I am working out of a DE textbook and am in the Matrix Exponential method for first order ODEs. I want to confirm if I am thinking about this correctly.
I am using the definition: $e^{At}$ = $\sum\limits_{k=0}^{\infty }A^{k}\frac{t^{k}}{k!}$
For my problem I have: X' = $\begin{bmatrix}4& 3\\-4& -4\end{bmatrix}$X
(X' = A X)
Here, $k = 1$
$\sum\limits_{k=0}^{\infty }A^{k}\frac{t^{k}}{k!}$ = I + At = $\begin{bmatrix}1& 0\\0& 1\end{bmatrix}$ + $\begin{bmatrix}4& 3\\-4& -4\end{bmatrix}$t = $\begin{bmatrix}1+4t& 3t\\-4t& 1-4t\end{bmatrix}$
The general solution using this method for a homogeneous equation is: X = $e^{At}$C
X = $\begin{bmatrix}1+4t\\-4t\end{bmatrix}$C1 + $\begin{bmatrix}3t\\1-4t\end{bmatrix}$C2
When I use the method of finding Eigenvalues with corresponding Eigenvectors I get:
X = C1 $\begin{bmatrix}-3\\2\end{bmatrix}$$e^{2t}$ + C2 $\begin{bmatrix}-1\\2\end{bmatrix}$$e^{-2t}$ (This is the solution provided by the book.)
Can you please let me know where I am making my mistake, or if they are equivalent answers?
$\exp(At)$ my matrix diagonalization. If $D^{-1} A D= \Lambda$, $\Lambda=diag(\lambda_1, \lambda_2)$ then $$A=D \Lambda D^{-1}, f(A)=D f(\Lambda) D^{-1}$$. In your case $$\Lambda=\begin{pmatrix} -2 & 0 \\ 0 &2 \end{pmatrix}, D= \begin{pmatrix} -1 & -3 \\ 2 & 2 \end{pmatrix}, e^{\Lambda t}= \begin{pmatrix} e^{2t} & 0 \\0 & e^{-2t}\end{pmatrix}$$ Then $$ e^{At}=\frac{1}{4}\begin{pmatrix} 2(-e^{-2t}+3e^{2t}) & -3e^{-2t}+3e^{2t} \\ 4(e^{-2t}+e^{2t}) & 2(3e^{-2t}-e^{2t}) \end{pmatrix}.$$ Does this help?