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).

I think you are fine with
$$
f''=−2+2f^2+4xff'\text{ resp. } y''=−2+2y^2+4xyy'
$$
if there are no further requirement to the second order ODE. You could add multiples of the original ODE for variety, but that is not necessary.
Substituting the initial value for $f(0)=y(0)$ in the initial equation gives you $f'(0)=y'(0)$, inserting in this equation allows to compute $f''(0)=y''(0)$ and thus the required Taylor polynomial.
