Question
Solve the following differential equation $y x \frac{ d y}{ d x}= x ^2+2 y ^2$

Answer

$ y x \frac{ d y}{ d x}= x ^2+2 y ^2$
$\therefore \frac{ d y}{ d x}=\frac{x^2+2 y^2}{x y}\ldots(i) $
Put $y=v x$$\ldots(ii)$
Differentiating w.r.t. $x$, we get
$\frac{ d y}{ d x}= v +x \frac{ dv }{ d x}\ldots(iii)$
Substituting (ii) and (iii) in (i), we get
$ v +x \frac{ dv }{ d x}=\frac{x^2+2 v ^2 x^2}{x( v x)}$
$\therefore v +x \frac{ dv }{ d x}=\frac{x^2\left(1+2 v ^2\right)}{x^2 v }$
$\therefore x \frac{ dv }{ d x}=\frac{1+2 v ^2}{ v }- v$
$=\frac{1+ v ^2}{ v }$
$\therefore \frac{ v }{1+ v ^2} dv =\frac{1}{x} d x $
Integrating on both sides, we get
$\frac{1}{\int} \frac{2 v }{1+ v ^2} dv =\int \frac{ dv }{x}$
$\therefore \frac{1}{2} \log \left|1+ v ^2\right|=\log | x |+\log | c |$
$\therefore \log \left|1+ c ^2\right|=2 \operatorname{og}| x |+2 \log | c |$
$=\log \left| x ^2\right|+\log \left| c ^2\right|$
$\therefore \log \left|1+ v ^2\right|=\log \left| c ^2 x ^2\right|$
$\therefore 1+ v ^2= c ^2 x ^2$
$\therefore 1+\frac{y^2}{x^2}= c ^2 x ^2$
$\therefore x ^2+ y ^2= c ^2 x ^4$

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