I have the differential equation $$y^2(y')+2xy'=y$$ The solution goes like this:
We let $q=\frac{dx}{dy}$ and we differentiate $$ 2x=yq-y^2q^{-2}\\ 2q=q-2yq^{-2}+(y+2y^2q^{-3})q'\\ q^3=-2y \text{ or }q=yq' $$
I really don't understand the last part. How did we get $q^3=-2y \text{ or }q=yq'$?
Thanks in advance!
Subtract $2q$ from the RHS and factor
$$ y(1 + 2yq^{-3})q' - (q+2yq^{-2}) = 0 $$
multiply through by $q^3$
$$ y(q^3 + 2y)q' - q(q^3+2y) = 0 $$ $$ (yq'-q)(q^3+2y) = 0 $$
Therefore $yq'-q = 0$ or $q^3+2y=0$