<s>[INST] 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 .
 [/INST] $$x^2ydx = ( x^3 + y^3)dy$$
$$\frac{dy}{dx}=\frac{x^2y}{x^3+y^3}$$
$$y'=\frac{y/x}{1+y^3/x^3}$$
now assume $y=ux$ or $u=\frac{y}{x}$
$$y'=\frac{u}{1+u^3}$$
this is first order homogeneous equation
$$y'=u+x\frac{du}{dx}$$
$$u+x\frac{du}{dx}=\frac{u}{1+u^3}$$
$$x\frac{du}{dx}=\frac{u}{1+u^3}-u$$
$$x\frac{du}{dx}=\frac{u-u-u^4}{1+u^3}$$
$$x\frac{du}{dx}=-\frac{u^4}{1+u^3}$$
$$-\frac{1+u^3}{u^4}du=\frac{dx}{x}$$
$$-(1/u^4+1/u)du=\frac{dx}{x}$$
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