MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 3

The solution of

$\dfrac{dy}{dx}=\dfrac{x^{2}+4x-9}{x+2}$ is:

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$y=(x+2)^{2}-13log\left | x+2 \right |+c$
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$y=(x+2)^{2}-5log\left | x+2 \right |+c$
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$y=\frac{x^{2}}{2}+2x+13log\left | x+2 \right |+c$
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$y=\frac{x^{2}}{2}+2x-13log\left | x+2 \right |+c$

Explanation

$\dfrac{dy}{dx}=\dfrac{(x+2)^{2}-13}{x+2}$

$\dfrac{dy}{dx}=(x+2)-\dfrac{13}{x+2}$

$\int dy=\int (x+2)dx-\int \dfrac{13}{x+2}dx$

$\Rightarrow y=\dfrac{x^{2}+4x}{2}-13\ log(x+2)+c$

The solution of

$(x^{2}-y^{2}x^{2})\dfrac{dy}{dx}+(y^{2}+x^{2}y^{2})=0$ is:

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

Explanation

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

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

$\displaystyle \Rightarrow \dfrac { 1 }{ y } +y=-\dfrac { 1 }{ x } +x+c\Rightarrow \left( x-y \right) -\left( \dfrac { 1 }{ x } +\dfrac { 1 }{ y } \right) =c$

The solution of

$\frac{dy}{dx}=e^{3x-2y}+x^{2}e^{-2y}$

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$y=log(e^{3x}+x^{2})+c$
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$y=2log(e^{3x}+x^{2})+c$
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$2e^{2x}=3(e^{3y}+y^{3})+c$
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$3e^{2y}=2(e^{3x}+x^{3})+c$

Explanation

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

$\Rightarrow \int e^{2y}dy=\int \left ( e^{3x}+x^{2} \right )dx+c$

$\Rightarrow \dfrac{e^{2}}{2}=\dfrac{e^{3x}}{3}+\ ^{x\ 3}/_{3}+c$

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

The solution of

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

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$e^{\sqrt{y-1}}=cx$
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$\sqrt{y-x}=ex$
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$\dfrac{1}{\sqrt{y-1}}=cx$
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$x(\sqrt{y-x})=c$

Explanation

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

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

$\Rightarrow \dfrac{(y-1)^{1}/_{2}}{^{1}/_{2}}=2log\ x+c$

$\Rightarrow 2(y-1)^{^{1}/_{2}}=2log\ x+log\ c$

$\Rightarrow (y-1)^{^{1}/_{2}}=log\ x\ c$

$\Rightarrow e^{\sqrt{y-1}}=cx$

The solution of

$x^{2}\dfrac{dy}{dx}=\sqrt{4-y^{2}}$ is:

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$sin^{-1}\left ( \dfrac{y}{2} \right )+\dfrac{1}{x}=c$
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$\dfrac{x^{3}}{3}+sin^{-1}\left ( \dfrac{y}{4} \right )=c$
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$\sin^{-1}\left ( \dfrac{y}{2} \right )+2x=c$
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$\sqrt{4-y^{2}}+2x=c$

Explanation

$\Rightarrow \dfrac{dy}{\sqrt{4-y^{2}}}=\dfrac{dx}{x^{2}}$

$\displaystyle\Rightarrow \int \dfrac{dy}{\sqrt{4-y^{2}}}=\int \dfrac{dx}{x^{2}}+c$

$\Rightarrow sin^{-1}({y/{2}})=\ {-1}/{x}+c$

$\Rightarrow sin^{-1}(\dfrac{y}{2})+{1}/{x}=c$

The solution of

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

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$\tan y=c(e^{x}-1)$
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$\tan ^{2}y=(e^{x}-1)c$
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$\cot y=c(e^{x}-1)^{2}$
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$\tan y=c(e^{x}-1)^{3}$

Explanation

$3e^{x}\cos^{2}ydx=(e^{x}-1)\cot\ ydy$

$\Rightarrow \dfrac{3e^{x}}{e^{x}-1}dx=\dfrac{1}{\tan y}\sec^{2}ydy$

Put

$e^{x}-1=t$ ;

$tan\ y=k$

$\Rightarrow e^{x}dx=dt$ ;

$\sec^{2}ydy=dk$

$\Rightarrow \int \dfrac{3dt}{t}+log\ c=\int \dfrac{dk}{k}$

$\Rightarrow log\ t^{3}c=log\ k$

$\Rightarrow k=ct^{3}$

$\Rightarrow \tan y=c(e^{x}-1)^{3}$

The solution of

$xcos^{2}ydx=ycos^{2}xdy$ is:

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$\tan x \tan y =c$
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$y \tan y =x \tan x + c$
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$\tan x \cos y = \tan y \cos x +c$
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$y \tan y - x \tan x + log \left( \dfrac{cosy}{cosx} \right )=c$

Explanation

$\Rightarrow \int x\ sec^{2}x\ dx=\int y\ sec^{2}ydy+c$

$\Rightarrow x\ tan\ x-\int tan\ x\ dx=y\ tan\ y-\int tan\ y+c$

$\Rightarrow x\ tan\ x-log\ sec\ x=y\ tan\ y-log\ sec\ y+c$

$\Rightarrow y\ tan\ y-x\ tan\ x+log\frac{cos\ y}{cos\ x}=c$

The solution of

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

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$ytan^{-1}x=c$
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$xtan^{-1}x=0$
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$y(1+x^{2})=c$
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$y^{2}(1+x^{2})=c$

Explanation

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

let

$tan^{-1}x=t$

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

$\Rightarrow \int \dfrac{dt}{t}+\int \dfrac{dy}{y}=log\ c$

$\Rightarrow log\ yt=log\ c$

$\Rightarrow yt=c$

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

The solution of

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

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

Explanation

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

Let

$1+x^{2}=t ; 1+e^{y}=k$

$\Rightarrow 2xdx=dt$ and

$e^{y}dy=dk$

$\Rightarrow\displaystyle \int {\frac{dk}{k}}=\int {\frac{dt}{t}}$

$\Rightarrow\displaystyle \log\frac{k}{t}=\log C$

Where,

$\log C$ is the constant of integration.

$\Rightarrow \dfrac kt = C$

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

Hence, option A.

Solution of

$(e^{x}+1)ydy+(y+1)dx=0$ is:

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$(y+1)(1+e^{-x})=ce^{y}$
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$(e^{x}+1)y=c$
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$(1+e^{x})(y+1)=c$
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$(e^{x}+1)x=c$

Explanation

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

$\Rightarrow \left ( \dfrac{y}{y+1} \right )dy+\left ( \dfrac{e^{-x}}{1+e^{-x}} \right )dx=0$

$\Rightarrow \int \left ( 1-\dfrac{1}{y+1} \right )dy-\int \left ( \dfrac{-e^{-x}}{1+e^{-x}} \right )dx=-c$

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

$y+c=log(y+1)(1+e^{-x})$

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

The solution of

$x\sqrt{1+y^{2}}dx+y\sqrt{1+x^{2}}dy=0$ is:

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$sinh^{-1}+sinh^{-1}y=c$
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$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$
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$(1+x^{2})(1+y^{2})=c$
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$\sqrt{\frac{1+x^{2}}{1+y^{2}}}=c$

Explanation

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

let

$t=\sqrt{1+x^{2}}$ ;

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

$\Rightarrow dt=\dfrac{x}{\sqrt{1+x^{2}}}dx$ ;

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

$\Rightarrow \int dt+\int dk=0$

$\Rightarrow t+k=c$

Hence, Option B is correct

Equation of the curve whose polar sub tangent

$r^{2}\displaystyle \frac{d\theta}{dr}$ is constant

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$r(\theta+c)+k=0$
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$r^{2}(\theta+c)=2k$
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$r(\theta-c)=k^{2}$
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$r\theta=c$

Explanation

Let the constant length of the polar subtangent be

$a$
Therefore as given in the question

$\displaystyle { r }^{ 2 }\frac { d\theta }{ dr } =a$
which is the differential equation of the required curve
To solve, put it into the form

$\displaystyle d\theta =a\left( \frac { 1 }{ { r }^{ 2 } } \right) dr\Rightarrow \int { d\theta } =\int { a\left( \frac { 1 }{ { r }^{ 2 } } \right) dr }$

$\displaystyle \Rightarrow \theta +c=-\frac { a }{ r } \Rightarrow r\left( \theta +c \right) +a=0$

The solution of

$ydx+xdy=dx+dy$ is:

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$xy=x+y+c$
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$x-y\dfrac{x}{y}+c=0$
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$xy-x+y=c$
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$x+y\dfrac{x}{y}+c=0$

Explanation

$\Rightarrow ydx+xdy=dx+dy$

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

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

$\Rightarrow log(y-1)=-log(x-1)+loge$

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

$\Rightarrow (x-1)(y-1)=c$

$xy=x+y+c$

The solution of

$\displaystyle \frac{dy}{dx}=\sqrt{1+x+y+xy}$ is:

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$\sqrt{1+y}=\sqrt{1+x}+c$
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$\frac{2}{3}\sqrt{1+y}=\left ( 1+x \right )^{\frac{1}{2}}+c$
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$3\sqrt{1+y}=\left ( 1+x \right )^{\frac{3}{2}}+c]$
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$\sqrt{1+x}=\sqrt{1+y}=c$

Explanation

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

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

$\Rightarrow 2\sqrt{y+1}=\dfrac{2}{3}(x+1)^{3/2}+c$

$\Rightarrow 3\sqrt{y+1}=(x+1)^{3/2}+c$

The solution of:

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

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

Explanation

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

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

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

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

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

The solution of

$\dfrac{dy}{dx}=\dfrac{ax+h}{by+k}$ represents a parabola when :

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$a\ne0,b=0$
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$a=1,b=2$
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$a=0,b\neq 0$
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$a=2,b=1$

The solution of

$cosec^{2}x\dfrac{dy}{dx}=\dfrac{1}{y}$ is:

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

Explanation

$\Rightarrow ydy=\sin^{2}xdx$

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

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

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

$\Rightarrow y^{2}=x-\sin x \cos x +c$
Hence, option 'A' is correct.

The solution of

$\displaystyle \frac{dy}{dx}=xy+2x-3y-6$ is:

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$(y+2)^{2}=c.e^{(x-3)^{2}}$
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$\log(y+2)=x^{2}-3x+c$
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$y+2=2(x-3)+c$
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$(y+2)(x-3)=c$

Explanation

$\dfrac{dy}{dx}-y(x-3)+2(x-3)$

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

$\Rightarrow \int \dfrac{dy}{y+2}=\int (x-3)dx+c_{1}$

$\Rightarrow log(y+2)=\dfrac{x^{2}}{2}-3x+c_{1}$

$\Rightarrow 2log(y+2)=(x-3)^{2}+c[c=c_{1}-9]$

$\Rightarrow (y+2)^{2}=e^{(x-3)^{2}}e^{c}$

$(y+2)^{2}=ce^{(x-3)^{2}}$

Solution of

$y-x\displaystyle \frac{dy}{dx}=3[1-x^{2}\frac{dy}{dx}]$ is:

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$(y+3)(1+3x)=cx$
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$(y-3)(1-3x)=cx$
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$(y-3)(1+3x)=cx$
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$(y+3)(1-3x)=cx$

Explanation

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

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

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

$\dfrac{dx}{(1-3x)}=\dfrac{dy}{y-3}$

$\Rightarrow \int \dfrac{3}{x-3x}+\dfrac{1}{x} dx=\int \dfrac{dy}{y-3} -log c$

$\Rightarrow -\log(1-3x)+\log x=\log(y-3)-\log c$

$\Rightarrow \log x c=\log(y-3)(1-3x)$

$\Rightarrow xc=(y-3)(1-3x)$

Solution of

$\displaystyle \frac{dy}{dx}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$ is:

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$\tan ^{ -1 }{ \left( \frac { 2x+1 }{ \sqrt { 3 } } \right) } +\tan ^{ -1 } (\frac { 2y+1 }{ \sqrt { 3 } } )=c$
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$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})+\sin^{-1}(\frac{2y+1}{\sqrt{3}})=c$
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$\displaystyle \sinh^{-1}(\frac{2x+1}{\sqrt{3}})+\sinh^{-1}(\frac{2y+1}{\sqrt{3}})=c$
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$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})=c$

Explanation

$\Rightarrow \dfrac{-dy}{(y+\dfrac{1}{2})^{2}+(\dfrac{\sqrt{3}}{2})^{2}} = \dfrac{dx}{(x+\dfrac{1}{2})^{2}+ (\dfrac{\sqrt{3}}{2})^{2}}$

$\Rightarrow -tan^{-1} \left(\dfrac{y+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right) = tan^{-1} \left (\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right) + c$

$\Rightarrow c= tan^{-1} \left(\dfrac{(2x+1)}{\sqrt{3}}\right) + tan^{-1} \left(\dfrac{2y+1}{\sqrt{3}}\right)$

The solution of

$e^{x-y}dx+e^{y-x}dy=0$ is:

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

Explanation

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

$\Rightarrow \int e^{2x}dx+ \int e^{2y}=c_{1}$

$\Rightarrow \dfrac{e^{2x}}{2}+\dfrac{e^{2y}}{2}=\dfrac{c_{1}}{2}$

$\Rightarrow e^{2x}+e^{2y}=2c_{1}=c$

General solution of

$\displaystyle \frac{dy}{dx}=\frac{1}{\log_{x}e}$ is given as

$y =$

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$\displaystyle \frac{1}{x}+c$
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$\displaystyle \frac{x^{2}}{2}+c$
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$x\log_{e}x-x+c$
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$\displaystyle \frac{x}{2}+c$

Explanation

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

$\Rightarrow {dy} = \dfrac{log x}{log e} dx$

$\Rightarrow \int dy = \int log_{e} x + c$

$\Rightarrow y = x (log_{e} x-1) + c$ $[\because\int logx=x(logx-1)]$

Solution of

$xe^{x^{2}+y}.dx=y.dy$ is:

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

Explanation

$xe^{x^{2}} dx = ye^{-y} dy$

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

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

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

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

$\frac{{dy}}{{dx}} = {e^{ - 2y}}$ and

$y = 0$ when

$x = 5$ , then value of x for

$y = 3$ is

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$e^5$
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$e^6+1$
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$\frac{{{e^6} + 9}}{2}$
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$log_e6$

Explanation

$\frac {dy}{dx}=e^{-2y}$

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

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

$\frac {2}{2}=5+c$

$c=\frac {-9}{2}$

$x=\frac {e^6+9}{2}$

The solution of

$\cos \mathrm{x}\cos \mathrm{y}\mathrm{d}\mathrm{x}+ \sin \mathrm{x} \sin \mathrm{y} d\mathrm{y} =0$ is

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$cosx=c\sin y$
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$\sin x=c\cos y$
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$\sec x-\sec y=c$
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$\tan x=c$

Explanation

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

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

$\Rightarrow log \sec y=- \log \sin x + \log c'$

$\Rightarrow log \sec y= \log cosec x + \log c'$

$\Rightarrow log \sec y= \log ( cosec x \times c')$

$\Rightarrow \dfrac{1}{\cos y} = \dfrac{1}{c' \sin x}$

$\sin x = c \cos y$

Solution of

$\displaystyle \frac{dy}{dx}=4+4x-3y-3xy$ is:

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$2\log(4-3y)+3x^{2}+6x=c$
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$\log(3-4y)+3y^{2}+6y=c$
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$(4-3y)(1+x)=c$
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$\log(4-3y)+x^{2}+3x=c$

Explanation

$\dfrac{dy}{dx} = 4(1+x)-3y(1+x)$

$\Rightarrow\displaystyle \int \dfrac{dy}{4-3y} = \int (1+x)dx + c$

$\Rightarrow -\dfrac{1}{3} log (\dfrac{4}{3}-y) = x+\dfrac{x^{2}}{2} + c$

$\Rightarrow c = \dfrac{2x+x^{2}}{2} + \dfrac{1}{3} log (\dfrac{4-3y}{3})$

$\Rightarrow c= 6x+3x^{2} + 2log (4-3y)$

The solution of

$\displaystyle \frac{dy}{dx}=xy+x+y+1$

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$y=\displaystyle \frac{x^{2}}{2}+xc$
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$x=\displaystyle \frac{y^{2}}{2}+y+c$
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$y+1=c.e(\displaystyle \frac{x^{2}+2x}{2})$
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$y+1=x(x+1)$

Explanation

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

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

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

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

The solution of

$\sin^{-1}ydx+\displaystyle \dfrac{x}{\sqrt{1-y^{2}}}dy=0$ is:

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$y\sin^{-1}x=c$
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$y=c\sin^{-1}x$
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$y=\displaystyle \sin(\frac{c}{x})$
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$x=c\mathrm{s}in\mathrm{y}$

Explanation

Given $\sin^{-1}ydx+\displaystyle \dfrac{x}{\sqrt{1-y^{2}}}dy=0$

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

$let \sin ^{-1}y = t$

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

$\Rightarrow -\int \dfrac{dx}{x} = \int \dfrac{dt}{t} - log c$

$\Rightarrow log c = log xt$

$\Rightarrow c=xt$

$\Rightarrow t = \dfrac{c}{x}$

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

$\Rightarrow y = sin (\dfrac{c}{x})$

lf the primitive of

$\displaystyle \frac{1}{f(x)}$ is equal to

$\log\{f(x)\}^{2}+c$ , then

$f(x)$ is:

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$x+d$
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$\displaystyle \frac{x}{2}+d$
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$\displaystyle \frac{x^{2}}{2}+d$
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$x^{2}+d$

Explanation

Given

$\int { \frac { 1 }{ f\left( x \right) } } dx={ \log { \{ f(x)\} } }^{ 2 }+C$

$\displaystyle \Rightarrow \int { \frac { 1 }{ f\left( x \right) } } dx=2{ \log { \{ f(x)\} } }+C$

$\displaystyle\Rightarrow \frac{1}{f(x)}=2 \times \frac{1}{f(x)} \times f'(x)$

$\Rightarrow f'(x)=\dfrac{1}{2}$

$\displaystyle \Rightarrow f(x) =\frac{1}{2}\int dx$

$\displaystyle \Rightarrow f(x) =\frac{x}{2}+d$

Find the solution of

$\displaystyle \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}$ .

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$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{3}+c$
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$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{4}+c$
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$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{4}+c$
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$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{3}+c$

Explanation

Given

$\displaystyle \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}$

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

$\Rightarrow \displaystyle e^{y}dy=\left ( e^{x}+x^{2} \right )dx.$
Now integrating both sides.

$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{3}+c.$

Find the solution of

$\displaystyle \left ( e^{y}+1 \right )\cos x dx+e^{y}\sin x dy=0$

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$\displaystyle \sin x\left ( e^{y}+1 \right )=c.$
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$\displaystyle \sin x\left ( e^{y}-1 \right )=c.$
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$\displaystyle \sin x\left ( 2e^{y}+1 \right )=c.$
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$\displaystyle \sin x\left ( 3e^{y}-1 \right )=c.$

Explanation

Given

$\displaystyle \left ( e^{y}+1 \right )\cos x dx+e^{y}\sin x dy=0$

$\displaystyle \therefore \frac{\cos x}{\sin x}dx+\frac{e^{y}}{e^{y}+1}dy=0.$

Integrating we get,

$\displaystyle \int\frac{\cos x}{\sin x}dx+\int\frac{e^{y}}{e^{y}+1}dy=0.$

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

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

Let

$f$ be the differentiable for all

$x$ , If

$f(1)=-2$ and

${f}'\left ( x \right )\geq 2$ for

$[1, 6]$ , then:

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$f(6)< 8$
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$f(6)\geq 8$
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$f(6)=5$
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$f(6)< 5$

Explanation

Let if possible

${f}'\left ( x \right )=2$ for

$\Rightarrow$

$f\left ( x \right )=2x+c$ (Integrating both side w.r.t. x)

$\therefore f(1)=2+c, -2=2+c$

$\Rightarrow c=-4$

$\therefore f(x)=2x-4$

$\therefore$

$f\left ( 6 \right )=2\times 6-4=8$

$\therefore$

$f\left ( 6 \right )\geq 8$

$\displaystyle x\cos ^{2}ydx=y\cos ^{2}x dy$

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$\displaystyle x\tan x+\log \sec x= y\tan y+\log \sec y+c.$
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$\displaystyle x\tan x+\log \sec x= y\tan y-\log \sec y+c.$
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$\displaystyle x\tan x-\log \sec x= y\tan y+\log \sec y+c.$
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$\displaystyle x\tan x-\log \sec x= y\tan y-\log \sec y+c.$

Explanation

Given

$\displaystyle x\cos ^{2}ydx=y\cos ^{2}x dy$

$\displaystyle x\sec 2 x dx=y \sec ^{2}y dy.$

$\displaystyle \int x\sec 2 x dx=\int y \sec ^{2}y dy.$
Integrate by parts, we get

$\displaystyle x\tan x-\int \tan x dx= y\tan y-\int \tan y dy+c$

$\Rightarrow \displaystyle x\tan x-\log \sec x= y\tan y-\log \sec y+c.$

$\displaystyle e^{x-y}dx+e^{^{y-x}}dy=0$
Solve the differential equations.

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$\displaystyle e^{2x}+e^{2y}=k$
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$\displaystyle e^{-2x}+e^{-2y}=k$
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$\displaystyle e^{-2x}+e^{-2y}=-k$
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$\displaystyle e^{2x}+e^{2y}=-k$

Explanation

$\displaystyle e^{x-y}dx+e^{^{y-x}}dy=0$
Multiplying by

$e^{x+y}$

$\Rightarrow e^{2x}dx+e^{2y}dy=0$
Now integrating we get,

$e^{2x}+e^{2y}=k$

Find the solution of

$\displaystyle \left ( e^{x}+1 \right )y dy=\left ( y+1 \right )e^{x}dx$ .

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$\displaystyle k\left ( y+1 \right )\left ( e^{x}+1 \right )=e^{y}$
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$\displaystyle k\left ( y-1 \right )\left ( e^{x}+1 \right )=e^{y}$
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$\displaystyle k\left ( y+1 \right )\left ( e^{x}-1 \right )=e^{y}$
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$\displaystyle k\left ( y-1 \right )\left ( e^{x}-1 \right )=e^{y}$

Explanation

Given

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

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

$\displaystyle \frac{1+y - 1}{1+y} dy=\frac{e^{x}}{1+e^x}dx$

$\displaystyle dy-\frac{1}{1+y} dy=\frac{e^{x}}{1+e^x}dx$

Integrating,

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

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

$\displaystyle k\left ( y+1 \right )\left ( e^{x}+1 \right )=e^{y}.$ Taking

$\displaystyle c= \log k.$

$e^{\frac{dy}{dx}}=x+1$ given that when x = 0, y = 3, if x=1, then

$y=\left ( 2 \right ).\ln \left (2 \right )+k$ , what is k?

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1
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2
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4
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5

Explanation

Given,

$\displaystyle { e }^{ \frac { dy }{ dx } }=x+1$

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

$\displaystyle \Rightarrow \int { \frac { dy }{ dx } dx } =\int { \log { \left( x+1 \right) } dx }$

$\Rightarrow y=-x+\log { \left( x+1 \right) +x\log { \left( x+1 \right) +c } }$
For

$x=0;y=3\Rightarrow 3=c$
For

$x=1;$

$y=-1+\log { 2 } +\log { 2 } +3=2\log { 2+\log { 2 } }$

$\therefore k=2$

The differential equation

$\displaystyle \frac{dy}{dx}=\frac{\sqrt{1-y^{2}}}{y}$ determines a family of circles with:

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variable radii and a fixed centre at $\displaystyle (1,1)$
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variable radii and a fixed centre at $\displaystyle (0, -1)$
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fixed radius 1 and variable centres along the x-axis
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fixed radius 1 and variable centres along the y-axis

Explanation

$\cfrac { dy }{ dx } =\cfrac { \sqrt { 1-y^{ 2 } } }{ y } \\ \Rightarrow \cfrac { dy }{ dx } \cfrac { y }{ \sqrt { 1-y^{ 2 } } } =1\\ \Rightarrow \int { \cfrac { y }{ \sqrt { 1-y^{ 2 } } } dy } =\int { dx } \\ \Rightarrow -\sqrt { 1-y^{ 2 } } =x+c\Rightarrow -y^{ 2 }+1=\left( x+c \right) ^{ 2 }\\ \Rightarrow \left( x+c \right) ^{ 2 }+y^{ 2 }=1$

Find the solution of

$\displaystyle \left ( 1-x \right )dy-\left ( 3+y \right )dx=0$

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$\displaystyle \left ( 3+y \right )\left ( 1-x \right )=k$
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$\displaystyle \left ( 3-y \right )\left ( 1-x \right )=k$
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$\displaystyle \left ( 3+y \right )\left ( 1+x \right )=k$
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$\displaystyle \left ( 3-y \right )\left ( 1+x \right )=k$

Explanation

Given,

$\displaystyle \left ( 1-x \right )dy-\left ( 3+y \right )dx=0$

$\Rightarrow \displaystyle \frac{dy}{3+y}=\frac{dx}{1-x}$
or

$\displaystyle \int \frac{dy}{3+y}=\int \frac{dx}{1-x}+c$
or

$\displaystyle \log \left ( 3+y \right )=-\log \left ( 1+x \right )+c,$
or

$\displaystyle \log \left ( 3+y \right )+\log \left ( 1-x \right )=c,$
or

$\displaystyle \log \left ( 3+y \right )\left ( 1-x \right )=c,or \log k.$

$\displaystyle \therefore \left ( 3+y \right )\left ( 1-x \right )=k$ is the required solution. c being arbitrary constant, can be put equal to

$\displaystyle \log k$ for the sake of simplification.

The solution of the equation

$\displaystyle \frac{d^{2}y}{dx^{2}}=e^{-2x}$ is:

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$\displaystyle \frac{1}{4}e^{-2x}$
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$\displaystyle \frac{1}{4}e^{-2x}+cx+d$
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$\displaystyle \frac{1}{4}e^{-2x}+cx^{2}+d$
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$\displaystyle \frac{1}{4}e^{-2x}+c+d$

Explanation

$\dfrac { d^{ 2 }y }{ dx^{ 2 } } =e^{ -2x }$
Integrating both sides w.r.t. x

$\dfrac { dy }{ dx } =\cfrac { e^{ -2x } }{ -2 } +c$
Again integrating w.r.t. x

$y=\dfrac { e^{ -2x } }{ 4 } +cx+d$

Find the solution of

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

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

Explanation

Given

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

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

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

Integrating we get,

$\displaystyle \therefore \log x-\frac{x^{2}}{2}=-\frac{y^{2}}{2}-2y-2\log \left ( 1-y \right )+k$

$\displaystyle ydx-x dy=xy\:dx$

Then the solution is:

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$\displaystyle x=ky\:e^{x}.$
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$\displaystyle y=kx\:e^{y}.$
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$\displaystyle x=y\:e^{x}+k$
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$\displaystyle y=x\:e^{y}+k$

Explanation

$\displaystyle ydx-x dy=xy\:dx$

$\Rightarrow \displaystyle y\left ( 1-x \right )dx=xdy$

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

Integrating we get,

$\ln x-x = \ln y+\ln k =\ln ky$

$\Rightarrow x=\ln\left(\tfrac{x}{ky}\right)$

$\Rightarrow \displaystyle x=ky\:e^{x}.$

Solve the given differential equation

$\displaystyle \left ( xy^{2}+x \right )dx+\left ( yx^{2}+y \right )dy=0.$

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$\displaystyle \left ( x^{2}+1 \right )\left ( y^{2}+1 \right )=c$
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$log\left( x^{ 2 }+1 \right) log\left( y^{ 2 }+1 \right) =c$
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$\displaystyle \left ( x^{2}+1 \right )+\left ( y^{2}+1 \right )=c$
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none of these

Explanation

Given,

$\displaystyle \left ( xy^{2}+x \right )dx+\left ( yx^{2}+y \right )dy=0.$

$\displaystyle \frac{2y}{1+y^2}dy =-\cfrac{2x}{a+x^2}dx$

Integrating we get,

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

$\Rightarrow \displaystyle \left ( x^{2}+1 \right )\left ( y^{2}+1 \right )=c$

Find the solution of

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

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$\displaystyle y-x=\log \frac{kx}{y}$
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$\displaystyle y+x=\log \frac{ky}{x}$
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$\displaystyle y+x=\log \frac{y}{kx}$
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$\displaystyle y-x=\log \frac{ky}{x}$

Explanation

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

$\displaystyle \frac{y+1}{y}dy=\frac{x+1}{x}dx,$
or

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

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

$\displaystyle y+\log y=x+\log x+\log k$

$\displaystyle \therefore y-x=\log \frac{kx}{y}$

Solve the diffrential equation:

$\displaystyle \log \frac{dy}{dx}=ax+by$

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$\displaystyle -\frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$
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$\displaystyle \frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$
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$\displaystyle -\frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$
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$\displaystyle \frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$

Explanation

$\displaystyle \log \frac{dy}{dx}=ax+by$

$\displaystyle \frac{dy}{dx}=e^{ax+by}=e^{ax}.e^{by}$
or

$\displaystyle e^{-by}dy=e^{ax}dx$
integrating we get,

$\displaystyle -\frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$

Find the solution of

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

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$\displaystyle \frac{1}{2}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$
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$\displaystyle \frac{1}{4}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$
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$\displaystyle \frac{1}{2}\log \left ( 1-y^{2} \right )=\log x+\tan ^{-1}x+c$
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$\displaystyle \frac{1}{4}\log \left ( 1-y^{2} \right )=\log x+\tan ^{-1}x+c$

Explanation

Given,

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

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

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

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

Find the solution of

$\displaystyle a\left ( x\frac{dy}{dx}+2y \right )=xy\frac{dy}{dx}$ .

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$\displaystyle yx^{2}=ke^{y/a}$
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$\displaystyle xy^{2}=ke^{y/a}$
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$\displaystyle yx^{3}=ke^{y/a}$
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$\displaystyle xy^{3}=ke^{y/a}$

Explanation

Given

$\displaystyle a\left ( x\frac{dy}{dx}+2y \right )=xy\frac{dy}{dx}$
or

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

$\displaystyle \frac{a}{y}dy+\frac{2a}{x}dx=dy$
Integrating,we get

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

$a \displaystyle \log y+2a \log x=y+c$
or

$\displaystyle \log yx^{2}=\frac{y+c}{a};$

$\displaystyle \therefore yx^{2}=e^{\left ( y+c \right )/a}=ke^{y/a}$

Find the solution of

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

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$\displaystyle \log \tan \frac{y}{4}=c-2\sin \frac{x}{2}$
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$\displaystyle \log \tan \frac{y}{2}=c+2\sin \frac{x}{2}$
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$\displaystyle \log \tan \frac{y}{2}=c-2\sin \frac{x}{2}$
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$\displaystyle \log \tan \frac{y}{4}=c+2\sin \frac{x}{2}$

Explanation

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

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

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

$\displaystyle \int cosec\frac{y}{2}dy=\int -2\cos \frac{x}{2}dx$
or

$\displaystyle \log \tan \frac{y}{4}=c-2\sin \frac{x}{2}$

Find the solution of

$\displaystyle \left ( x^{2}-yx^{2} \right )\frac{dy}{dx}+\left ( y^{2}+xy^{2} \right )=0$

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$\displaystyle \log \frac{x}{ky}=\frac{x+y}{xy}$
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$\displaystyle \log \frac{x}{ky}=\frac{x-y}{xy}$
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$\displaystyle \log \frac{x}{ky}=\frac{x-y}{y}$
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$\displaystyle \log \frac{x}{ky}=\frac{x+y}{y}$

Explanation

$\displaystyle x^{2}\left ( 1-y \right )\frac{dy}{dx}+y^{2}\left ( 1+x \right )=0$
or

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

$\displaystyle \left ( \frac{1}{y^{2}}-\frac{1}{y} \right )dy+\left ( \frac{1}{x^{2}}+\frac{1}{x} \right )dx=0.$
Integrating, we get

$\displaystyle \log x-\log y-\left ( \frac{1}{x}+\frac{1}{y} \right )=c=\log k$ (say)
or

$\displaystyle \log \frac{x}{ky}=\frac{x+y}{xy}$

$\displaystyle \frac{dy}{dx}+\sqrt{\left ( \frac{1-y^{2}}{1-x^{2}} \right )}=0.$
Solve the equation.

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

Explanation

$\displaystyle \frac{dy}{dx}+\sqrt{\left ( \frac{1-y^{2}}{1-x^{2}} \right )}=0.$

$\Rightarrow \displaystyle \frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}$
Integrating we get,

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

$y=ae^{-1/x}+b$ is a solution of

$\displaystyle\frac{dy}{dx}=\frac{y}{x^{2}}$ when

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$a=1,b=0$
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$a=3,b=1$
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$a=1,b=1$
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$a=2,b=2$

Explanation

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

$\Rightarrow \dfrac{dy}{y}=\dfrac{dx}{x^2}=x^{-2}dx$
Integrating both sides, we get

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

$\Rightarrow y = e^{-1/x+c}=e^ce^{-1/x}=ae^{-1/x}$
Where

$e^c = a=$ any constant.
Comparing it with given solution we get,

$b=0, a\in R$

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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