Solve the differential equation : (x^2+y^2 ) d x-2 x y d y=0 - Vidyadip

Question

Solve the differential equation : $\left(x^2+y^2\right) d x-2 x y d y=0$

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Answer

Differential equation is
$(x^2 – y^2) dx + 2xy dy = 0$
$\Rightarrow \frac{d y}{d x}=\frac{-\left(x^2-y^2\right)}{2 x y}...(1)$
Which is a Homogeneous differential equation then put $y = Vx$ in $(1)$
$ \text { By (1) } \frac{d}{d x}(V x)=\frac{(V x)^2-x^2}{2 x(V x)}$
$ \Rightarrow V+x \frac{d V}{d x}=\frac{V^2-1}{2 V}$
$ \Rightarrow x \frac{d V}{d x}=\frac{V^2-1}{2 V}-V$
$ \Rightarrow x \frac{d V}{d x}=\frac{V^2-1-2 V^2}{2 V}$
$ \Rightarrow x \frac{d V}{d x}=\frac{-V^2-1}{2 V}$
$\Rightarrow \frac{2 V}{V^2+1} d V=-\frac{d x}{x}$
Now integrating both sides
$ \int \frac{2 V}{V^2+1} d V=-\int \frac{d x}{x}$
$ \Rightarrow \log \left( V ^2+1\right)=-\log x +\log c$
$ \Rightarrow \log \left(V^2+1\right)=\log \left(\frac{C}{x}\right)$
$ \Rightarrow V ^2+1=\frac{C}{x}$
$ \Rightarrow \frac{y^2}{x^2}+1=\frac{C}{x}$
$\Rightarrow y^2+x^2=C x$

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