how to make use of t' in tx = Ax vs general case

Question

I ran into this problem on my differential equations homework set. Previously, the questions were in the form x' = Ax, and they were relatively straight forward and easy to complete.

I was gifted a hint that this problem revolves around variable coefficients.

I'm uncertain how to approach this problem.

My guess is we work backwards from the eigenvalue, to acquire r. And we can get the equations in terms of those variables? Thank you

Answer 1

If $tx'(t)=Ax(t)$ and $x(t)=t^r \xi$, then $tx'(t)=rt^r \xi$, hence

$rt^r \xi=Ax(t)=t^r A \xi$. This gives $A \xi=r \xi$. If $ \xi \ne 0$, then $r$ is an eigenvalue of $A$ with corresponding eigenvector $ \xi$.

Answer 2

Set $t=e^s$, then $u(s)=x(e^s)$ satisfies $$ u'(s)=e^sx'(e^s)=Ax(e^s)=Au(s) $$ and you are back to a linear system with constant coefficients.

This is the same trick you can use to reduce Euler-Cauchy equations to linear ODE with constant coefficients.