## 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.)