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CGPET 2019 Solved Paper

Section: Mathematics

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Question : 116 of 150

Marks: +1, -0

The solution of the differential equation

\frac{d y}{d x}=\frac{x y}{x^{2}+y^{2}}

is

$[$
$c'^{2}=c]$
$c y^{2}=e^{\frac{x^{2}}{y^{2}}}$
$c y=e^{\frac{x}{y}}$

Solution:

\frac{d y}{d x}=\frac{x y}{x^{2}+y^{2}}

.......(i)

Put

y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}

From Eq. (i)

v+x \frac{d v}{d x}=\frac{(x)(v x)}{x^{2}+v^{2} x^{2}}=\frac{x^{2} v}{x^{2}+v^{2} x^{2}}

=\frac{x^{2} v}{x^{2}(1+v^{2})}

\Rightarrow x \frac{d v}{d x}=\frac{v}{1+v^{2}}-v=\frac{v-v-v^{3}}{1+v^{2}}

=\frac{-v^{3}}{1+v^{2}}

\Rightarrow x \frac{d v}{d x}=\frac{-v^{3}}{1+v^{2}}

\Rightarrow \frac{1+v^{2}}{v^{3}} d v=-\frac{d x}{x}

On integration both sides

\Rightarrow \int \frac{1+v^{2}}{v^{3}} d v=-\int \frac{d x}{x}

\Rightarrow \int \frac{1}{v^{3}} d v+\int \frac{1}{v} d v=-\int \frac{d x}{x}

\Rightarrow -\frac{1}{2 v^{2}}+\log v=-\log x-\log c'

\Rightarrow \frac{-x^{2}}{2 y^{2}}+\log \frac{y}{x}=-\log x-\log c'

\Rightarrow \frac{-x^{2}}{2 y^{2}}+\log \frac{y}{x}+\log x=-\log c'

\Rightarrow \frac{-x^{2}}{2 y^{2}}+\log\left(\frac{y}{x} \cdot x\right)=-\log c'

\{\because \log m+\log n=\log m n\}

\Rightarrow \log y+\log c'=\frac{x^{2}}{2 y^{2}} \Rightarrow \log (y \cdot c')

=\frac{x^{2}}{2 y^{2}}

\Rightarrow y c'=e^{\frac{x^{2}}{2 y^{2}}} \Rightarrow \log (y \cdot c')

=\frac{x^{2}}{2 y^{2}}

\Rightarrow y^{2} c'^{2}=e^{\frac{x^{2}}{y^{2}}}

Let

\Rightarrow y^{2} c=e^{\frac{x^{2}}{y^{2}}}

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