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I am looking for an algorithm that gives the solution to the system of ODE $\vec{x}\,'=A\vec{x}.$ My text doesn't give one. I have an idea, but I would like some feedback in case I'm wrong.

(1) Find the eigenvalues $\lambda_i$ by solving $\det(A-\lambda I)=0$.

(2) (a) If $\lambda_i$ has mutiplicity $m$ as well as the eigenspace = $\text{span}(v_1,\ldots,v_m)$, then the corresponding term in the solution is $x_i=(c_1v_1+\cdots+c_mv_m)e^{\lambda_i}$.

(b) If $\lambda=a\pm bi$, then I can find an eigenvector $v$ for $\lambda = a+bi$ and the corresponding term in the solution is $A\text{Re}(ve^{\lambda t})+B\text{Im}(ve^{\lambda t})$

(c) If $\lambda$ has multiplicity $m$ but the corresponding eigenspace has dimension less than $m$, then I can find generalized eigenvectors and apply the method given by Git Gud in Finding a solution basis to find the corresponding term of the solution.

(3) Once I exhaust all eigenvalues, the general solution the sum of all the terms $x_i$ described above.

Am I correct?

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cap
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  • The missing phrase is Jordan Normal Form. – Will Jagy Feb 18 '16 at 04:44
  • @WillJagy I am not sure how that applies here. Do you know of any books or links that show how to solve 3x3 systems of ODE? – cap Feb 18 '16 at 05:01
  • see my answer at your other question, it is a beginning. For a single Jordan block, the diagonal part and the nilpotent part commute. Here is an inexpensive book that shows this: http://store.doverpublications.com/0486658465.html – Will Jagy Feb 18 '16 at 05:05

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