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If $p(x)$ is a polynomial with real coefficients, consider the following ODE:

$$A'(t)=p(A)$$ where $A$ is an $n \times n$ matrix.

My question is, can we have a general solutions for this kind of ODE? (For example, if $n=1$ we can definitely use the separation of variables)

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1 Answer 1

The solutions we get in the case $n=1$ tend to be implicit (not solved for $A$), and usually do not exist for all times. If this does not bother you, you can get something similar for general $n$. Indeed, let $F(z)=\int \frac{1}{p(z)}\,dz$ (this makes sense locally; in a larger scale one has to worry about branches of logarithms). Since $A'$ is of the form $p(A)$, it commutes with $A$. This justifies carrying out the usual chain rule: $\frac{d}{dt}\,F(A) = F'(A)\,A' = (p(A))^{-1}\,p(A)=I$. Therefore, $F(A)=tI+F(A_0)$ gives the solution of the IVP $A'=p(A)$, $A(0)=A_0$.

Illustrating on a simple example: $A'=A^2$. Here $F(A)=-A^{-1}$, hence $A(t)=(A_0^{-1}-tI)^{-1}=A_0(I-tA_0)^{-1}$. (The latter version works even if $A_0$ is noninvertible).

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