General solution of differntial equation $(x^2)(y)dx = ( (x^3) + (y^3))dy$ I tried this problem by many ways , i took $dy$ to other side and tried to form bernaulli equation but it is not helping and not giving me proper answer .
 A: Write it as
$$
\frac{dy}{dx}=\frac{x^2\,y}{x^3+y^3}.
$$
It is an equation of homogeneous type. Try the change
$$
y=x\,z.
$$
A: $$x^2ydx = ( x^3 + y^3)dy$$
$$\frac{dy}{dx}=\frac{x^2y}{x^3+y^3}$$
$$y'=\frac{y/x}{1+y^3/x^3}$$
now assume $y=ux$ or $u=\frac{y}{x}$
$$y'=\frac{u}{1+u^3}$$
this is first order homogeneous equation
$$y'=u+x\frac{du}{dx}$$
$$u+x\frac{du}{dx}=\frac{u}{1+u^3}$$
$$x\frac{du}{dx}=\frac{u}{1+u^3}-u$$
$$x\frac{du}{dx}=\frac{u-u-u^4}{1+u^3}$$
$$x\frac{du}{dx}=-\frac{u^4}{1+u^3}$$
$$-\frac{1+u^3}{u^4}du=\frac{dx}{x}$$
$$-(1/u^4+1/u)du=\frac{dx}{x}$$
A: Notice, we have $$x^2ydx=(x^3+y^3)dy$$
$$\frac{dy}{dx}=\frac{x^2y}{x^3+y^3}$$
$$\frac{dy}{dx}=\frac{\frac{y}{x}}{1+\frac{y^3}{x^3}}$$
Now, let $y=vx\implies \frac{dy}{dx}=v+x\frac{dv}{dx}$
$$v+x\frac{dv}{dx}=\frac{v}{1+v^3}$$
$$x\frac{dv}{dx}=\frac{-v^4}{1+v^3}$$
$$\frac{1+v^3}{v^4}\ dv=-\frac{dx}{x}$$
$$\left(\frac{1}{v^{4}}+\frac{1}{v}\right)\ dv=-\frac{dx}{x}$$
$$\int\left(\frac{1}{v^{4}}+\frac{1}{v}\right)\ dv=-\int \frac{dx}{x}$$
$$\frac{-1}{3v^3}+\ln|v|=-\ln|x|+C$$
$$\frac{x^3}{3y^3}-\ln\left|\frac{y}{x}\right|=\ln|x|+C_1$$
