I have a question about the transformation from a non-autonomous ODE into an autonomous ODE. In my lecture there were two examples:
I) The two dimensional ODE
\begin{align} y_1'' – y_1 + y_z &= cos(x), \\
y_2'' + x^2y_1' – y_1 y_2 ' &= 0.
\end{align} Write this as an autonomous ODE $y' = g(y)$ of first order and define $g$.
To this end they said $x' = 1, y_1' = v_1, y_2' = v_2$ and defined $g$ by
\begin{align}
g(x,y_1,y_2,v_1,v_2) \mapsto (1, v_1, v_2, y_1-y_2 + cos(x), y_1v_2 – x^2 v_1)^t.
\end{align}
The second example was
II) $x'' -a(1-x^2)x' +x = 0$, where $a > 0$. Here they said $y_1 = x, y_2 = x'$ and defined the function $g$ by
\begin{align}
g(y_1,y_2)= (y_2, a(1-y_1^2)y_2 -y_1)^t.
\end{align}
I don't see how these two techniques are somewhat similar? I that in the transformation one always ones $1$ in the first entry of the vector like in I).
Best Answer
In both cases the transformed system is a first order autonomous system.
You are correct in having a $1$ in the first spot of the first system because $x$ is an independent variable so $x'=1$, but in the second one $ x$ is a dependent variable.
That is why in the second system, $y_1=x$ which makes $y'_1=x'=y_2\ne 1.$