Solution to a system of ODE:sConsider a system of linear ODE:s:x′(t)=Ax(t)+CHere A and C are coefficient matrices. If C=0, the solution is x(t)=exp⁡(Ax(t))x(0), where exp represents the matrix exponential. Can we write a similar solution for a general C?

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Pedro Taylor · Accepted Answer

Step 1Using the matrix exponential as integrating factor you get(e−Atx(t))′=e−AtCwhich can be integrated to (at first only for regular A, but in the end for all A)Step 2e−Atx(t)−x(0)=−A−1(e−At−I)C⇒x(t)=eAtx(0)+tϕ1(At)Cwhere ϕ1 is the matrix version of the function ϕ1(z)=ez−1z, continued with ϕ1(0)=1 as per its power series. Note that ϕ1, despite its singular definition, is an analytical or entire function, similar to the exponential.These modified exponentials, matrix phi functions or whatever name they got in-between, ϕ2(z)=ez−1−zz2 etc. occur also in exponential Runge-Kutta methods, thus they are also implemented in good numerical linear algebra libraries along the matrix exponential.

Solution to a system of ODE:s Consider a system of linear

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