# Transform into autonomous ODE

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).

• In I), the highest order derivatives $y''_1$ and $y''_2$ are functions of $y'_1$, $y'_2$, $y_1$, $y_2$ and $x$, hence the need for the variables $$u_0=x\quad u_1=y_1\quad u_2=y_2\quad u_3=y'_1\quad u_4=y'_2$$ In II), the highest order derivative $x''$ is a function of $x'$ and $x$ hence the need for the variables $$u=x\quad v=x'$$
– Did
Jul 5, 2018 at 17:47
• @Did Ah, okay. And the first entry in II is $y_2$ since $y_1' = y_2$?.
– user397268
Jul 5, 2018 at 17:52

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 $y'_1=x'=y_2\ne 1.$