Given the equation
$$
\begin{cases}
(t^2+1)x'(t) – y(t) = 0 \\
x(t) + (t^2+1)y'(t) = 0
\end{cases}
$$
Find an appropiate change of variables to transform the system into an autonomous one. In a previous part of the excercise we find that the non-null solutions are non-periodic, if that helps at all.
I've tried several different changes of variable, including polar coordinates, $(u,v) = (tx,ty), (t^2x,t^2y), (x/t,y/t)$ but none of them seem to work. I suspect that the change itself has to necessarily depend on $t$ because otherwise I can't imagine how the $t$'s would fade, but other than that I'm clueless.
Any hints are much appreciated.
Best Answer
$$\begin{cases} (t^2+1)x'(t) - y(t) = 0 \\ x(t) + (t^2+1)y'(t) = 0 \end{cases}$$ Try to substitute $u=\arctan t$ $$x'(t)=\dfrac {dx}{dt}=\dfrac {dx}{du}\dfrac {du}{dt}=\dfrac 1{t^2+1}\dfrac {dx}{du}$$
Then the system becomes: $$\begin{cases} x'(\tan u) - y(\tan u) = 0 \\ x(\tan u) + y'(\tan u) = 0 \end{cases}$$ Here $x'(u) $ means $\dfrac {dx}{du}$. You can also rewrite the system as: $$\pmatrix {x \\ y}'=\pmatrix {0 & 1 \\ -1 & 0}\pmatrix {x \\y}$$ $$X'=AX$$