Wow. Cool problem. Putzer's algorithm is not that old, so it is not that well known.
But, having researched matrix exponentials, I think I may be able to help a wee bit.
By Putzer's algorithm, \(\displaystyle e^{At}=p_{1}(t)M_{0}+p_{2}(t)M_{1}\)
Where \(\displaystyle M_{0}=I, \;\ M_{1}=\begin{bmatrix}a-{\lambda}_{1}&b\\c&d-{\lambda}_{1}\end{bmatrix}\)
\(\displaystyle p_{1}(t)=e^{{\lambda}_{1}t}, \;\ p_{2}(t)=\frac{1}{{\lambda}_{1}-{\lambda}_{2}}\left(e^{{\lambda}_{1}t}-e^{{\lambda}_{2}t}\right)\)
Of course, I assume you know that the 2 lambdas represent the eigenvalues of matrix A.
The characteristic polynomial of said matrix is \(\displaystyle {\lambda}^{2}+4\). Which means the eigenvalues are \(\displaystyle {\lambda}_{1}=2i, \;\ {\lambda}_{2}=-2i\)
Now, proceeding. I am not going to go into every derivation and proof of why and/or how this works. You can probably find all that somewhere if you need it.
Therefore, from all of the above:
\(\displaystyle e^{At}=\begin{bmatrix}e^{{\lambda}_{1}t}+\frac{a-{\lambda}_{1}}{{\lambda}_{1}-{\lambda}_{2}}\left(e^{{\lambda}_{1}t}-e^{{\lambda}_{2}t}\right)&\frac{b}{{\lambda}_{1}-{\lambda}_{2}}\left(e^{{\lambda}_{1}t}-e^{{\lambda}_{2}t}\right)\\ \frac{c}{{\lambda}_{1}-{\lambda}_{2}}\left(e^{{\lambda}_{1}t}-e^{{\lambda}_{2}t}\right)& e^{{\lambda}_{1}t}+\frac{d-{\lambda}_{1}}{{\lambda}_{1}-{\lambda}_{2}}\left(e^{{\lambda}_{1}t}-e^{{\lambda}_{2}t}\right)\end{bmatrix}\)
Plug in a,b,c,d and the eigenvalues.
The matrix exponential is used to solve DE's. So, I suppose this is all that is needed because I see no initial values.
