Solve $$y'=\begin{pmatrix} 1 &2\\ -2&-3 \\ \end{pmatrix}y$$
I have found the eigenvalue $\lambda=-1$ (algebraic multiplicity of 2) and the eigenvector is $\begin{pmatrix} -1\\ 1 \\ \end{pmatrix}$ how should I approach the solution?
A system of ODE with algebraic multiplicity
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$\begingroup$
ordinary-differential-equations
1 Answers
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The solution of $y'=Ay$ is given by $$y(x)=e^{Ax}y(0)$$ where $e^{Ax}$ is a square matrix that is usually calculated from extracting Jordan form, i.e. $$A=PJP^{-1}\implies e^{Ax}=P\,e^{Jx}P^{-1}$$ and $P$ can be obtained from generalized eigenvectors, one of which you have already calculated: $$P=\pmatrix{v_1&v_2}=\pmatrix{-1&-1/2\\1&0}$$ Therefore: $$J=\pmatrix{-1&1\\0&-1}\implies e^{Jx}=\pmatrix{1&x\\0&1}e^{-x}$$