Question

$(y^3-2x^{2}y)dx+(2{xy}^2-x^3)dy=0$

Answer 1

$(y^3-2x^{2}y)dx+(2x{y}^2-x^3)dy=0$

$\frac{dy}{dx}=-\frac{y(y^2-2x^2)}{x(2y^2-x^2)}$

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

Putting $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=-v\left[\frac{v^2-2}{2v^2-1}\right]$

$x\frac{dv}{dx}=\frac{-v^3+2v}{2v^2-1}-v$

$x\frac{dv}{dx}=\frac{-3v^3+3v}{2v^2-1}$

$-\frac{1}{3}\int \frac{2v^2-1}{v^3-v}dv=\int \frac{dx}{x}$

$\int \frac{2v^2-1}{v\left(v+1\right)\left(v-1\right)}dv=-3\int \frac{dx}{x}$

Let, $\frac{2v^2-1}{v\left(v+1\right)\left(v-1\right)}=\frac{A}{v}+\frac{B}{v+1}+\frac{C}{v-1}$

$2v^2-1=A\left(v+1\right)\left(v-1\right)+Bv\left(v-1\right)+Cv\left(v+1\right)$

Put $v=0$ we get $A=1$

Put $v=1$ we get $C=\frac{1}{2}$

Put $v=-1$ we get $B=\frac{1}{2}$

So, $\int \left(\frac{1}{v}+\frac{\frac{1}{2}}{v+1}+\frac{\frac{1}{2}}{v-1}\right)dv=-3\int \frac{dx}{x}$

$\log v+\frac{1}{2}\log |v+1|+\frac{1}{2}\log |v-1|=-3\log x+\log C$

$\log v\sqrt{v^2-1}=\log \frac{C}{x^3}$

$v\sqrt{v^2-1}=\frac{C}{x^3}$

Put $v=\frac{y}{x}$

We get, $xy\sqrt{y^2-x^2}=C$