Change of variable to transform non-autonomous linear ODE to autonomous

Question

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.

Answer 1

$$ \begin{cases} (t^2+1)x'(t) - y(t) = 0 \\ x(t) + (t^2+1)y'(t) = 0 \end{cases}\quad\implies\quad \begin{cases} \frac{dx}{dt} = \frac{y}{t^2+1} \\ \frac{dy}{dt} = -\frac{x}{t^2+1} \end{cases}\quad\implies\quad \frac{dy}{dx} =-\frac{x}{y} $$ $$ydy+xdx=0\quad\implies\quad d(y^2+x^2)=0$$ $$y^2+x^2=C=\text{constant}$$ The real solutions are the circles with center $(0,0)$.

If you want $x(t)$ solve $\frac{dx}{dt} = \frac{\pm\sqrt{C-x^2}}{t^2+1}$ which is separable. An similary for $y(t)$.

Answer 2

$$\begin{cases} (t^2+1)x'(t) - y(t) = 0 \\ x(t) + (t^2+1)y'(t) = 0 \end{cases}$$ 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: $$\pmatrix {x \\ y}'=\pmatrix {0 & 1 \\ -1 & 0}\pmatrix {x \\y}$$ $$X'=AX$$