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Q. If $x\left(y-x\right) \frac{dy}{dx}=y\left(y+x\right)$, then

Differential Equations

Solution:

$\frac{dx}{dy}$$=\frac{y\left(y+x\right)}{x\left(y-x\right)}$
Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$
Now, given equation becomes
$\frac{xdv}{dx}+v=\frac{v\left(v+1\right)}{\left(v-1\right)}$
$\Rightarrow \frac{xdv}{dx}=\frac{v\left(v+1\right)}{v-1}-v=\frac{2v}{v-1}$
$\Rightarrow \frac{dx}{x}=\left(\frac{v-1}{2v}\right)dv$
$\Rightarrow 2\,ln\,x=v-ln\,v+ln\,c$
$\Rightarrow ln\,xy=\frac{y}{x}+ln\,c$
$\Rightarrow xy=ce^{\frac{y}{x}}$.