I'm currently trying to solve this particular ODE:
$$(x^3 + xy^2)y' = y^3$$
but am having trouble finding the way to do so. I'm currently going through the textbook Advanced Engineering Mathematics 10e (Kreyszig) and the methods that I've studied are separating variables, using integrating factors, and using the Bernoulli equation. I've tried all three with this particular ODE and have been unsuccessful.
My initial thought was to change it into its differential form:
$$-y^3dx + (x^3 + xy^2)dy = 0$$
so that we can find the integrating factor to solve it. However, I've observed that finding the integrating factor isn't quite working out the way that the textbook describes (i.e. integrating factor $F$ should be of one variable rather than two) and am puzzled as to how to proceed from here.
Any tips or advice on how to solve this would be greatly appreciated. Thank you.
Best Answer
The terms that can combine to the derivative of a monomial are $-y^3\,dx+xy^2\,dy=x^2y^2\,d(y/x)$. This suggests to insert $y=xu$ into the ODE to get $$ x^3(1+u^2)(xu'+u)=x^3u^3\implies x(1+u^2)u'+u=0 $$ which is now separable. You can then, if you want, reconstruct the integrating factor that produces the integrable form in one step. This is more confirmation than solution method.
Trying to match the list of "classical methods", a change of dependency, $x(y)$ instead of $y(x)$, and $x'(y)=1/y'(x)$, gives the equation $$ y^3x'=y^2x+x^3, $$ which is now a Bernoulli equation. Set $u=x^{-2}$ to get the linear DE $$ y^3u'(y)=-2y^2u-2\\~\\ (y^2u(y))'=-\frac{2}y\implies \frac{y^2}{x^2}=-2\ln|y|+c $$