Can somebody please help me in finding the general solution of $2x^3dy=y(y^2+3x^2)dx$ using another method?
My attempt for the first method:
$2x^3y^{-3}dy-3x^2y^{-2}dx=dx$
Let $v=y^{-2}$; $dv=-2y^{-3}dy$
The equation now becomes:
$-x^3dv-3x^2vdx=dx$
$\frac{dv}{dx}+\frac{3v}{x}=-\frac{1}{x^3}$
$P(x)=\frac{3}{x}$, I.F.$=e^{3\int x^{-1}dx}=x^3$
$vx^3=-\int \frac{x^3}{x^3}dx$
$vx^3=-x+C$
$x^3=y^2(C-x)$
\begin{gather*} \frac{dy}{dx} =\frac{y^{3}}{2x^{3}} +\frac{3y}{2x}\\ \\ Let\ y=tx\\ \frac{dy}{dx} =t+x\frac{dt}{dx}\\ t+x\frac{dt}{dx} =\frac{t^{3}}{2} +\frac{3t}{2}\\ x\frac{dt}{dx} =\frac{t^{3} +t}{2}\\ \\ \int \frac{2dt}{t^{3} +t} =\int \frac{dx}{x}\\ \\ Now,\ \int \frac{2dt}{t^{3} +t} =\int \frac{2dt}{t\left( t^{2} +1\right)} =\int \frac{2tdt}{t^{2}\left( t^{2} +1\right)}\\ \\ Let\ t^{2} =u\\ 2tdt=du\\ \\ \int \frac{2dt}{t^{3} +t} =\int \frac{du}{u( u+1)} =\ln\frac{u}{u+1} =\ln\frac{t^{2}}{t^{2} +1} +C\\ \\ So,\ form\ the\ original\ differential\ equation,\\ \ln\frac{t^{2}}{t^{2} +1} +C_{1} =\ln x+C_{2}\\ \ln\frac{y^{2}}{y^{2} +x^{2}} -\ln x=C\\ \\ So,\ the\ general\ solution\ is\ \\ y^{2} =e^{c} x\left( x^{2} +y^{2}\right) \end{gather*} Hope this answers your question!