Solve $2y+(x^2 y + 1)xy'=0$. The book that contains this problem suggests a way of solving such differential equations - do a substitution $y=z^m$, $dy=m z^{m-1}dz$ with such $m$ that the differential equation becomes homogenous. Here $m=-2$ works. However if I substitute $y=z^{-2}$ I lose all information about solutions with $y \leq 0$.
Let's say I do this substitution nonetheless.
$$2z^{-2} + (x^2z^{-2}+1)x(-2)z^{-3}z'=0$$
$$z^{-2}dx = (x^3z^{-5} +x z^{-3})dz$$
Now it's homogenous. Substitute $t=\frac{x}{z}, dx=tdz+zdt$ $$tdz+zdt = (t^3+t)dz$$ $$\frac{dz}{z}=\frac{dt}{t^3}$$ Note that we lost $t=x=0$, but it's not a solution. $$\ln{|z|} = -\frac{1}{2t^2} + C,$$ Substitute back: $$\ln{\frac{1}{\sqrt{y}}} = -\frac{1}{x^2 y} + C.$$
Check whether $y=0$ is a solution - it is. But the book containing this problem says the solutions are $x^2 y \ln{Cy}=1$ and $y=0$. It allows for $y$ to be negative, but our solution does not. How to go about it?
Making $z = \frac{1}{x^2y}$ and substituting we obtain
$$ 2y+(x^2y+1)x y'= 0 \equiv 2z+x(1+z)z'=0 $$
and now
$$ \frac{(z+1)}{2z}dz = - \frac{dx}{x} $$
etc.