_2 - c_1}{I}$ , where

$I$ is the integration factor.

Similarly,

$w(x) - u(x) =$

$\dfrac{c_3 - c_1}{I}$ .

Dividing, both the equations, we get

$\dfrac{c_2 - c_1}{ c_3 - c_1} = k \, (\text{say})$ .

Hence, the ratio is a constant.

Find the general solution of the differential equation:
$xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx$

$\cfrac { y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x } =cx$

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

Verify that $y=4sin 3x$ is the solution of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} + 9y = 0$ .

$\begin{array}{l}y = 4\sin 3x\\\frac{{dy}}{{dx}} = 12\cos 3x\\\frac{{{d^2}y}}{{d{x^2}}} = - 36\sin 3x\\ = - 9x\left( {4\sin 3x} \right)\\\frac{{{d^2}y}}{{d{x^2}}} = - 9y\\\frac{{{d^2}y}}{{d{x^2}}} + 9y = 0\end{array}$

Find order and degree of the following differential equations .
(i) $\dfrac{dy}{dx} + y = \dfrac{1}{\dfrac{dy}{dx}}$
(ii) $_e\left[\dfrac{dy}{dx}-\dfrac{d^3y}{dx^3}\right] = ln \left[\dfrac{d^5y}{dx^5}+ 1\right]$
(iii) $\left[(\dfrac{dy}{dx})^{1/3} + y \right]^2 = \dfrac{d^2y}{dx^2}$

(i)

$\displaystyle \frac{dy}{dx}+y=\frac{1}{\left ( \frac{dy}{dx} \right )}$

$\displaystyle \left ( \frac{dy}{dx} \right )^{2}+y\left ( \frac{dy}{dx} \right )=1$

Order =

$1,$ Degree =

$2$

(ii)

$\displaystyle _{e}\left[ \frac{dy}{dx}-\frac{d^{3}y}{dx^{3}} \right ]=ln\left [ \frac{d^{5}y}{dx^{5}}+1 \right ]$

Order and Degree not defined.

(iii)

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

Order =

$2,$ Degree =

$1$

Find the particular solution of the difference equation
$(1-y^{2})(1+\log x)dx+2xydy=0$
given that $y=0$ when $x=1$

Consider the following equation.

$(1-{{y}^{2}})(1+logx)dx+2xydy=0$ and $y=0\,\,\,\,\And \,\,\,\,x=1$

$\dfrac{(1+logx)dx}{x}=\dfrac{ydy}{(1-{{y}^{2}})}$

On Integrate both sides, we get

$\int_{{}}^{{}}{\dfrac{(1+logx)dx}{x}}=\int_{{}}^{{}}{\dfrac{-2y}{(1-{{y}^{2}})}}dy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( 1 \right)$

Let $u=\log x\,\,\,\,\,\,\,\,\,\,\,\,\,\,v=\left( 1-{{y}^{2}} \right)$

$xdu=dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,dy=\dfrac{1}{-2y}dv$

$\int_{{}}^{{}}{\dfrac{(1+u)xdu}{x}}=\int_{i}^{{}}{\dfrac{-2y}{-2yv}}dv$

$\int_{{}}^{{}}{\left( 1+u \right)du=\int_{{}}^{{}}{\dfrac{1}{v}}}\,dv$

$u+\dfrac{{{u}^{2}}}{2}=\ln v+C$

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

${{\left( \log 0 \right)}^{2}}+2\log 0+{{\left( 1 \right)}^{2}}-1=C$

${{\left( \log x \right)}^{2}}+2\log x+{{y}^{2}}-1=0$

Hence this is the required answer.

Solve: $\cos^2x\dfrac{dy}{dx}+y=\tan x$ .

Given equation is

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

$\Rightarrow \dfrac{dy}{dx}+\sec^{2}xy=\tan x.sec^{2}x$

The genral solution of the equation,

$\dfrac{dy}{dx}+f(x).y=g(x)$

$y=(e^{-\int f(x).dx})(\int g(x)e^{\int f(x).dx}.dx+C)$

In the equation,

$f(x)= sec^{2}x, g(x)= \tan x - \sec^{2}x$

The general equation is,

$y=(e^{-\int \sec^{2}xdx})(\int \tan x. \sec^{2}x e^{\tan x}dx+C)$

$\int \tan x. \sec^{2}x e^{\tan x}dx= \tan x.e^{\tan x}-\int \int \sec^{2}x.e^{\tan x}$

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

$\therefore$ The solution is

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

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

Find the order of the differential equation obtained by eliminating the arbitrary constants $b$ and $c$ from $xy=ce^{x}+be^{-x}+x^{2}$ .

Since no. of arbitory constant

$=2$

So, order of differential equation formed

$=2$ .

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

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

$x+y=v\rightarrow \dfrac{dy}{dx}=\dfrac{dv}{dx}-1$

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

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

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

$\int \dfrac{dv}{1+\frac{1-\tan ^2\frac{v}{2}}{1+\tan ^2\frac{v}{2}}+\frac{2\tan \frac{v}{2}}{1+\tan ^2\frac{v}{2}}}=\int dx$

$\int \dfrac{\sec ^2\frac{v}{2}}{2\left ( 1+\tan \frac{v}{2} \right )}=\int dx$

$\ln \left ( 1+\tan \dfrac{v}{2} \right )=x+c$

$\ln \left ( 1+\tan \dfrac{x+y}{2} \right )=x+c$

If $y=(cos^{-1}x)^2$ , prove that $(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-2=0$ . Hence find $y_2$ when $x=0$ .

$y=(\cos ^{ -1 } )^{ 2 }\\ \cfrac { dy }{ dx } =\cfrac { d }{ dx } [\cos ^{ -1 } x]^{ 2 }\\ \cfrac { dy }{ dx } =2\cos ^{ -1 } x\cfrac { d }{ dx } (\cos ^{ -1 } x)\\ \cfrac { dy }{ dx } =2\cos ^{ -1 } x\cfrac { (-1) }{ \sqrt { ( } 1-x^{ 2 }) } \\ \cfrac { d^{ 2 } }{ dx^{ 2 } } =\cfrac { d }{ dx } (\cfrac { dy }{ dx } )=\cfrac { d }{ dx } { [2\cos ^{ -1 } x(-1)] }{ \sqrt { 1-x^{ 2 } } }\\ \cfrac { d^{ 2 }y }{ dx^{ 2 } } ={ -2\frac { d }{ dx } }(\cos ^{ -1 } x)\times \cfrac { 1 }{ \sqrt { 1-x^{ 2 } } } +2\cos ^{ -1 } x\cfrac { d }{ dx } [{ -1 }{ \sqrt { 1-x^{ 2 } } }]\\ \cfrac { d^{ 2 }y }{ dx^{ 2 } } =(-2)\cfrac { (-1) }{ \sqrt { 1-x^{ 2 } } } +2\cos ^{ -1 } x\cfrac { d }{ dx } [-\cfrac { 1 }{ \sqrt { 1-x^{ 2 } } } ]\\ \cfrac { d^{ 2 }y }{ dx^{ 2 } } =\cfrac { (-2)(-1) }{ \sqrt { 1-x^{ 2 } } } +x\cos ^{ -1 } x\cfrac { 1 }{ 2 } (1-x^{ 2 })^{ \frac{3}{2} }.(-2x)\\ (1-x^{ 2 })\cfrac { d^{ 2 }y }{ dx^{ 2 } } =(1-x^{ 2 })[\cfrac { 2 }{ (1-x^{ 2 }) } +\cfrac { (-2x)\cos ^{ -1 } x }{ (1-x^{ 2 })^{ \frac{3}{2} } } ]=2+\cfrac { \cos ^{ -1 } x(-2x) }{ \sqrt { 1-{ x }^{ 2 } } } +x\cfrac { dy }{ dx } =\cfrac { 2\times \cos ^{ -1 } x }{ \sqrt { 1-x^{ 2 } } } \\ (1-x^{ 2 })\cfrac { d^{ 2 } }{ dx^{ 2 } } -x\cfrac { dy }{ dx } -x\cfrac { dy }{ dx } -2\\ =\cfrac { 2+\cos ^{ -1 } x(-2x) }{ \sqrt { 1-x^{ 2 } } } -\cfrac { 2\times \cos ^{ -1 } (x) }{ \sqrt { 1-x^{ 2 } } } -2\\ =0$

Hence proved

$y_2(x=0)=(\cfrac{d^2y}{dx^2})_(x=0)=\cfrac{2}{(1-0^2)}+\cfrac{(-2\times0)\cos^{-1}(10)}{(1-0^2)^{\frac{3}{2}}}=2\\ \therefore y_2=0 \quad when\quad x=0$

If ${ x }^{ y }={ e }^{ x-y }$ , show that $\dfrac { dy }{ dx } =\dfrac{y\log{x}}{x\left(\log{x}+1\right)}$

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

$\Rightarrow y\log{x}=\left(x-y\right)$ by taking

$\log$ on both sides

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

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

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

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

Write the order of the differential equation $1+\left(\dfrac{dy}{dx}\right)^2=7\left(\dfrac{d^2y}{dx^2}\right)^3$ .

The order of the differential equation

$1+\left(\dfrac{dy}{dx}\right)^2=7\left(\dfrac{d^2y}{dx^2}\right)^3$ is

$2$

Show that $xy=ae^x+be^{-x}+x^2$ is a solution of the differential equation $x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}-xy+x^2-2=0$ .

$xy=ae^x+be^{-x}+x^2$

Differentiating w.r.t. x, we get,

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

Differentiating w.r.t. x, we get,

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

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

Therefore,

$x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}-xy+x^2-2=(ae^x+be^{-x}+2)-(ae^x+be^{-x}+x^2)+x^2-2=0$

Thus,

$xy=ae^x+be^{-x}+x^2$ is a solution of the given differential equation.

Verify that $y = A\cos 2x - B\sin 2x$ is the general solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} + 4y = 0$ .

1671642_1706810_ans_48b24400a9b34da680087473dc761d22.jpeg

Verify that $y = ce^{\tan^{-1}x}$ is a solution of the differential equation $(1 + x^{2}) \dfrac {d^{2}y}{dx^{2}} + (2x - 1) \dfrac {dy}{dx} = 0$ .

1671652_1706825_ans_149f286e210849dfaa5b5c39fac41ea4.jpeg

The number of arbitrary constants in the general solution of a differential equation of order three is ______ .

There are three arbitrary constant.

Solve the following differential equation:
$x^{2} \dfrac {dy}{dx} = y^{2} + 2xy$
Given that : $y = 1$ , when $x = 1$ .

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

Rearranging ,

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

Since the differential equation is of the form

$\dfrac{dy}{dx} = f(y/x)$

we can use separation of variables theory

Let

$y = kx$ substitute

$y = kx$ in the equation ...(i)

$\Rightarrow \dfrac{dy}{dx} = k^2 + 2k$ ...(2)

and, since

$y = k x \Rightarrow \dfrac{dy}{dx} = \left[k + x \left(\dfrac{dk}{dx} \right) \right]$ ....(3)

equating (2) & (3)

$k^2 + 2k = k + x \dfrac{dk}{dx} \Rightarrow x \dfrac{dk}{dx} = k^2 + k \Rightarrow \dfrac{dk}{k^2 + k} = \dfrac{dx}{x}$ ...(4)

$\Rightarrow \dfrac{dk}{k (k + 1)} = \left(\dfrac{A}{k} + \dfrac{B}{k + 1} \right) dk = \left(\dfrac{A (k + 2) + B(k)}{k (kH)} \right) dk = \left(\dfrac{A + Ak + B k}{k (kH)} \right) dk$

$\Rightarrow A = 1 , A + B = 0$ (equating coefficients on LHS & RHS)

$\Rightarrow A = 1 , B = -1$

$\Rightarrow \displaystyle \int \dfrac{dk}{k(k + 1)} = \int \dfrac{dk}{k} - \int \dfrac{dk}{k + 1} = \dfrac{dx}{x} \Rightarrow \log (k) - \log (k + 1) = \log x + \log c$

$\Rightarrow \log \left(\dfrac{k}{k + 1} \right) = \log (cx) \Rightarrow \dfrac{k}{k + 1} = cx \Rightarrow \dfrac{\left[\dfrac{(y)}{(x)} \right]}{\left[\left(\dfrac{y}{x} \right) + 1 \right]} = cx$

when

$x = 1, y = 1 \Rightarrow \dfrac{1}{(1 + 1)} = 1(c) \Rightarrow c = \dfrac{1}{2}$

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

Find the particular solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x(2logx + 1)}{sin y + y cos y}$ given that $y = \dfrac{\pi}{2}$ where x = 1.

Seperating terms of

$y$ and

$x$

$\left(\sin{y}+y\cos{y}\right)dy=x\left(2\log{x}+1\right)dx$

By integrating both sides, we get

$\int{\left(\sin{y}+y\cos{y}\right)dy}=\int{x\left(2\log{x}+1\right)dx}$

$\int{\sin{y}dy}+\int{y\cos{y}dy}=2\int{x\log{x}dx}+\int{x dx}$

$-cos{y}+\int{y\cos{y}dy}=2\int{x\log{x}dx}+\dfrac{{x}^{2}}{2}$ ......

$\left(1\right)$

Consider

$\int{y\cos{y}dy}$

Let

$u=y\Rightarrow du=dy$ and

$dv=\cos{y}dy\Rightarrow v=\sin{y}$

We have

$\int{u.dv}=uv-\int{v.du}$

$\int{y\cos{y}dy}=y\sin{y}-\int{\sin{y}dy}$

$\int{y\cos{y}dy}=y\sin{y}-\left(-\cos{y}\right)$

$\int{y\cos{y}dy}=y\sin{y}+\cos{y}$ ......

$\left(2\right)$

Consider

$\int{x\log{x}dx}$

Let

$u=\log{x}\Rightarrow du=\dfrac{1}{x}dx$ and

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

We have

$\int{u.dv}=uv-\int{v.du}$

$\Rightarrow \int{x\log{x}dx}=\log{x}\dfrac{{x}^{2}}{2}-\int{\dfrac{{x}^{2}}{2}\times \dfrac{1}{x}dx}$

$\Rightarrow \int{x\log{x}dx}=\dfrac{{x}^{2}\log{x}}{2}-\int{\dfrac{x}{2}dx}$

$\Rightarrow \int{x\log{x}dx}=\dfrac{{x}^{2}\log{x}}{2}-\dfrac{{x}^{2}}{4}$ ......

$\left(3\right)$

Substituting

$\left(2\right)$ and

$\left(3\right)$ in

$\left(1\right)$ we get

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

$-\cos{y}+y\sin{y}+\cos{y}=2\left[\dfrac{{x}^{2}\log{x}}{2}-\dfrac{{x}^{2}}{4}\right]+\dfrac{{x}^{2}}{2}+c$ where

$c$ is the constant of integration.

$\Rightarrow y\sin{y}={x}^{2}\log{x}-\dfrac{{x}^{2}}{2}+\dfrac{{x}^{2}}{2}+c$

$\Rightarrow y\sin{y}={x}^{2}\log{x}+c$

$x=1,y=\dfrac{\pi}{2}$ we have

$\Rightarrow \dfrac{\pi}{2}\sin{\dfrac{\pi}{2}}={1}^{2}\log{1}+c$ where

$\log{1}=0$

$\Rightarrow c=\dfrac{\pi}{2}$ where

$\sin{\dfrac{\pi}{2}}=1$

$\therefore y\sin{y}={x}^{2}\log{x}+\dfrac{\pi}{2}$

Differential Equations - Class 12 Engineering Maths - Extra Questions

Find the order and degree of the differential equations.
a) $y ^ { \prime \prime } + 3 y ^ { \prime ^ { 2 } } + y ^ { 3 } = 0$
b) $\left( \dfrac { d y } { d x } \right) ^ { 2 } + \dfrac { 1 } { d y d x } = 2$
c) $\dfrac { d ^ { 2 } y } { d x ^ { 2 } } + 4 y = 0$

$\cos \left( \frac { d y } { d x } \right) = a ( a \in \mathbf { R } ) ; y = 1$ when $x = 0$ .Find $y$ ?

Show that $y=cx+\dfrac{a}{c}$ is a solution of the differential equation $y=x\dfrac{dy}{dx}+\dfrac{a}{\dfrac{dy}{dx}}$ .

Write the integrating factor of the differential equation:
$\cos x\dfrac{dy}{dx}+y=\sin x; 0\le x <\dfrac{\pi}{2}$ .

Find the integrating factor of $\cfrac { dy }{ dx } +\cfrac { y }{ x } ={ x }^{ 2 }$ .

Find the differential coefficient of $\tan ^{ -1 }{ x }$ w.r to $x$ .

Find the general solution of the differential equation:
$xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx$

Verify that $y=4sin 3x$ is the solution of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} + 9y = 0$ .

Find order and degree of the following differential equations .
(i) $\dfrac{dy}{dx} + y = \dfrac{1}{\dfrac{dy}{dx}}$
(ii) $_e\left[\dfrac{dy}{dx}-\dfrac{d^3y}{dx^3}\right] = ln \left[\dfrac{d^5y}{dx^5}+ 1\right]$
(iii) $\left[(\dfrac{dy}{dx})^{1/3} + y \right]^2 = \dfrac{d^2y}{dx^2}$

Find the particular solution of the difference equation
$(1-y^{2})(1+\log x)dx+2xydy=0$
given that $y=0$ when $x=1$

Solve: $\cos^2x\dfrac{dy}{dx}+y=\tan x$ .

Find the order of the differential equation obtained by eliminating the arbitrary constants $b$ and $c$ from $xy=ce^{x}+be^{-x}+x^{2}$ .

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

If $y=(cos^{-1}x)^2$ , prove that $(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-2=0$ . Hence find $y_2$ when $x=0$ .

If ${ x }^{ y }={ e }^{ x-y }$ , show that $\dfrac { dy }{ dx } =\dfrac{y\log{x}}{x\left(\log{x}+1\right)}$

Write the order of the differential equation $1+\left(\dfrac{dy}{dx}\right)^2=7\left(\dfrac{d^2y}{dx^2}\right)^3$ .

Show that $xy=ae^x+be^{-x}+x^2$ is a solution of the differential equation $x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}-xy+x^2-2=0$ .

Verify that $y = A\cos 2x - B\sin 2x$ is the general solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} + 4y = 0$ .

Verify that $y = ce^{\tan^{-1}x}$ is a solution of the differential equation $(1 + x^{2}) \dfrac {d^{2}y}{dx^{2}} + (2x - 1) \dfrac {dy}{dx} = 0$ .

The number of arbitrary constants in the general solution of a differential equation of order three is ______ .

Solve the following differential equation:
$x^{2} \dfrac {dy}{dx} = y^{2} + 2xy$
Given that : $y = 1$ , when $x = 1$ .

Find the particular solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x(2logx + 1)}{sin y + y cos y}$ given that $y = \dfrac{\pi}{2}$ where x = 1.

Class 12 Engineering Maths Extra Questions

Differential Equations - Class 12 Engineering Maths - Extra Questions

Find the order and degree of the differential equations.a) y′′+3y′2+y3=0y ^ { \prime \prime } + 3 y ^ { \prime ^ { 2 } } + y ^ { 3 } = 0b) (dydx)2+1dydx=2\left( \dfrac { d y } { d x } \right) ^ { 2 } + \dfrac { 1 } { d y d x } = 2c) d2ydx2+4y=0\dfrac { d ^ { 2 } y } { d x ^ { 2 } } + 4 y = 0

cos(dydx)=a(a∈R);y=1\cos \left( \frac { d y } { d x } \right) = a ( a \in \mathbf { R } ) ; y = 1 when x=0x = 0.Find yy?

Show that y=cx+acy=cx+\dfrac{a}{c} is a solution of the differential equation y=xdydx+adydxy=x\dfrac{dy}{dx}+\dfrac{a}{\dfrac{dy}{dx}}.

Write the integrating factor of the differential equation:cosxdydx+y=sinx;0≤x<π2\cos x\dfrac{dy}{dx}+y=\sin x; 0\le x <\dfrac{\pi}{2}.

Find the integrating factor of dydx+yx=x2\cfrac { dy }{ dx } +\cfrac { y }{ x } ={ x }^{ 2 }.

Find the differential coefficient of tan−1x\tan ^{ -1 }{ x } w.r to xx.

Find the general solution of the differential equation:xdy−ydx=√x2+y2dxxdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx

Verify that y=4sin3xy=4sin 3x is the solution of the differential equation d2ydx2+9y=0\frac{{{d^2}y}}{{d{x^2}}} + 9y = 0.

Find the particular solution of the difference equation (1−y2)(1+logx)dx+2xydy=0(1-y^{2})(1+\log x)dx+2xydy=0given that y=0y=0 when x=1x=1

Solve: cos2xdydx+y=tanx\cos^2x\dfrac{dy}{dx}+y=\tan x.

Find the order of the differential equation obtained by eliminating the arbitrary constants bb and cc from xy=cex+be−x+x2xy=ce^{x}+be^{-x}+x^{2}.

Solve: dydx=cos(x+y)+sin(x+y)\dfrac{dy}{dx}=cos(x+y)+sin(x+y)

If y=(cos−1x)2y=(cos^{-1}x)^2, prove that (1−x2)d2ydx2−xdydx−2=0(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-2=0. Hence find y2y_2 when x=0x=0.

If xy=ex−y{ x }^{ y }={ e }^{ x-y }, show that dydx=ylogxx(logx+1)\dfrac { dy }{ dx } =\dfrac{y\log{x}}{x\left(\log{x}+1\right)}

Write the order of the differential equation 1+(dydx)2=7(d2ydx2)31+\left(\dfrac{dy}{dx}\right)^2=7\left(\dfrac{d^2y}{dx^2}\right)^3.

Show that xy=aex+be−x+x2xy=ae^x+be^{-x}+x^2 is a solution of the differential equation xd2ydx2+2dydx−xy+x2−2=0x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}-xy+x^2-2=0.

Verify that y=Acos2x−Bsin2xy = A\cos 2x - B\sin 2x is the general solution of the differential equation d2ydx2+4y=0\dfrac {d^{2}y}{dx^{2}} + 4y = 0.

Verify that y=cetan−1xy = ce^{\tan^{-1}x} is a solution of the differential equation (1+x2)d2ydx2+(2x−1)dydx=0(1 + x^{2}) \dfrac {d^{2}y}{dx^{2}} + (2x - 1) \dfrac {dy}{dx} = 0.

The number of arbitrary constants in the general solution of a differential equation of order three is ______ .

Solve the following differential equation:x2dydx=y2+2xyx^{2} \dfrac {dy}{dx} = y^{2} + 2xyGiven that : y=1y = 1, when x=1x = 1.

Find the particular solution of the differential equation dydx=x(2logx+1)siny+ycosy\dfrac{dy}{dx} = \dfrac{x(2logx + 1)}{sin y + y cos y} given that y=π2y = \dfrac{\pi}{2} where x = 1.

Class 12 Engineering Maths Extra Questions

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Find the order and degree of the differential equations.
a) $y ^ { \prime \prime } + 3 y ^ { \prime ^ { 2 } } + y ^ { 3 } = 0$
b) $\left( \dfrac { d y } { d x } \right) ^ { 2 } + \dfrac { 1 } { d y d x } = 2$
c) $\dfrac { d ^ { 2 } y } { d x ^ { 2 } } + 4 y = 0$

$\cos \left( \frac { d y } { d x } \right) = a ( a \in \mathbf { R } ) ; y = 1$ when $x = 0$ .Find $y$ ?

Show that $y=cx+\dfrac{a}{c}$ is a solution of the differential equation $y=x\dfrac{dy}{dx}+\dfrac{a}{\dfrac{dy}{dx}}$ .

Write the integrating factor of the differential equation:
$\cos x\dfrac{dy}{dx}+y=\sin x; 0\le x <\dfrac{\pi}{2}$ .

Find the integrating factor of $\cfrac { dy }{ dx } +\cfrac { y }{ x } ={ x }^{ 2 }$ .

Find the differential coefficient of $\tan ^{ -1 }{ x }$ w.r to $x$ .

Find the general solution of the differential equation:
$xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx$

Verify that $y=4sin 3x$ is the solution of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} + 9y = 0$ .

Find the particular solution of the difference equation
$(1-y^{2})(1+\log x)dx+2xydy=0$
given that $y=0$ when $x=1$

Solve: $\cos^2x\dfrac{dy}{dx}+y=\tan x$ .

Find the order of the differential equation obtained by eliminating the arbitrary constants $b$ and $c$ from $xy=ce^{x}+be^{-x}+x^{2}$ .

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

If $y=(cos^{-1}x)^2$ , prove that $(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-2=0$ . Hence find $y_2$ when $x=0$ .

If ${ x }^{ y }={ e }^{ x-y }$ , show that $\dfrac { dy }{ dx } =\dfrac{y\log{x}}{x\left(\log{x}+1\right)}$

Write the order of the differential equation $1+\left(\dfrac{dy}{dx}\right)^2=7\left(\dfrac{d^2y}{dx^2}\right)^3$ .

Show that $xy=ae^x+be^{-x}+x^2$ is a solution of the differential equation $x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}-xy+x^2-2=0$ .

Verify that $y = A\cos 2x - B\sin 2x$ is the general solution of the differential equation $\dfrac {d^{2}y}{dx^{2}} + 4y = 0$ .

Verify that $y = ce^{\tan^{-1}x}$ is a solution of the differential equation $(1 + x^{2}) \dfrac {d^{2}y}{dx^{2}} + (2x - 1) \dfrac {dy}{dx} = 0$ .

Solve the following differential equation:
$x^{2} \dfrac {dy}{dx} = y^{2} + 2xy$
Given that : $y = 1$ , when $x = 1$ .

Find the particular solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x(2logx + 1)}{sin y + y cos y}$ given that $y = \dfrac{\pi}{2}$ where x = 1.