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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!

Conjecture
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1 Answers1

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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.

mathcounterexamples.net
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  • Thanks for useful link. I guess the condition $\forall\ t, s\ in\ \mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ? – Conjecture Feb 09 '18 at 14:11
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    Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards. – mathcounterexamples.net Feb 09 '18 at 15:01