<s>[INST] How can we apply the forward Euler method to $x''=x^2$? If we want to apply the forward Euler method to $x''=x$ with $x(0)=0, x'(0)=1$, we can introduce a new function $$u:=\begin{bmatrix}x'\\x \end{bmatrix}$$
then $$u'=\begin{bmatrix}0&1\\1&0\\ \end{bmatrix}u \ \ \text{and} \ \ u(0)=\begin{bmatrix}1\\0 \end{bmatrix}.$$
So then first step of the forward Euler would be
\begin{align*} 
u_1 & = u(0)+hu'(0)
 \\ & =  \begin{bmatrix}1\\0 \end{bmatrix} + h \begin{bmatrix}0&1\\1&0\\ \end{bmatrix}\begin{bmatrix}1\\0 \end{bmatrix}
 \\ & = \begin{bmatrix}1\\h \end{bmatrix}.
\end{align*}
And if we translate it back we get $x_1=h$.
How can we do a similar thing with the equation $x''=x^2$ (with $x(0)=0, x'(0)=1$)? It doesn't look like it can be written as a system, so there should be some other trick.
 [/INST] Let $u = \begin{bmatrix} x \\ x'\end{bmatrix}$. Then, $u' = \begin{bmatrix} x' \\ x''\end{bmatrix} = \begin{bmatrix} x' \\ x^2\end{bmatrix}$ with $u(0) = \begin{bmatrix} 0 \\ 1\end{bmatrix}$. 
So, taking forward Euler steps yields: 
$u(h) \approx u(0) + hu'(0) = \begin{bmatrix} 0 \\ 1\end{bmatrix} + h\begin{bmatrix} 1 \\ 0^2\end{bmatrix} = \begin{bmatrix} h \\ 1\end{bmatrix}$
$u(2h) \approx u(h) + hu'(h) = \begin{bmatrix} h \\ 1\end{bmatrix} + h\begin{bmatrix} 1 \\ h^2\end{bmatrix} = \begin{bmatrix} 2h \\ 1+h^3\end{bmatrix}$
$u(3h) \approx u(2h) + hu'(2h) = \begin{bmatrix} 2h \\ 1+h^3\end{bmatrix} + h\begin{bmatrix} 1+h^3 \\ (2h)^2\end{bmatrix} = \begin{bmatrix} 3h+h^4 \\ 1+5h^3\end{bmatrix}$
and so on. The only difference is that $u'$ is a non-linear function of $u$. 
</s>