Compute a fundamental matrix for a linear system of first-order differential equations

Let D = \begin{bmatrix} 1 & 2 & -4 & 4 \\ 2 & -1 & 4 & -8 \\ 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 3 \end{bmatrix}. Find a solution of the differential equation \frac{d}{dt} Y = DY which satisfies y_i(0) = 1 for i \in \{1,2,3,4\}.


We showed in this previous exercise that \mathsf{exp}(Dt)C is such a solution, where C = [1\ 1\ 1\ 1]^\mathsf{T}.

We will use the strategy we outlined in this previous exercise. Note that if P = \begin{bmatrix} 0 & 1 & 2 & 0 \\ 2 & 0 & -2 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}, then P^{-1}DP = B = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} is in Jordan canonical form.

Now \mathsf{exp}(Dt) = \mathsf{exp}(PBP^{-1}t) = P\mathsf{exp}(Bt)P^{-1} = P \left(\mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} t \oplus \mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} t \right) P^{-1} = P \begin{bmatrix} e^t & te^t & 0 & 0 \\ 0 & e^t & 0 & 0 \\ 0 & 0 & e^t & te^t \\ 0 & 0 & 0 & e^t \end{bmatrix} P^{-1}.

Evidently then Y = e^t \begin{bmatrix} 2t+1 \\ -4t+1 \\ -t+1 \\ t+1 \end{bmatrix} is such a solution. (Computations performed by WolframAlpha.)

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