Help with a solution to $y'' = y^2$

Question

I think I have a solution to the ode $y'' = y^2$ but when I substitute my answer in it doesn't work out too well. My ansatz is that $y = e^{pt}$ from there we get:

$p^2e^{pt}-e^{2pt} = 0$

$e^{pt}(p^2-e^{pt}) = 0$

$e^{pt}$ cannot equal zero so $p^2-e^{pt}=0$

We can $p$ the subject of the formula with lambert's function.

$p=-2W_n(-t/2)/t$

Substituting this in we get $e^{-2W_n(-t/2)}$ but if we substitute this in the differential equation $y^2$ will not be equal to $y''$.

I did see that somebody else asked this question here:Solve $y''=y^2$ But I would like to know why the way I went about doing it is not right?

Answer 1

When you differentiated $e^{pt}$ you assumed that $p$ is costant with respect to $t$, then you can't substitute it with a function in $t$.

If $p(t)$ is a two-times differentiable function in $t$ and $y(t)=e^{p(t)}$ you have $$ y'(t)=p'(t)e^{p(t)}\\ y''(t)=p''(t)e^{p(t)}+[p'(t)]^2e^{p(t)} $$ and you have $$ p''(t)+[p'(t)]^2=e^{p(t)} $$

In other words you forgot the second-order term, for this reason your approach didn't work.

Answer 2

This is not an answer to the question, but a comment too long to be edited in the comments section.

$$y''=y^2$$ $$y''y'=y^2y'$$ $$(\frac12y'^2)'=(\frac13y^3)'$$ $$\frac12y'^2=\frac13y^3+\text{constant}$$ $$y'=\pm\sqrt{\frac23y^3+c}$$ $$dt=\pm\frac{dy}{\sqrt{\frac23y^3+c}}$$ $$t=\pm\int\frac{dy}{\sqrt{\frac23y^3+c}}$$ This is an elliptic integral. The inverse function $y(t)$ is a Weierstrass elliptic function : https://mathworld.wolfram.com/WeierstrassEllipticFunction.html