Transforme $y'=-2x+2xy^2$ in an ODE of second oreder. I have the following Cauchy problem
$$(C):\begin{cases}y'=-2x+2xy^2\\ y(0)=\frac{1}{2}\end{cases}$$
1) Let $f$ a solution of $(C)$. Write an ODE of second order such that $f$ is solution.
2) Write the taylor expansion at second order of $f$.
My attempt
1) I have no idea how to do. I tried do a substitution like $z=1-y^2$ or $z=y^2$, but nothing. After, I was thinking that if$$f'=-2x+2xf^2$$ then $$f''=-2+2f^2+4xff'$$
which is on ODE of second order, but same as before, I didn't find an easy substitution that take of the term $f^2$. Any idea ?
I insist on the fact that I'm not looking for solving the ODE, I'm only looking for the answer of 1). Indeed, to solve it it's easy since $$(C)\iff \frac{y'}{(1-y)(1+y)}=2x$$
an thus, we can simply do a decomposition to conclude.
2) I think it's a consequence of the 1).

Let $y=\tanh \frac z 2$, then $$dy=\frac{dz}{1+\cosh z}$$ on the other hand $$(1+\cosh z) \left(1-\tanh ^2\frac{z}{2}\right)=2$$ therefore the equation $dy=-2x(1-y^2)dx$ becomes $$dz=-4xdx$$ or $$z''+4=0.$$
