Convert a nonlinear coupled system of equations to a linear system of equations

Question

Suppose we want to convert,
\begin{align*}
y_1’’ &= t^2-y_1’-y_2^2 &\quad \quad y_1(0)=0\quad y_1’(0)=1\\
y_2’’ &= t-y_2’-y_1^3 &\quad \quad y_2(0)=1\quad y_2’(0)=0
\end{align*}
into a first-order autonomous linear system of ODEs.

To convert to first-order and autonomous you can define,
\begin{align*}
x_1 &= y_1 \\
x_2 &= y_1’ \\
x_3 &= y_2 \\
x_4 &= y_2’ \\
x_5 &= t
\end{align*}
and we end up with
\begin{align*}
x_1’ &= x_2 \\
x_2’ &= x_5^2-x_2-x_3^2 \\
x_3’ &= x_4 \\
x_4’ &= x_5+x_4-x_1^3 \\
x_5’ &= 1
\end{align*}
But it remains to convert it to a linear system. How does one accomplish that? Can one simply define $x_6=y_2^2$ and $x_7=y_1^2$ and $x_8=y_1^3$?

Answer 1

It is not so easy. The thing with monomials in a function is that it's differential: $$(y(t)^n)' = y'(t)\cdot ny(t)^{n-1}$$

This is product between function and it's derivative. It is non-linear.

I think you need to find some representation where function concatenation with polynomial is linear. This is possible with for example Carleman matrices.

So our function y is represented $\bf M_y$, squaring represented by $\bf M_2$, Differentiation $\bf M_D$

Now we can write $(y(t)^2)'$ as $$\bf M_D M_2M_y$$

But I am not sure that such $\bf M_D$ exists for these Carleman matrices.