Question

Find an implicit solution to the ODE. ("Homogeneous") y' = y^2x/y^3 +x^3 + y/x

Answer 1

I'm going to guess that you meant to include parentheses somewhere, so that the ODE is supposed to be

$y'=\dfrac{y^2x}{y^3+x^3}+\dfrac yx$

Then substitute $y(x)=xv(x)$ so that $y'(x)=xv'(x)+v(x)$. Then

$xv'+v=\dfrac{x^3v^2}{x^3v^3+x^3}+v$

$xv'=\dfrac{v^2}{v^3+1}$

which is separable as

$\dfrac{v^3+1}{v^2}\,\mathrm dv=\dfrac{\mathrm dx}x$

Integrate both sides: on the left,

$\displaystyle\int\frac{v^3+1}{v^2}\,\mathrm dv=\int\left(v+\frac1{v^2}\right)\,\mathrm dv=\dfrac12v^2-\dfrac1v$

The other side is trivial. We end up with

$\dfrac12v^2-\dfrac1v=\ln|x|+C$

Solve in terms of $y(x)$:

$\boxed{\dfrac{y^2}{2x^2}-\dfrac xy=\ln|x|+C}$