MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 4

Solution of the given differential equation

$\displaystyle \left ( 1-x^{2} \right )\frac{dy}{dx}+xy=xy^{2}$ is

0%
$y(1-y)=c\displaystyle \sqrt{1-x^2}$
0%
$y(y-1)=c\displaystyle \sqrt{1-x^2}$
0%
$x(1-x)=c\displaystyle \sqrt{1-y^2}$
0%
$x(x-1)=c\displaystyle \sqrt{1-y^2}$

Which of the following functions are homogeneous?

0%
$x \sin y + y \sin x$
0%
$x\, e^{y/x}\, +\, y\, e^{x/y}$
0%
$x^2\, -\, xy$
0%
$arc \sin (xy)$

Explanation

Homogeneous equation by definition is

$f(\lambda x)=\lambda^{n}f(x)$

$A)f(x,y)=xsiny+ysinx$
Now

$f(\lambda x,\lambda y)=\lambda x sin(\lambda y)+\lambda y sin(\lambda x)$

$=\lambda(xsin\lambda y+ysin(\lambda x))$
But

$xsin\lambda y+ysin(\lambda x)\neq f(x,y)$
Hence it is not a homogeneous function.

$B)f(x,y)=xe^{y/x}+ye^{x/y}$

$f(\lambda x,\lambda y)= \lambda xe^{\lambda y/\lambda x}+\lambda ye^{\lambda x/\lambda y}$

$=\lambda(e^{x/y}+e^{y/x})$

$=\lambda f(x,y)$
Hence it is a homogeneous function.

$C)f(x,y)=x^{2}-xy$

$f(\lambda x,\lambda y)=\lambda^{2} x^{2}-\lambda^{2}(xy)$

$=\lambda^{2}(x^{2}-xy)$

$=\lambda^{2}f(x,y)$
Hence it is a homogeneous function.

$D)f(x,y)=sin^{-1}(xy)$

$f(\lambda x,\lambda y)=sin^{-1}(\lambda^{2}(xy))$

$\neq f(x,y)$
Hence it is not a homogeneous equation.

Find general solution of

$\sqrt{1 + 4x^{2}} dy = y^{3} xdx$ .

0%
$\displaystyle - \frac{1}{2y^{2}} = \frac{1}{4}\sqrt{1 + 4x^{2}} + k$
0%
$\displaystyle \frac{1}{2y^{2}} = \frac{1}{4}\sqrt{1 + 4x^{2}} + k$
0%
$\displaystyle \frac{1}{2y^{2}} = -\frac{1}{4}\sqrt{1 + 4x^{2}} + k$
0%
$\displaystyle \frac{1}{2y^{2}} = -\frac{1}{4}\sqrt{1 - 4x^{2}} + k$

Explanation

$\sqrt { 1+4{ x }^{ 2 } } dy={ y }^{ 3 }xdx\\$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { x{ y }^{ 3 } }{ \sqrt { 1+4{ x }^{ 2 } } } \\ \Rightarrow \cfrac { \cfrac { dy }{ dx } }{ { y }^{ 3 } } =\cfrac { x }{ \sqrt { 1+4{ x }^{ 2 } } }$

Integrating both sides w.r.t x

$\displaystyle \int { \cfrac { \frac { dy }{ dx } }{ { y }^{ 3 } } dx } =\displaystyle \int { \cfrac { x }{ \sqrt { 1+4{ x }^{ 2 } } } dx }$

$\displaystyle \int { \cfrac { 1 }{ { y }^{ 3 } } dy } =\displaystyle \int { \cfrac { x }{ \sqrt { 1+4{ x }^{ 2 } } } dx }$

$\Rightarrow -\cfrac { 1 }{ 2{ y }^{ 2 } } =\cfrac { 1 }{ 4 } \sqrt { 1+4{ x }^{ 2 } } +c$

The solution of the differential equation

$\displaystyle \left( 1+\cos { x } \right) \frac { dy }{ dx } =1-\cos { x }$ is

0%
$\displaystyle y=\tan { \frac { x }{ 2 } } +x+c$
0%
$\displaystyle y=2\tan { \frac { x }{ 2 } } -x+c$
0%
$\displaystyle y=\tan { \frac { x }{ 2 } } -x+c$
0%
$\displaystyle y=x-2\tan { \frac { x }{ 2 } } +c$

Explanation

$\displaystyle \left( 1+\cos { x } \right) \frac { dy }{ dx } =1-\cos { x }$

$\displaystyle \Rightarrow \int { dy } =\int { \frac { 1-\cos { x } }{ 1+\cos { x } } dx }$

$\displaystyle \Rightarrow \int { dy } =\int { \tan ^{ 2 }{ \frac { x }{ 2 } } dx } +c$

$\displaystyle \Rightarrow y=\int { \left( \sec ^{ 2 }{ \frac { x }{ 2 } } -1 \right) dx } +c$

$\displaystyle \Rightarrow y=2\tan { \frac { x }{ 2 } } -x+c$

The equation of the curve in which sub-normal varies as the square of the ordinate is (

$k$ is constant of proportionality)

0%
$y=Ae^{2kx}$
0%
$y=e^{kx}$
0%
$y^{2}/2+kx=A$
0%
$y^{2}+kx^{2}=A$

Explanation

$|y\dfrac{dy}{dx}|=ky^{2}$

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

$\dfrac{dy}{y}=k.dx$

Integrating both sides

$lny=kx+c$

$y=e^{kx}.e^{c}$
Or

$y=Ce^{kx}$ where

$C=e^{c}$ .

Solve :

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

0%
$\sec y = -2 \cos x + C$
0%
$\sec y = 2 \cos x + C$
0%
$\sec y = - \cos x + C$
0%
$\sec y = \cos x + C$

Explanation

Given,

$(\tan y)\displaystyle \frac{dy}{dx} = \sin(x + y) + \sin(x - y)=2\sin x\cos y$

$\Rightarrow (\sec y \tan y)dy = \sin x dx$

$\Rightarrow d(\sec y) =- 2d(\cos x)$

Integrating on both sides we get,

$\sec y = -2\cos x+c$

Find general solution of

$\displaystyle \frac{2dy}{dx} = \frac{y(x + 1)}{x}$ .

0%
$\log y^{2} = x + \log |x| + k$
0%
$\log y^{2} = x + \log (x) + k$
0%
$\log y = x + \log (x) + k$
0%
$2\log y = x - \log |x| + k$

Explanation

Given,

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

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

Integrating both sides w.r.t

$x$

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

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

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

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

The solution to the differential equation

$y\ln y \, +\, xy'\, =\, 0\,$ where

$\, y(1)\, =\, e$ , is:

0%
$x(\ln y)\, =\, 1$
0%
$xy(\ln y)\, =\, 1$
0%
$(\ln y )^2\, =\, 2$
0%
$\ln y\, +\, \left ( \displaystyle \frac {x^2}{2} \right )\, y\, =\, 1$

A solution of the differential equation,

$\left ( \displaystyle \dfrac {dy}{dx} \right ) ^2\, -\,x\, \displaystyle \dfrac {dy}{dx}\, +\, y\, =\, 0$ is

0%
$y\, =\, 2$
0%
$y\, =\, 2x$
0%
$y\, =\, 2x\, \, -4$
0%
$y\, =\, 2x^2\, \,- 4$

Explanation

$y=-{ \left( \cfrac { dy }{ dx } \right) }^{ 2 }+x\cfrac { dy }{ dx } \\ \Rightarrow \cfrac { dy }{ dx } =x\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\cfrac { dy }{ dx } -2\cfrac { dy }{ dx } \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\ \Rightarrow \cfrac { dy }{ dx } =\cfrac { dy }{ dx } +\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } \left( x-2\cfrac { dy }{ dx } \right) \\ \Rightarrow \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } \left( x-2\cfrac { dy }{ dx } \right) =0$

$\Rightarrow \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0$ or

$x-2\cfrac { dy }{ dx } =0$

$\Rightarrow y=cx-c^{ 2 }$ or

$y=\cfrac { { x }^{ 2 } }{ 4 }$

The solution to the differential equation

$(x + 1) \displaystyle \frac{dy}{dx} - y = e^{3x} (x + 1)^2$ is

0%
$y = (x + 1)e^{3x} + c$
0%
$3y = (x + 1) + e^{3x} + c$
0%
$\displaystyle \frac{3y}{x+ 1} = e^{3x} + c$
0%
$ye^{-3x} = 3 (x + 1) + c$

Explanation

The given equation is

$\displaystyle \dfrac{dy}{dx} - \dfrac{y}{x + 1} = e^{3x} (x+ 1)$
I.F.

$= \displaystyle e^{\displaystyle \int - \dfrac{1}{x + 1} dx} = e^{\displaystyle - log (x + 1)} = \dfrac{1}{x + 1}$
The solution is

$\displaystyle y \left ( \dfrac{1}{x +1} \right ) = \int e^{3x} (x +1). \dfrac{1}{x + 1} dx + a$

$\Rightarrow \displaystyle \dfrac{y}{x + 1} = \int e^{3x} dx+ a = \dfrac{e^{3x}}{3} + a$

$\Rightarrow \displaystyle \dfrac{3y}{x + 1} = e^{3x} + c, c = 3a$

The solution of

$\displaystyle \frac {dy}{dx}\, =\, \displaystyle \frac {ax\, +\, h}{by\, +\, k}$ represent a parabola if:

0%
$a\, =\, 2,\, b\, =\, 0$
0%
$a\, =\, 2,\, b\, =\, 2$
0%
$a\, =\, 0,\, b\, =\, 2$
0%
$a\, =\, 0,\, b\, =\, 0$

Explanation

$y'=\cfrac { ax+h }{ by+k } \\ \Rightarrow \cfrac { dy }{ dx } \left( k+by \right) =h+ax\\ \Rightarrow \int { \cfrac { dy }{ dx } \left( k+by \right) dx } =\int { \left( h+ax \right) dx } \\ \Rightarrow \cfrac { b{ y }^{ 2 } }{ 2 } +ky=\cfrac { a{ x }^{ 2 } }{ 2 } +hx+c$
Represent parabola only when

$a=0,b=2$ or

$a=2,b=0$

The solution of differential equation

$\displaystyle \frac { dy }{ dx } =-\left( \frac { y+\sin { x } }{ x } \right)$ satisfying condition

$\displaystyle y(0) = 1$ , is

0%
$\displaystyle \cos x = xy - 1$
0%
$\displaystyle \cos x = xy + 1$
0%
$\displaystyle \cos x = xy$
0%
$\displaystyle \cos x = x + 1$

Explanation

$\displaystyle xdy=-ydx-\sin { x } dx$

$\displaystyle \Rightarrow -\sin { x } dx=xdy+ydx$

$\displaystyle \Rightarrow -\sin { x } dx=d\left( xy \right)$

$\displaystyle \Rightarrow \int { -\sin x { dx } } =\int { d\left( xy \right) }$

$\displaystyle \Rightarrow \cos { x } =xy+c$

$\displaystyle y(0)=1\Rightarrow c=1$

$\displaystyle \Rightarrow \cos { x } =xy+1$

The solution of the equation

$\displaystyle \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ x }+{ e }^{ -x }$ is-
Note : (where c & d are arbitrary constants in the given options)

0%
$\displaystyle y={ e }^{ x }-{ e }^{ -x }+cx+d$
0%
$\displaystyle y={ e }^{ x }+{ e }^{ -x }+cx+d$
0%
$\displaystyle y=-{ e }^{ x }+{ e }^{ -x }+cx+d$
0%
None of these

Explanation

Given,

$\displaystyle \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ x }+{ e }^{ -x }$
Integrating gives

$\Rightarrow \displaystyle \frac { dy }{ { dx } } ={ e }^{ x }-{ e }^{ -x }-c$
Again integrating,

$\displaystyle \Rightarrow y={ e }^{ x }+{ e }^{ -x }-cx+d$

Solution of the differential equation

$\displaystyle \left( 2x-y+2 \right) dx+\left( 4x-2y-1 \right) dy=0$ is

0%
$\displaystyle 2x-y={ ce }^{ -\left( x+2y \right) }$
0%
$\displaystyle 2x+y={ ce }^{ \left( 2x+y \right) }$
0%
$\displaystyle x-2y={ ce }^{ -\left( x+2y \right) }$
0%
$\displaystyle 2x+y={ ce }^{ \left( x+2y \right) }$

Explanation

Given equation is

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

Put

$\displaystyle 2x-y=v\Rightarrow \displaystyle 2-\frac { dy }{ dx } =\frac { dv }{ dx }$

$\Rightarrow\displaystyle -\left( \frac { dv }{ dx } -2 \right) =-\left( \frac { v+2 }{ 2v-1 } \right)$

$\displaystyle \Rightarrow \frac { dv }{ dx } =\left( \frac { v+2 }{ 2v-1 } +2 \right) \Rightarrow \frac { dv }{ dx } =\frac { 5v }{ 2v-1 }$

$\displaystyle \Rightarrow \int { \frac { 2v-1 }{ 5v } } dv=\int { dx }$

$\displaystyle \Rightarrow \frac { 2 }{ 5 } v-\frac { 1 }{ 5 } \log { v } =x+\log { c }$

$\displaystyle \Rightarrow \frac { 2 }{ 5 } \left( 2x-y \right) -\frac { 1 }{ 5 } \log { \left( 2x-y \right) } =x+\log { c }$

$\displaystyle \Rightarrow 2\left( 2x-y \right) -\log { \left( 2x-y \right) } =5x+5\log { c }$

$\displaystyle \Rightarrow 4x-2y-\log { \left( 2x-y \right) } =5x-\log { c }$

$\displaystyle \Rightarrow \log { c } -\log { \left( 2x-y \right) } =\left( x+2y \right)$

$\displaystyle\Rightarrow \log { \left( \frac { c }{ 2x-y } \right) } =\left( x+2y \right)$

$\displaystyle \Rightarrow\left( \frac { c }{ 2x-y } \right) ={ e }^{ \left( x+2y \right) }$

$\displaystyle\Rightarrow c.{ e }^{ -\left( x+2y \right) }=\left( 2x-y \right)$

Solution of differential equation

$\displaystyle \frac { dx }{ dy } =\tan { x } \left( 1+ysinx \right)$ is given by -

0%
$\displaystyle cosec x= -y+1+{ Ce }^{ -y }$
0%
$\displaystyle y=\tan { x } +{ Ce }^{ X }$
0%
$\displaystyle \sin { x } { e }^{ y }=1+y+C$
0%
$\displaystyle cosecx=y+{ Ce }^{ y }$

Explanation

$\displaystyle \cot { x }* \csc x\frac { dx }{ dy } = \csc x + y$
Put

$\csc x=t$

$\displaystyle \Rightarrow -\csc x \ * \cot x\frac { dx }{ dy } =\frac { dt }{ dy }$
Now D.E is

$\displaystyle \frac { dt }{ dy } =-t-y$

I.F. =

$\displaystyle { e }^{\int dy }{ ={ e }^{ y } }$
solution is

$\displaystyle t.{ e }^{ y }=\int { -y{ e }^{ y }dy } =-ye^y+e^y+c$

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

Find the equation of the curve for which the normal at any point (x, y) passes through the origin. The curve represents a :

0%
ellipse
0%
rectange
0%
circle
0%
hyperbola

Explanation

The equation of the normal at any point

$(x,y)$ of the curve is

$\cfrac { dy }{ dx } \left( Y-y \right) +\left( X-x \right) =0$
Since the required curve passes through origin

$\cfrac { dy }{ dx } \left( 0-y \right) +\left( 0-x \right) =0\Rightarrow 2\left( -y \right) dy+2\left( -x \right) dx=0$
Integrating

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

Find general solution of

$y-x\, \cfrac {dy}{dx}\, =\, b(1+x^2\cfrac {dy}{dx})$ is:

0%
$\quad b+kx=y(1+bx)$
0%
$\quad b+ky=x(1+bx)$
0%
$\quad b+ky=x(1+by)$
0%
$\quad b+kx=x(1+by)$

Explanation

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

$\Rightarrow\quad y- b = (x+bx^2)\, \displaystyle \frac {dy}{dx}\, \Rightarrow\, \displaystyle \frac {dx}{x(1+bx)}\, =\, \displaystyle \frac {dy}{y-b}$

$\Rightarrow\quad \int \left ( \displaystyle \frac {1}{x}\, -\, \displaystyle \frac {b}{1+bx} \right )\, dx\, =\, \log\, (y-b)\, +\, \log\, c$

$\Rightarrow \log x - \log (1 + bx) = \log (y - b) + \log c$

$\Rightarrow\quad \displaystyle \frac {x}{c}\, =\, (1+bx)(y-b)$

$\Rightarrow\quad b+ \left (\displaystyle \frac {1}{c}\, +\, b^2 \right ) x = y (1+bx)$

$\Rightarrow\quad b+kx=y(1+bx)$

Which of the following differential equations has y = x as one of its particular solution?

0%
$\cfrac{d^2y}{dx^2}-x^2\cfrac{dy}{dx}+xy=x$
0%
$\cfrac{d^2y}{dx^2}+x\cfrac{dy}{dx}+xy=x$
0%
$\cfrac{d^2y}{dx^2} -x^2\cfrac{dy}{dx}+xy=0$
0%
$\cfrac{d^2y}{dx^2}+x\cfrac{dy}{dx}+xy=0$

Explanation

$y=x\\ \dfrac{dy}{dx}=1\\ \dfrac{d^2y}{dx^2}=0$
Now putting these values in given options then we will find that
Option C is correct

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

Which of the following differential equations has

$y = c_1 e^x + c_2 e^{x}$ as the general solution?

0%
$\dfrac{d^2y}{dx^2}+y = 0$
0%
$\dfrac{d^2y}{dx^2}-y = 0$
0%
$\dfrac{d^2y}{dx^2}+1 = 0$
0%
$\dfrac{d^2y}{dx^2}-1 = 0$

Explanation

$y=c_1e^x+c_2e^x\\ \dfrac{dy}{dx}=c_1e^x+c_2e^x\\ \dfrac{d^2y}{dx^2}=c_1e^x+c_2e^x\\ \therefore \dfrac{d^2y}{dx^2}-y=0$

The solution of the differential equation

$\dfrac { dy }{ dx } =\dfrac { x-y+3 }{ 2\left( x-y \right) +5 }$ is

0%
$2\left( x-y \right) +\log { \left( x-y \right) } =x+c$
0%
$2\left( x-y \right) -\log { \left( x-y+2 \right) } =x+c$
0%
$2\left( x-y \right) +\log { \left( x-y+2 \right) } =x+c$
0%
None of the above

Explanation

Given differential equation is

$\dfrac { dy }{ dx } =\dfrac { x-y+3 }{ 2\left( x-y \right) +5 }$

Let

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

$\therefore \dfrac { dv }{ dx } =\dfrac { v+2 }{ 2v+5 }$

$\Rightarrow \displaystyle\int { \dfrac { 2v+5 }{ v+2 } dv } =\displaystyle\int { dx }$

$\Rightarrow \displaystyle\int { \left( 2+\dfrac { 1 }{ v+2 } \right) dv } =\displaystyle\int { dx }$

$\Rightarrow 2v+\log { \left( v+2 \right) } =x+c$

$\Rightarrow 2\left( x-y \right) +\log { \left( x-y+2 \right) } =x+c$

The general solution of the differential equation

$\log _{ e }{ \left( \cfrac { dy }{ dx } \right) } =x+y$ is:

0%
${ e }^{ x }+{ e }^{ -y }=C$
0%
${ e }^{ x }+{ e }^{ y }=C$
0%
${ e }^{ y }+{ e }^{ x }=C$
0%
${ e }^{ -x }+{ e }^{ -y }=C$

Explanation

Given

$\log _{ e }{ \left( \cfrac { dy }{ dx } \right) } =x+y$

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

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

On integrating, we get

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

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

Find the general solution of

$dy=y \sec x dx$ .

0%
$y=C(\sec x -2\tan x)$
0%
$y=C(\sec x +\tan x)$
0%
$y=C(2\sec x +\tan x)$
0%
None of these

Explanation

Given Differential Equation is

$dy=y\sec { x } dx$

Integrating with respect to

$x$ gives,

$\int { \cfrac { 1 }{ y } dy } =\int { \sec { x } } dx$

$\ln { y } =\ln { \left( \sec { x } +\tan { x } \right) } +\ln { c }$

Where

$c$ is the integration constant,

$\therefore y=c\left( \sec { x } +\tan { x } \right)$ .

The solution of the differential equation

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

0%
$xy=c$
0%
$x+y=c$
0%
$log\, x\, log\, y = c$
0%
$x^2+y^2=c$

Explanation

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

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

integrating,

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

$= lny = -lnx+c$

$\Rightarrow lnx.y = c \Rightarrow \boxed {xy = constant}$

1117076_460561_ans_a188d41a9eb84908baf206da79e79040.JPG

The general solution of

$\cfrac{dy}{dx}=\cfrac{2x-y}{x+2y}$ is

0%
${x}^{2}-xy+{y}^{2}=c$
0%
${x}^{2}-xy-{y}^{2}=c$
0%
${x}^{2}+xy-{y}^{2}=c$
0%
${x}^{2}-x{y}^{2}=c$

Explanation

Given

$\cfrac{dy}{dx}=\cfrac{2x-y}{x+2y}$
Put

$y=vx$

$\cfrac{dy}{dx}=v+x\cfrac{dv}{dx}$

$\therefore$

$v+x\cfrac { dv }{ dx } =\cfrac{2-v}{1+2v}$

$\Rightarrow$

$x\cfrac { dv }{ dx } =\cfrac{2-v-v(1+2v)}{1+2v}$

$\Rightarrow$

$x\cfrac { dv }{ dx } =\cfrac{2-2v-2{v}^{2}}{1+2v}$

$\Rightarrow$

$\int { \cfrac { 1+2v }{ 2(1-v-{ v }^{ 2 }) } } =\int { \cfrac { 1 }{ x } } dx$

$\Rightarrow$

$\log{a}-\cfrac{1}{2}\log{1-v-{v}^{2})}=\log{x}$

$\Rightarrow$

$2\log {a}-\log{(1-v-{v}^{2}}=2\log {x}$

$\Rightarrow$

$\log {c}=\log{[{x}^{2}(1-v-{v}^{2}]}$ [Let

${a}^{2}=c]$

$\Rightarrow$

$c={x}^{2}(1-v-{v}^{2})$

$\Rightarrow$

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

$\Rightarrow$

$c={x}^{2}(\cfrac{{x}^{2}-xy-{y}^{2}}{{x}^{2}})$

$\Rightarrow$

${x}^{2}-xy-{y}^{2}=c$

The order and degree of differential equation

${ \left( 1+3\cfrac { dy }{ dx } \right) }^{ 2/3 }=4\cfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } }$ , are

0%
$1,\cfrac{2}{3}$
0%
$3,1$
0%
$3,3$
0%
$1,2$

Explanation

Given

${ \left( 1+3\cfrac { dy }{ dx } \right) }^{ 2/3 }=4\cfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } }$

On taking cube on both sides, we get

${ \left( 1+3\cfrac { dy }{ dx } \right) }^{ 2 }={ \left( 4\cfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 3 }$

Now,

Order: Order of a differential equation is the order of the highest order derivative (also known as differential coefficient) present in the equation

Here Order is

$3$

Degree: The degree of the differential equation is represented by the power of the highest order derivative in the given differential equation.

Here Degree is

$3$

The general solution of the differential equation

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

0%
$xy = C$
0%
$x = Cy^2$
0%
$y=Cx$
0%
$y=Cx^2$

Explanation

$\dfrac{ydx-xdy}{y}=0\\ dx-\dfrac{x}{y}dy=0\\ dx=\dfrac{x}{y}dy\\ \dfrac{dx}{x}=\dfrac{dy}{y}\\ lnx=lny+lnc\\ x=yc$

The general solution of the differential equation

$\displaystyle \frac { dy }{ dx } +\frac { 1+\cos { 2y } }{ 1-\cos { 2x } } =0$ is given by:

0%
$\tan { y } +\cot { x } =c$
0%
$\displaystyle \tan { y } -\cot { x } =c$
0%
$\displaystyle \tan { x } -\cot { y } =c$
0%
$\displaystyle \tan { x } +\cot { x } =c$

Explanation

$\displaystyle \frac { dy }{ dx } +\frac { 1+\cos { 2y } }{ 1-\cos { 2x } } =0$

$\displaystyle \Rightarrow \frac { dy }{ dx } = - \frac { \cos ^{ 2 }{ y } }{\sin ^{ 2 }{ x } }$

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

$\int \sec^2y dy = - \int \csc^2x dx$

$\tan y +c = \cot x +C$

$\tan y - \cot x = c$

Which of the following are solutions of the differential equation

$y^{\prime\prime}-y=0$ ?

0%
$y=Ce^x$
0%
$y=C \dfrac {e^x}{2}$
0%
$y=Ce^{-x}$
0%
None of these

Explanation

Given

$\ddot { y } -y=0$

Put

$y=e^{mx}$ , we get

$m^{2}e^{mx}-e^{mx}=0$

$\Rightarrow e^{mx}(m^{2}-1)=0$

$\Rightarrow m=\pm1$

Therefore

$y=Ce^{x}$ and

$y=Ce^{-x}$ are solutions of given differential equation

Therefore options

$A$ and

$C$ are correct

Find a particular solution for the following differential equation.

$y'-4y'-12y=te^{4t}$

0%
$y(t)=\dfrac{1}{32}(3t+1)e^{4t}$
0%
$y(t)=-\dfrac{1}{18}(3t+1)e^{4t}$
0%
$y(t)=-\dfrac{1}{36}(3t+1)e^{4t}$
0%
None of these

Explanation

Given, ${ y }'-4{ y '}-12y=t{ e }^{ 4t }$

$\Rightarrow -{ 3y }'-12y=t{ e }^{ 4t }\\ \Rightarrow { y }'+4y=-\dfrac{t{ e }^{ 4t }}{3}\\ \Rightarrow \dfrac { dy }{ dt } +4y=-\dfrac{1}{3}t{ e }^{ 4t }$

This is a linear form of differential equation with

$\displaystyle{\text{I.F.}={ e }^{ \int 4dt }={ e }^{ 4t }}$

Solution of the differential equation is

$\displaystyle{y(\text{I.F.})=\int { Q( } \text{I.F.})dt\\ y{ e }^{ 4t }=\int { \dfrac { -t{ e }^{ 4t } }{ 3 } { e }^{ 4t } } dt\\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \int { t{ e }^{ 8t } } dt\\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ t\int { { e }^{ 8t } } -\int { \dfrac { d }{ dt } (t)\int { { e }^{ 8t } } } \right\} \\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ \dfrac { t{ e }^{ 8t } }{ 8 } -\int { \dfrac { { e }^{ 8t } }{ 8 } } \right\} \\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ \dfrac { t{ e }^{ 8t } }{ 8 } -\dfrac { { e }^{ 8t } }{ 64 } \right\} \\ y{ e }^{ 4t }=\dfrac { -{ e }^{ 8t } }{ 3 } \left\{ \dfrac { 8t-1 }{ 64 } \right\} \\ y=\dfrac { -{ e }^{ 4t } }{ 192 } (8t-1)}$

So, option D is correct.

If the general solutions of a differential equation is

$(y+c)^2=cx$ , where

$c$ is an arbitrary constant, then the order and degree of differential equation are:

0%
$1,2$
0%
$2,1$
0%
$1,1$
0%
None of these

Explanation

Given general solution is

${(y+c)}^{2}=cx$

Now differentiate it with respect to

$x$ on both sides

we get

$2(y+c)\frac{dy}{dx} = c$ , Let

$\frac{dy}{dx}$ be

$Y$

$\Rightarrow c=\frac { 2yY }{ 1-2Y }$

By substituting

$c$ in general solution , we get

${ (y+\frac { 2yY }{ 1-2Y } ) }^{ 2 }=\frac { 2yY }{ 1-2Y } x$

$\Rightarrow { y }^{ 2 }=2yxY(1-2Y)=2yxY-4yx{ Y }^{ 2 }$

Therefore the order of equation is

$1$ and the degree of equation is

$2$

Therefore option

$A$ is correct

Which of the following are true?

0%
Particular solution is a solution of a differential equation containing no arbitrary constants.
0%
Particular Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.
0%
General solution is a solution of a differential equation containing no arbitrary constants.
0%
General Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.

Verify that

$y=Cx^3$ is a solution of the differential equation

$xy'-3y=0$ for any value of C. Then

find the particular solution determined by the initial condition

$y=2$ when

$x=-3$ .

0%
$y=\dfrac{2}{27}x^2$
0%
$y=-\dfrac{2}{27}x^3$
0%
$y=-\dfrac{2}{25}x^3$
0%
None of these

Explanation

Given that

$y=C{ x }^{ 3 }$ is the solution of the differential equation

$x{ y }^{ ' }=3y$

So to get a particular solution we need to find the value of

$C$ for a given initial value condition,

i.e.,

$y=2$ when

$x=-3$ ,

Substituting this in the general solution gives,

$2=C(-27)$

$\Longrightarrow C=-\cfrac { 2 }{ 27 }$

$\therefore$ the particular solution is

$y=-\cfrac { 2 }{ 27 } { x }^{ 3 }$

The solution for the differential equation

$\cfrac { dy }{ y } +\cfrac { dx }{ x } =0$ is:

0%
$\cfrac { 1 }{ y } +\cfrac { 1 }{ x } =c$
0%
$\log { x } .\log { y } =c$
0%
$xy=C$
0%
$x+y=c$

Explanation

We have,

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

Integrate both sides,

$\displaystyle \int\left(\cfrac { dy }{ y } +\cfrac { dx }{ x } \right)=\int 0$

$\displaystyle\Rightarrow \int\cfrac { dy }{ y } +\int \cfrac { dx }{ x } =c$

, since differentiation any constant is 0

$\Rightarrow \ln x+\ln y=c$ , use basic formula

$\Rightarrow \ln(xy)=c$ , product rule of logarithm

$\log a+\log b=\log(ab)$

$\Rightarrow xy=e^c=C$

The solution of differential equation

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

0%
$y = \dfrac {x^{2} + C}{4x^{2}}$
0%
$y = \dfrac {x^{2}}{4} + C$
0%
$y = \dfrac {x^{4} + C}{x^{2}}$
0%
$y = \dfrac {x^{4} + C}{4x^{2}}$

Explanation

We have,

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

Divide each sides by

$x$

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

Integrating factor

$=e^{ \int \frac{2}{x}dx}=e^{2\ln x}=e^{\ln x^2}=x^2$

Hence solution is given by,

$\displaystyle y(x^2)=\int x^2\cdot x dx+c$

$\displaystyle \Rightarrow x^2y=\int x^3dx+C=\dfrac{x^4}{4}+c$

$\Rightarrow y=\dfrac{x^4+4c}{x^2}=\dfrac{x^4+C}{4x^2}$

The solution of the differential equation

$y\sin\left(\dfrac{x}{y}\right)dx=\left(x \sin\left(\dfrac{x}{y} \right)-y \right) dy$ satisfying

$y(\dfrac{\pi}{4})=1$ is

0%
$\cos\dfrac{x}{y}=-\log_ey+\dfrac{1}{\sqrt{2}}$
0%
$\sin\dfrac{x}{y}=\log_ey+\dfrac{1}{\sqrt{2}}$
0%
$\sin\dfrac{x}{y}=\log_ex-\dfrac{1}{\sqrt{2}}$
0%
$\cos\dfrac{x}{y}=-\log_ex-\dfrac{1}{\sqrt{2}}$

Explanation

Given differential equation is

$\cfrac { y\sin { \frac { x }{ y } } }{ x\sin { \frac { x }{ y } } -y } =\cfrac { dy }{ dx }$

Now replace

$x=vy$ such that,

$\cfrac { dx }{ dy } =v+y\cfrac { dv }{ dy }$

$\Longrightarrow v+y\cfrac { dv }{ dy } =\cfrac { v\sin { v } -1 }{ \sin { v } }$

$\Longrightarrow y\cfrac { dv }{ dy } =\cfrac { -1 }{ \sin { v } }$

Separating the variables and integrating gives,

$\int { \sin { v } } dv=\int { \cfrac { 1 }{ y } } dy$ ,

$\Longrightarrow -\cos { v } +c=\ln { y }$

Given

$y\left( \cfrac { \pi }{ 4 } \right) =1$

$\Longrightarrow c=\cfrac { 1 }{ \sqrt { 2 } }$

$\therefore$ the particular solution is ,

$\cos { \cfrac { x }{ y } } =-\ln { y } +\cfrac { 1 }{ \sqrt { 2 } }$

The integrating factor of linear differential equation

$\cfrac { dy }{ dx } +y\sec { x } =\tan { x }$ is:

0%
$\sec { x } -\tan { x }$
0%
$\sec { x } .\tan { x }$
0%
$\sec { x } +\tan { x }$
0%
$\sec { x } .\cot { x }$

Explanation

Given,

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

Here

$P=\sec { x }$

$\therefore$ Integrating factor

$={ e }^{ \int { P } dx }$

$={ e }^{ \int { \sec { x } dx } }$

$={ e }^{ \log { \left( \sec { x } +\tan { x } \right) } }$

$=\sec { x } +\tan { x }$

The integrating factor of the differential equation

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

0%
$\sec{x}$
0%
$\cos{x}$
0%
${e}^{\tan{x}}$
0%
$\cot{x}$

Explanation

We have,

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

We know that the general equation

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

So,

$P=\tan x, Q=\cos x$

We know that

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

Therefore,

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

$I.F=e^{-\ln |\cos x|}$

$I.F=e^{\ln \frac{1}{\sec x}}$

$I.F= \dfrac{1}{\sec x}$

$I.F=\cos x$

Hence, this is the answer.

The solution of

$\dfrac {d^{2}x}{dy^{2}} - x = k$ , where

$k$ is a non-zero constant, vanishes when

$y = 0$ and tends of finite limit as

$y$ tends to infinity, is

0%
$x = k(1 + e^{-y})$
0%
$x = k(e^{y} + e^{-y} - 2)$
0%
$x = k(e^{-y} - 1)$
0%
$x = k(e^{y} - 1)$

Explanation

We can write given differential equation as,

$(D^{2} - 1)x = k ... (i)$
where,

$D = \dfrac {d}{dy}$
Its auxiliary equation is

$m^{2} - 1 = 0$ , so that

$m = 1, -1$
Hence,

$CF = C_{1}e^{y} + C_{2}e^{-y}$
where

$C_{1}, C_{2}$ are arbitrary constants
Now, also

$PI = \dfrac {1}{D^{2} - 1}k$

$= k.\dfrac {1}{D^{2} - 1} e^{0.y}$

$= k.\dfrac {1}{0^{2} - 1}e^{0.y} = -k$
So, solution of eq. (i) is

$x = C_{1}e^{y} + C_{2}e^{-y} - k ... (ii)$
Given that

$x = 0$ , when

$y = 0$
So,

$0 = C_{1} + C_{2} - k$ (From (ii))

$\Rightarrow C_{1} + C_{2} = k .... (iii)$
Multiplying both sides of eq. (ii) by

$e^{-y}$ , we get

$x.e^{-y} = C_{1} + C_{2}e^{-2y} - ke^{-y} ... (iv)$
Given that

$x\rightarrow m$ when

$y\rightarrow \infty, m$ being a finite quantity.
So, eq (iv) becomes

$x\times 0 = C_{1} + C_{2} \times 0 - (k\times 0)$

$\Rightarrow C_{1} = 0 ....(v)$
From eqs. (iv) and (v), we get

$C_{1} = 0$ and

$C_{2} = k$
Hence, eq. (ii) becomes

$x = ke^{-y} - k = k(e^{-y} - 1)$ which is the required solution.

For the differential equation

${ \left( \cfrac { dy }{ dx } \right) }^{ 2 }-x\left( \cfrac { dy }{ dx } \right) +y=0$ , which one of the following is not its solution?

0%
$y=x-1$
0%
$4y={x}^{2}$
0%
$y=x$
0%
$y=-x-1$

Explanation

Given differential equation is

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

To solve this we just need to substitute the options,

$A)$

$y=x-1\Longrightarrow \dfrac { dy }{ dx } =1$ ,

$1-x+x-1=0$ ,

Therefore

$A$ is a solution of the given differential equation.

$B)$

$y=\dfrac { { x }^{ 2 } }{ 4 } \Longrightarrow \dfrac { dy }{ dx } =\dfrac { x }{ 2 }$ ,

$\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { { x }^{ 2 } }{ 2 } +\dfrac { { x }^{ 2 } }{ 4 } =0$ ,

Therefore

$B$ is a solution to the given differential equation.

$C)$

$y=x\Longrightarrow \dfrac { dy }{ dx } =1\\ \Longrightarrow 1-x+x=1\neq 0$ ,

Therefore

$C$ is not a solution of the given differential equation.

$D)$

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

$1+x-1-x=0$ ,

Therefore

$D$ is a solution of the given differential equation.

The solution of the differential equation

$\dfrac {dy}{dx} = \dfrac {yf'(x) - y^{2}}{f(x)}$ is:

0%
$f(x) = y + C$
0%
$f(x) = y(x + C)$
0%
$f(x) = x + C$
0%
None of the above

Explanation

$\dfrac {dy}{dx} = \dfrac {yf'(x) - y^{2}}{f(x)}$

$\Rightarrow yf'(x) dx - f(x) dy = y^{2}dx$

$\Rightarrow \dfrac {yf'(x) dx - f(x) dy}{y^{2}} = dx$

$\Rightarrow d\left \{\dfrac {f(x)}{y}\right \} = dx$
On integration, we get

$\dfrac {f(x)}{y} = x + C$

$\Rightarrow f(x) = y(x + C)$

Solution of

$\cfrac { dx }{ dy } +mx=0$ ,

$m< 0$ is

0%
$x=c{ e }^{ my }$
0%
$x=c{ e }^{ -my }$
0%
$x=my+c$
0%
$x=c$

Explanation

Given differential equation is

$\dfrac{dx}{dy}+mx=0$

$\Rightarrow \dfrac{dx}{x}=-mdy$

Now integrate both sides,

$\ln x=-my+C$

$\Rightarrow x=e^{-my+C}=e^{-my}\cdot e^C$

$\Rightarrow x=ce^{-my}$ , where

$c=e^C$

What is the general solution of the differential equation

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

0%
$\sin { y } =c\left( 1-{ e }^{ x } \right)$ where $c$ is the constant of integration
0%
$\cos { y } =c\left( 1-{ e }^{ x } \right)$ where $c$ is the constant of integration
0%
$\cot { y } =c\left( 1-{ e }^{ x } \right)$ where $c$ is the constant of integration
0%
None of the above

Explanation

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

$\Rightarrow \dfrac { { e }^{ x } }{ 1-{ e }^{ x } } dx+\dfrac {\sec ^{ 2 }{ y } }{ \tan y } dy=0$ .....

$(i)$

Now, to calculate

$\displaystyle \int { \dfrac { { e }^{ x } }{ 1-{ e }^{ x } } dx }$

Put

$1-{ e }^{ x }=t\Rightarrow -{ e }^{ x }dx=dt$

$\therefore \displaystyle \int { \frac { { e }^{ x } }{ 1-{ e }^{ x } } dx } =\int { -\frac { dt }{ t } } =-\ln t =-\ln(1-{ e }^{ x })$

Also,

$\displaystyle \int { \frac {\sec ^{ 2 }{ y } }{ \tan y } dy }$

Put

$p=\tan y\Rightarrow dp=\sec^{ 2 }{ y } dy$

$\therefore \displaystyle \int { \frac {\sec ^{ 2 }{ y } }{ \tan y } dy } =\int { \frac { dp }{ p } } =\ln p=\ln(\tan y)$

On integrating

$(i)$ , we get

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

Hence, D is correct.

The solution of

$\dfrac{dy}{dx}=\sqrt{1-x^2-y^2+x^2y^2}$ is:
where c is an arbitrary constant

0%
${\sin}^{-1}y={\sin}^{-1}x+c$
0%
$2{\sin}^{-1}y=\sqrt{1-x^2}+{\sin}^{-1}x+c$
0%
$2{\sin}^{-1}y=x \sqrt{1-x^2}+{\sin}^{-1}x+c$
0%
$2{\sin}^{-1}y=x \sqrt{1-x^2}+{\cos}^{-1}x+c$

Explanation

Solution:

Given,

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

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

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

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

or,

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

On integrating both sides, we get

$\int$

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

or,

$\sin^{-1}y=\cfrac12[x\sqrt{1-x^2}+\sin^{-1}x]+A$

or,

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

Hence, C is the correct option.

What is the general solution of the differential equation

${x}^{2}dy+{y}^{2}dx=0$ ?

0%
$x+y=c$ where $c$ is the constant of integration
0%
$xy=c$ where $c$ is the constant of integration
0%
$c(x+y)=xy$ where $c$ is the constant of integration
0%
None of the above

Explanation

${ x }^{ 2 }dy{ +y }^{ 2 }dx=0$

$\Rightarrow \dfrac { dy }{ { y }^{ 2 } } +\dfrac { dx }{ { x }^{ 2 } } =0$

On integrating, we get

$\Rightarrow -\dfrac { 1 }{ y } -\dfrac { 1 }{ x } +\dfrac { 1 }{ c } =0\\ \Rightarrow c(x+y)=xy$
Option C is correct

What is the solution of

$\frac{dy}{dx}=2y-1$ is :

0%
$y=\frac{1-e^{-2x}}{2}$
0%
$y=\frac{1+e^{2x}}{2}$
0%
$y=1+e^x$
0%
$y=\frac{1+e^{x}}{2}$

Explanation

Given :

$\cfrac{dy}{dx}=2y-1$ .....

$(i)$

Let

$2y-1=t$ .....

$(ii)$

Differentiating both sides with respect to

$x$ , we get

$2\cfrac{dy}{dx}=\cfrac{dt}{dx}$

Substituting the value of

$\cfrac{dy}{dx}$ in the original equation, we get

$\cfrac{dt}{2 dx}=t$

Rearranging terms, we get

$\cfrac{dt}{t}=2dx$

Integrating both sides, we get

$\int \cfrac{dt}{t}=\int 2dx$

$\Rightarrow \ln{t}=2x$

$\Rightarrow t={e}^{2x}$

Substituting

$t$ in

$(ii),$ we get

$2y-1={e}^{2x}$

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

What is the number of arbitrary constants in the particular solution of differential equation of third order ?

0%
0
0%
1
0%
2
0%
3

Explanation

$\textbf{Step 1: Defining}$

$\text{We know that a particular solution of a differential equation contains no constants}$

$\text{This is applicable for all order of differential equations}$

$\textbf{Hence, Number of arbitrary constants in the particular solution of a differential equation }$

$\textbf{of third order is 0}$

What is D equal to ?

0%
-1
0%
1
0%
-2
0%
None of the above

Explanation

${\textbf{Step-1: Finding the value of D. }}$

${\text{Given,}}$

$\Rightarrow(x^2+x+1)dy+(y^2+y+1)dx=0$

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

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

${\text{Integrating both sides}}$

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

${\text{By using perfect square method}}$

$\Rightarrow\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}=-\int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$

$\Rightarrow \int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}+\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$

${\text{use formula}}$

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

$\Rightarrow \dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2x+1}{2}]+\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2y+1}{2}]+\dfrac{2}{\sqrt3} tan^{-1}C=0$

${\text{[.... Where C is the integral constant]}}$

${\text{use formula}}$

$tan^{-1}A+tan^{-1}B$

$\Rightarrow\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{\dfrac{2x+1}{3}+\dfrac{2y+1}{\sqrt3}}{1-(\dfrac {2x+1}{\sqrt3})(\dfrac{2y+1}{\sqrt3})}]=-\dfrac{2}{\sqrt3}tan^{-1} C$

$\Rightarrow \dfrac{2x+1+2y+1}{\sqrt 3}\times \dfrac{\sqrt3 \sqrt3}{3-4xy-2x-2y-1}=C$

$\Rightarrow \dfrac{\sqrt3(2x+2y+2)}{2-4xy-2x-2y}=C$

$\therefore \sqrt3\times2(x+y+1)=c\times 2(1-2xy-x-y)$

$\therefore (x+y+1)=\dfrac{C}{\sqrt3}(1-x-y-2xy)$

${\text{ Compare this equation with the given equation gives,}}$

$\therefore D=-2$

${\textbf{Hence option C is correct }}$

What is B equal to ?

0%
-1
0%
1
0%
2
0%
None of the above

Explanation

${\textbf{Step-1: Finding the value of D. }}$

${\text{Given,}}$

$\Rightarrow(x^2+x+1)dy+(y^2+y+1)dx=0$

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

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

${\text{Integrating both sides}}$

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

${\text{By using perfect square method}}$

$\Rightarrow\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}=-\int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$

$\Rightarrow \int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}+\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$

${\text{use formula}}$

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

$\Rightarrow \dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2x+1}{2}]+\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2y+1}{2}]+\dfrac{2}{\sqrt3} tan^{-1}C=0$

${\text{[.... Where C is the integral constant]}}$

${\text{use formula}}$

$tan^{-1}A+tan^{-1}B$

$\Rightarrow\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{\dfrac{2x+1}{3}+\dfrac{2y+1}{\sqrt3}}{1-(\dfrac {2x+1}{\sqrt3})(\dfrac{2y+1}{\sqrt3})}]=-\dfrac{2}{\sqrt3}tan^{-1} C$

$\Rightarrow \dfrac{2x+1+2y+1}{\sqrt 3}\times \dfrac{\sqrt3 \sqrt3}{3-4xy-2x-2y-1}=C$

$\Rightarrow \dfrac{\sqrt3(2x+2y+2)}{2-4xy-2x-2y}=C$

$\therefore \sqrt3\times2(x+y+1)=c\times 2(1-2xy-x-y)$

$\therefore (x+y+1)=\dfrac{C}{\sqrt3}(1-x-y-2xy)$

${\text{ Compare this equation with the given equation gives,}}$

$\therefore B=-1$

${\textbf{Hence option A is correct }}$

What is the equation of a curve passing through (0, 1) and whose differential equation is given by dy = y tan x dx ?

0%
y = cos x
0%
y = sin x
0%
y = sec x
0%
y = cosec x

Explanation

Given :

$dy=y \ \tan x \ dx$

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

On integrating both sides, we get

$\Rightarrow \ln y \ = \ -\ln \cos x \ +c$

The equation satisfies

$(0,1)$

$\Rightarrow 0=0+c$

$\Rightarrow c=0$

We get

$y=\cfrac{1}{\cos x}$

$y=\sec x$

What is C equal to ?

0%
1
0%
-1
0%
2
0%
None of the above

Explanation

${\textbf{Step-1: Finding the value of D. }}$

${\text{Given,}}$

$\Rightarrow(x^2+x+1)dy+(y^2+y+1)dx=0$

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

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

${\text{Integrating both sides}}$

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

${\text{By using perfect square method}}$

$\Rightarrow\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}=-\int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$

$\Rightarrow \int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}+\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$

${\text{use formula}}$

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

$\Rightarrow \dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2x+1}{2}]+\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2y+1}{2}]+\dfrac{2}{\sqrt3} tan^{-1}C=0$

${\text{[.... Where C is the integral constant]}}$

${\text{use formula}}$

$tan^{-1}A+tan^{-1}B$

$\Rightarrow\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{\dfrac{2x+1}{3}+\dfrac{2y+1}{\sqrt3}}{1-(\dfrac {2x+1}{\sqrt3})(\dfrac{2y+1}{\sqrt3})}]=-\dfrac{2}{\sqrt3}tan^{-1} C$

$\Rightarrow \dfrac{2x+1+2y+1}{\sqrt 3}\times \dfrac{\sqrt3 \sqrt3}{3-4xy-2x-2y-1}=C$

$\Rightarrow \dfrac{\sqrt3(2x+2y+2)}{2-4xy-2x-2y}=C$

$\therefore \sqrt3\times2(x+y+1)=c\times 2(1-2xy-x-y)$

$\therefore (x+y+1)=\dfrac{C}{\sqrt3}(1-x-y-2xy)$

${\text{ Compare this equation with the given equation gives,}}$

$\therefore C=-1$

${\textbf{Hence option B is correct}}$

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.