MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 2

$\dfrac{dy}{dx}=\dfrac{4x+2y+1}{x-2y+3}$ is a differential equation of the type:

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variable separable
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exact
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Homogeneous
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non homogeneous

Explanation

Homogeneous differential equations involve only derivatives of y and terms involving y, and they’re set to 0, as in this equation:

$\dfrac{d^4}{dx^4}+x\dfrac{d^2y}{dx^2}+y^2=02$

Non homogeneous differential equations are the same as homogeneous differential equations, except they can have terms involving only x (and constants) on the right side, as in this equation

$\dfrac{d^4}{dx^4}+x\dfrac{d^2y}{dx^2}+y^2=6x+3$

The solution of

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

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$y -1= c (x - 2)^{3}$
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$y +1= c (x - 2)^{3}$
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$y +1= c (x - 2)^{2}$
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$y -1= c (x - 2)^{2}$

Explanation

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

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

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

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

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

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

The solution of

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

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

Explanation

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

$\int (y-y^{2})dy=\int (x+x^{2})dx+c$

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

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

The solution of the differential equation

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

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$x + y = log \left (\displaystyle \frac{cy}{cx} \right )$
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$x + y = \log (cxy)$
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$x - y = \log \left ( \displaystyle \frac{cx}{y} \right )$
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$y - x = \log \left ( \displaystyle \frac{cx}{y} \right )$

Explanation

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

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

$\displaystyle \int (1+{1}/{y})dy=\int (1+{1}/{x})dx+c_{1}$

$\displaystyle \Rightarrow y+log\ y=x+log\ x+c_{1}$

$\displaystyle (y-x)+log{y}/{x}=c_{1}$

$\displaystyle \Rightarrow (y-x)=-log\ c_{2}-log{y}/_{x}$ ,

$c_{1}=-log\ c_{2}$

$\displaystyle =-log(\ {c_{2}y}/{x})$

$\displaystyle (y-x)=log(\ {c_{3}x}/{y})$ ,

$\displaystyle c_{3}={1}/{c_{2}}$

The solution of

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

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$\log \sin x = 2x + c$
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$\log \sin y = x + c$
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$\log \sin y = 2x + c$
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$\log \sin x = 2 y + c$

Explanation

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

$\int \cot\ y\ dy=\int dx+c$

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

The solution of the differential equation

$(1+x^{2})\displaystyle \frac{dy}{dx}=2 x \cot y$ , is:

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$\sec y = c (1+x^{2})$
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$\cos y = c (1+x^{2})$
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$\tan y = c (1+x^{2})$
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$\cot y = c (1+x^{2})$

Explanation

Given:

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

Rearranging this equation

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

Integrating both sides, we get

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

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

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

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

where,

$c$ is the constant of integration.

Hence, option A.

The solution of

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

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

Explanation

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

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

$e^{y}=e^{x}+c_{1}$

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

$e^{y}=e^{x}+c_{1}$

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

The solution of

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

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

Explanation

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

$\int ydy =\int (\dfrac{e^{x}}{1+e^{x}})dx+c$

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

The solution of

$\displaystyle \frac{dy}{dx}+y \tan x = 0$ is:

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$y = a \cos x$
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$y = a \sin x$
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$y = \log \cos x + c$
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$y = a \tan x + c$

Explanation

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

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

$\log\ y=\log\ c\ \cos\ x$

$\Rightarrow y=c\ \cos\ x$

The solution of

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

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

Explanation

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

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

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

$\Rightarrow \displaystyle tan^{-1}(\frac{y-x}{1+xy})=c_{1}$

$\displaystyle \Rightarrow tan\ c_{1}=(\frac{y-x}{1+xy})$

$let \ tan\ c_{1}=c$

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

$\displaystyle \Rightarrow y-x=(1+xy)c$

The solution of

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

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$\tan^{-1}y=\cos x+c$
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$\tan^{-1}y=\sin x+c$
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$y^{3}=\csc x +c$
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$\log\left ( 1+y^{2} \right )=\cos x+c$

Explanation

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

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

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

The solution of

$x^{3}dy-y^{3}dx=0$ is:

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$x^{4}-y^{4}=c$
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$x^{2}-y^{2}=c$
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$\dfrac{1}{x}-\dfrac{1}{y}=c$
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$\dfrac{1}{x^{2}}-\dfrac{1}{y^{2}}=c$

Explanation

$x^{3}dy-y^{3}dx=0$

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

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

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

The solution of

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

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

Explanation

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

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

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

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

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

The solution of

$\dfrac{d^{2}y}{dx^{2}}=xe^{x}+1$ is:

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$\displaystyle y=(x-1)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$
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$\displaystyle y=(x-2)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$
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$\displaystyle y=(x+2)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$
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$\displaystyle y=(x+2)e^{x}+C_{1}$

Explanation

$\dfrac{d}{dx}\left(\dfrac{dy}{dx} \right)=xe^{x}+1$

$\displaystyle \Rightarrow \int d\left(\dfrac{dy}{dx} \right)=\int (xe^{x}+1)dx+c_{1}$

$\displaystyle \Rightarrow \frac{dy}{dx}=xe^{x}-e^{x}+x+c_{1}$

$\displaystyle \int dy =\int ( xe^{x}-e^{x}+x+c_{1})dx+c_{2}$

$\displaystyle \Rightarrow \int dy =xe^{x}-e^{x}-e^{x}+\frac{x^{2}}{2}+c_{1}x+c_{2}$

$\displaystyle \Rightarrow y= xe^{x}-2e^{x}+\frac{x^{2}}{2}+c_{1}x+c_{2}$

$\displaystyle y=e^{x}(x-2)+\frac{x^{2}}{2}+c_{1}x+c_{2}$

The solution of

$x^{2}dy-y^{2}dx=0$ is:

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$\dfrac{1}{x}-\dfrac{1}{y}=c$
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$\dfrac{1}{x}+\dfrac{1}{y}=c$
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$x^{3}-y^{3}=c$
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$x^{2}-y^{2}=c$

Explanation

Given,

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

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

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

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

The solution of

$\dfrac{dy}{dx}=\text{cosech }y$ is:

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$x=\cosh y+c$
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$x=\text{sech}\ y+c$
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$x=\sinh y+c$
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$x=\coth y+c$

Explanation

$\dfrac{dy}{dx}=\text{cosech }y$

$\Rightarrow \dfrac{dy}{dx}=\dfrac1{\text{sinh }y}$

$\Rightarrow \sinh y\ dy =dx$

Integrating both sides, we get

$\displaystyle \int \sinh y\ dy=x+c$

$\Rightarrow \cosh y =x+c$

The solution of

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

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$x^{2}+y^{2}=c$
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$x^{2}-y^{2}=c$
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$x^{3}-y^{3}=c$
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$x^{3}+y^{3}=c$

Explanation

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

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

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

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

The solution of

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

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$1+y=c(1-x)$
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$(1-x)(1+y)=c$
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$(1-y)(1-x)=c$
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$(1+x)(1+y)=c$

Explanation

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

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

which is variable seprable form

take integration both side

we get,

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

$\log(1+y)(1-x)=logc$

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

The solution of

$\dfrac{dy}{dx}+y=1$ is:

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$x+\log\left | 1-y \right |=c$
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$y+\log\left | 1-x \right |=c$
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$x(1-y)=c$
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$y(1-x)=c$

Explanation

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

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

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

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

$\Rightarrow x+\log \left | 1-y \right |=c$

The solution of

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

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

Explanation

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

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

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

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

$\Rightarrow 3(x^{2}-y^{2})+2(x^{3}-y^{3})+c=0$

The solution of

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

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$sin^{-1}x+sin^{-1}y=c$
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$cot^{-1}x+cot^{-1}y=c$
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$Tan^{-1}x+Tan^{-1}y=c$
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$sinh^{-1}x+sinh^{-1}y=c$

Explanation

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

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

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

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

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

Solution of

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

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

Explanation

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

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

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

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

The solution of

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

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

Explanation

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

$\Rightarrow ydy=xdx$

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

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

The solution of

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

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$2x=log\left [ c\left ( 1+y^{2} \right ) \right ]$
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$x=cy^{2}$
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$c(1+y^{2})=x$
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$2y=log\left \{ c\left ( 1+x^{2} \right ) \right \}$

Explanation

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

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

$1+y^{2}= t \Rightarrow 2ydy=dt$

$\Rightarrow \dfrac{dt}{t}=2dx$

$\Rightarrow log t = 2x +c$

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

The solution of

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

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$e^{y}= c \sin x$
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$e^{y}= c \cos x$
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$e^{y}= c \csc x$
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$e^{y}= c \sec x$

Explanation

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

$dy=tanx.dx$

$dy=\dfrac{sinx}{cosx}.dx$

Let

$cosx=t$
Then

$sinx.dx=-dt$

Hence

$dy=\dfrac{-1}{t}dt$

$\int dy =\int \dfrac{-dt}{t}$

$y=-lnt+lnC$

$y=-ln(cosx)+lnC$

$y=ln(secx)+lnC$

$y=lnCsecx$

$e^{y}=C.secx$

The solution of

$x+y\dfrac{dy}{dx}=5$ is:

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$x^{2}-10x+y^{2}=c$
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$x^{2}-5x+y^{2}=c$
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$\dfrac{y^{2}}{2}=10x+x^{2}+c$
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$y^2=10x+x^{2}+c$

Explanation

Given,

$x+y\dfrac{dy}{dx}=5$

$\Rightarrow y dy=(5-x)dx$

$\Rightarrow \dfrac{y^{2}}{2}=5x-\dfrac{x^{2}}{2}+c$

$\Rightarrow y^{2}=10 x -x^{2} +c$

$\Rightarrow x^{2}-10x+y^{2}=0+c$

$\Rightarrow x^{2}-10x+y^{2}=c$

The solution of

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

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$e^{\displaystyle y}+e^{\displaystyle x}=c$
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$e^{\displaystyle -x}=e^{\displaystyle -y}+c$
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$e^{\displaystyle y-x}=c$
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$e^{\displaystyle y/x}=c$

Explanation

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

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

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

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

The solution of

$\displaystyle \dfrac{dy}{dx}=2y\tanh x$ is

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$cy=\sinh^{2}x$
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$cy=\sec h^{2}x$
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$cy=\cosh^{2}x$
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$cy=\coth^{2}x$

Explanation

$\dfrac{dy}{dx}=2 y \tan hx$

$\Rightarrow \int \dfrac{dy}{y}= 2 \int \tan hx dx$

$\Rightarrow \log y = 2 \log(\cos hx)-\log c$

$\Rightarrow \log cy=\log \cosh^2x$

$cy=\cosh^2x$

The solution of

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

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

Explanation

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

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

$\Rightarrow log c + 2 logx=log(1+y^{2})$

$\Rightarrow log Cx^{2}=log(1+y^{2})$

$\Rightarrow (1+y^{2})=Cx^{2}$

The solution of

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

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

Explanation

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

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

$\dfrac{2^{y}}{log _{2}}=\dfrac{2^{x}}{log_{2}}+c$

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

The solution of

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

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

Explanation

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

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

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

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

The solution of

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

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$log(xy)=\dfrac{1}{x}+\dfrac{1}{y}+c$
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$log y+\dfrac{1}{y}=x-\dfrac{1}{x}+c$
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$y-\dfrac{1}{y}=x-\dfrac{1}{x}+c$
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$log x+\dfrac{1}{x}=y-\dfrac{1}{y}+c$

Explanation

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

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

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

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

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

$\log { y } +\dfrac { 1 }{ y } =x-\dfrac { 1 }{ x } +c$
Hence, option 'B' is correct.

The solution of

$\tan x\ dy + \tan y\ dx$

$=$ 0

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$\tan x.\tan y= c$
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$\sec^{2}x+\sec^{2}y=c$
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$\sin x.\sin y$ $= c$
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$\cot x.\cot y$ $= c$

Explanation

$tan x dy + tan y dx = 0$

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

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

$\Rightarrow \dfrac{-dt}{t}= \dfrac{dk}{k}\ [t= sin\ y ; k= cos\ x]$

$\Rightarrow log\ C= log\ Kt$

$\Rightarrow C= sin\ x\ sin\ y$

The solution of

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

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

Explanation

$\dfrac{-e^{x}dx}{-e^{x}+1}= \dfrac{\sec^{2}ydy}{\tan\ y} ;\left [ \begin{matrix}put -(e^{x}-1)= K ; &\tan y = t \\ \Rightarrow -e^{x}dx= dK &\sec^{2}ydy= dt \end{matrix} \right ]$

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

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

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

The general solution of the differential equation

$\dfrac{dy}{dx}-\dfrac{2xy}{1+x^{2}}=0$ is

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

Explanation

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

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

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

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

Solution of

$\tan y.\sec^{2}xdx + \tan x.\sec^{2} ydy=0$ is:

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$\sec x \sec y=c$
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$\tan x. \tan y=c$
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$\sin x . \sin y=c$
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$\cos x .\cos y=c$

Explanation

$tan\ y\ sec^{2}xdx+tan\ x\ sec^{2}y\ dy= 0$

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

let

$\begin{matrix}tan \ x = t ; &tan \ y = k \\\Rightarrow sec^{2}x\ dx = dt &\Rightarrow sec^{2}y\ dy = k \end{matrix}$

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

$\Rightarrow log \ kt= c$

$\Rightarrow kt= c$

$\Rightarrow tan\ x\ tan\ y= c$

The solution of

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

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

Explanation

$let x= tan\theta ; y= tan \delta$

$\Rightarrow dx= \sec^{2}\theta\ d \theta ; dy = sec^{2}\delta\ d\ \delta$

$\Rightarrow tan\theta \ sec^{3}\theta d\theta +tan\delta sec^{2}\delta d \delta= 0$

let

$\begin{matrix}sec\theta =t ; &sec \delta =k \\\Rightarrow sec\theta d \theta = dt \ \ \ \ | &\Rightarrow sec\ \delta\ tan\ \delta\ d\ \delta = dk\end{matrix}$

$\Rightarrow \int t^{2}dt+\int k^{2}dk= c$

$\Rightarrow \dfrac {t^{3}+k^{3}}{3}= c$

$\Rightarrow sec^{3}\theta +sec^{3}\delta= c$

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

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

The solution of

$x^{2}+y^{2}\dfrac{dy}{dx}=4$ is

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$x^{2}+y^{2}=12x+c$
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$x^{2}+y^{2}=3x+c$
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$x^{3}+y^{3}=3x+c$
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$x^{3}+y^{3}=12x+c$

Explanation

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

$y^{2}dy = 4dx-x^{2}dx$

$\Rightarrow \dfrac{y^{3}}{3}= 4x-\dfrac{x^3}{3}+ \ ^{c}/_{3}$

$\Rightarrow x^{3}+y^{3}= 12x+c$

The solution of

$\dfrac{dy}{dx}=e^{2x-y}+x^{3}e^{-y}$ is:

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$e^{y}=e^{2x}+3x^{2}+c$
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$e^{y}=e^{2x}+\dfrac{x^{4}}{4}+c$
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$4e^{y}=2e^{2x}+x^{4}+c$
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$2e^{y}=e^{2x}+3x^{4}+c$

Explanation

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

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

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

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

Solve the differential equation:

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

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$x\sqrt{1+x^{2}}+y\sqrt{1+y^{2}}+log\left [ \left ( x+\sqrt{1+x^{2}} \right )\left ( y+\sqrt{1+y^{2}} \right ) \right ]=c$
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$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$
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$\dfrac{1}{\sqrt{1+x^{2}}}+\dfrac{1}{\sqrt{1+y^{2}}}=c$
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$log\left \{ \left ( \sqrt{1+x^{2}} \right )+\left ( \sqrt{1+y^{2}}\right ) \right \}=x+c$

Explanation

Given,

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

It is known that

$\int \sqrt{x^{2}\pm a^{2}}dx=\dfrac{1}{2}x\sqrt{x^{2}\pm a^{2}}\pm\dfrac{a^{2}}{2}ln|x+\sqrt{x^{2}\pm a^{2}}$

Applying this formula gives us

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

$\Rightarrow \dfrac{x}{2}\sqrt{x^{2}+1}+\dfrac{1}{2}ln|x+\sqrt{x^{2}+1}|=-[\dfrac{y}{2}\sqrt{y^{2}+1}+\dfrac{1}{2}ln|y+\sqrt{y^{2}+1}|]+C$

$\Rightarrow x\sqrt{x^{2}+1}+y\sqrt{y^{2}+1}+ln|x+\sqrt{x^{2}+1}|+ln|y+\sqrt{y^{2}+1}|=C$

$\Rightarrow x\sqrt{x^{2}+1}+y\sqrt{y^{2}+1}+ln|(x+\sqrt{x^{2}+1})(y+\sqrt{y^{2}+1})|=C$

The solution of

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

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

Explanation

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

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

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

The solution of

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

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

Explanation

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

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

$\therefore \frac{d}{dx}\sqrt{1-x^{2}}=\dfrac{-x}{\sqrt{1-x^{2}}}$

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

The solution of

$xdx+ydy=x^{2}ydy-xy^{2}dx$ is

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$x^{2}-1=c(1+y^{2})$
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$x^{2}+1=c(1-y^{2})$
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$x^{3}-1=c(1+y^{3})$
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$x^{3}+1=c(1-y^{3})$

Explanation

$\Rightarrow x\ dx+xy^{2}dx=x^{2}y\ dy-ydy$

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

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

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

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

The solution of

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

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$1+y^{2}=cx$
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$1-y^{2}=cx$
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$1+x^{2}=cy$
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$1-x^{2}=cy$

Explanation

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

$Let\ t\ =1+y^{2}$

$dt=2ydy$

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

$\Rightarrow log\ t\ =log\ x+log\ c$

$\Rightarrow t=x\ c$

$\Rightarrow (1+y^{2})=cx$

The solution of

$xcos^{2} y(dx) + tan y(dy)=0$ is:

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$x^{2}+sec^{2}y=c$
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$x^{2}+cot^{2}y=c$
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$x^{2}+sin^{2}y=c$
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$x^{2}+cos^{2}y=c$

Explanation

$x\ cos^{2}y\ dx+tan\ ydy=0$

$\Rightarrow xdx+tan\ y\ sec^{2}ydy=0$

Let tan y = t

$\Rightarrow sec^{2}y\ dy=dt$

$\int xdx+\int tdt=c_{1}$

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

$x^{2}+tan^{2}y=2c_{1}$

$x^{2}+sec^{2}y=2c_{1}+1$ put

$(2c_{1}+1)=c$

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

The solution of

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

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$be^{\displaystyle ax}+ae^{\displaystyle -by}=k$
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$e^{\displaystyle ax}+e^{\displaystyle -by}=c$
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$e^{\displaystyle ax+by}=c$
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$(ax+by)=cxy$

Explanation

$log\left( \dfrac { dy }{ dx } \right) =ax+by\\ \dfrac { dy }{ dx } ={ e }^{ ax+by }\\ \dfrac { dy }{ dx } ={ e }^{ ax }.{ e }^{ by }\\ \int { { e }^{ -by }dy } =\int { { e }^{ ax }dx } +c\\ \Rightarrow \dfrac { { e }^{ -by } }{ -b } =\dfrac { { e }^{ ax } }{ a } +c$

$\Rightarrow c= \dfrac{e^{ax}}{a}+\dfrac{e^{-by}}{b}$

$\Rightarrow be^{ax}+ae^{-by}= k$

The solution of

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

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$2y=x^{2}\left [ logx+\dfrac{1}{2} \right ]+c$
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$2y=x^{2}\left [ logx-\dfrac{1}{2} \right ]+c$
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$y=\dfrac{x^{2}}{2}(log2-x)+c$
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$y^{2}=x^{2} log x + x + c$

Explanation

$\Rightarrow \int dy=\int x\ log\ x+c$

$\int x\ log\ x\ dx=x^{2}(log\ x-1)-\int (x\ log\ x-x)dx$

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

$\int x\ log\ x\ dx=\dfrac{x^{2}(\dfrac{log\ x^{-1}}{2})}{2}$

$\Rightarrow \int dy=\int x\ log\ x+c$

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

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

Solution of

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

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$(x^{2}+1)(y^{2}+1)=c$
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$(xy+1)(xy-1)=c$
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$(x^{3}+1)(y^{3}+1)=c$
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$(1-x^{2})(1-y^{2})=c$

Explanation

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

$\Rightarrow \left ( \dfrac{2x}{x^{2}+1} \right )dx+\left ( \dfrac{2y}{y^{2}+1} \right )dy= 0$
let

$\begin{matrix}t= x^{2}+1 > &k= y^{2}+1 \\\Rightarrow dt= 2xdx &\Rightarrow dk=2ydy \end{matrix}$

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

$\Rightarrow kt = c$

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

The solution of

$\sqrt{1-x^{2}}sin^{-1}xdy+ydx=0$ is:

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$ytan^{-1}x=c$
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$ysin^{-1}x=c$
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$ycos^{-1}x=c$
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$xsin^{-1}x=c$

Explanation

$\sqrt{1-x^{2}}sin^{-1}xdy+ydx= 0$

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

let

$sin^{-1}x= t$

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

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

$\Rightarrow log \ yt= log\ c$

$yt= c$

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

The solution of

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

$y=0$ when

$x=0$ is:

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$e^{3x}+3e^{-y}=4$
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$e^{-y}=e^{3x}+4$
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$3e^{-y}=e^{3x}+12$
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$y=\dfrac{e^{3x+y}}{3}$

Explanation

$\int e\ ^{-y}dy=\int e\ ^{3x}dx+c$

$\Rightarrow \dfrac{e\ ^{-y}}{-1}=\dfrac{e\ ^{3x}}{3}+c$

$\Rightarrow 3e\ ^{-y}+e\ ^{3x}=c$
Put

$(x,y)=(0,0)$

$\Rightarrow 3+1=4=c$

$\Rightarrow 3e\ ^{-y}+e\ ^{3x}=4$

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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