Solution to Cauchy Problem

Question

I am trying to solve the following Cauchy Problem:

$y'(t) = A(t)y(t), A=\begin{pmatrix} t &-1 \\ 1 &t \end{pmatrix}, y(0)=y_0$

What I did:

I know that $ \forall\ t, s \in\ \mathbb{R}: A(s)A(t)=A(t)A(s)$

So the solution would be:

$y(t)= e^{ \begin{pmatrix} t^2/2 &-t \\ t &t^2/2 \end{pmatrix}} y_0$

Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.

Many thanks!

Answer 1

You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^\prime$ commute.