Say you have the following system:
$y'=Ay$
$y_{1,2}'(t) =\begin{bmatrix}a&b\\c&d\end{bmatrix}y_{1,2}(t)$
$\mathcal{L}(y')=\mathcal{L}(\begin{bmatrix}a&b\\c&d\end{bmatrix}y)$
$sY[s]-y(0)=\begin{bmatrix}a&b\\c&d\end{bmatrix}Y[s]$
$sY[s]-(\begin{bmatrix}a&b\\c&d\end{bmatrix})Y[s]=y(0)$
$(Is-\begin{bmatrix}a&b\\c&d\end{bmatrix})Y[s]=y(0)$
$Y[s]=(Is-\begin{bmatrix}a&b\\c&d\end{bmatrix})^{-1}y(0)$
$Y[s]=(Is-\begin{bmatrix}a&b\\c&d\end{bmatrix})^{-1}y(0)$
$Y[s]=(\begin{bmatrix}(s-a)&b\\c&(s-d)\end{bmatrix})^{-1}y(0)$
$Y[s]=\frac{1}{(s-d)(s-a)}\begin{bmatrix}(s-d)&-b\\-c&(s-a)\end{bmatrix}y(0)$
$y(t)=\mathcal{L}^{-1}(\frac{1}{(s-d)(s-a)}\begin{bmatrix}(s-d)&-b\\-c&(s-a)\end{bmatrix}y(0))$
If the eigen values, $\lambda_{1,2}$, were real,
Why do I not get something along the lines of $y_1(t)=c_1 e^{\lambda_1t}+c_2 e^{\lambda_2t}$ and instead I get $y_1(t)=(c_1 e^{\lambda_1t}+c_2 e^{\lambda_2t})y(0)$ where $c_1$ and $c_2$ are actual values provided by the inverse laplace.
since $y(0)$ is just an unknown initial value, as long as it is just a constant does that mean this is the same thing?
Are there any bad assumptions I might be making?