I am attempting to solve the following first order linear system of differential equations using the eigenvalue /eigenvector approach and then by the matrix exponential approach.
$$ \mathbf{x}^{\prime}(t) =
\begin{pmatrix}
6 & 1 \\
4 & 3
\end{pmatrix}
\mathbf{x} (t)$$
First, using the eigenvalue, eigenvector approach (and letting the solution be in the form $\mathbf{X(t)}=\mathbf{K}e^{\lambda t}$, where $\lambda$ is the eigenvalue and $\mathbf{K}$ is the corresponding eigenvector), I get solutions
$$ \mathbf{X_1}=
\begin{pmatrix}
1 \\
-4
\end{pmatrix} {e^{2t}} \text{ and } \mathbf{X_2}=
\begin{pmatrix}
1 \\
1
\end{pmatrix} {e^{7t}}$$
which gives the homogeneous solution as
$$\begin{align} \mathbf{X}(t) &=
C_{1}\begin{pmatrix}
1 \\
-4
\end{pmatrix} {e^{2t}} +
C_{2}\begin{pmatrix}
1 \\
1
\end{pmatrix} {e^{7t}} \\
&= \begin{bmatrix}
e^{2t} & e^{7t} \\
-4e^{2t} & e^{7t}
\end{bmatrix}
\begin{bmatrix}
C_1 \\
C_2
\end{bmatrix}
\end{align} \nonumber$$
However, when I try to solve the same equation using the matrix exponential approach, where $\mathbf{X}(t)=e^{\mathbf{A}t}\mathbf{c}$:
$$\begin{align} e^{\mathbf{A}t} &= \mathcal{L}^{-1}\{(s\mathbf{I}-\mathbf{A})^{-1}\} \\
&= \mathcal{L}^{-1} \left\{\begin{bmatrix}
s-6 & -1 \\
-4 & s-3
\end{bmatrix}^{-1}\right\} \\
&= \mathcal{L}^{-1} \left\{\frac{1}{(s-2)(s-7)}\begin{bmatrix}
s-3 & 1 \\
4 & s-6
\end{bmatrix}\right\} \\
&= \mathcal{L}^{-1} \left\{\begin{bmatrix}
\frac{s-3}{(s-2)(s-7)} & (\frac{1}{(s-2)(s-7)} \\
\frac{4}{(s-2)(s-7)} & \frac{s-6}{(s-2)(s-7)}
\end{bmatrix}\right\}
\end{align} \nonumber$$
But the inverse Laplace transform is not agreeing with my original solution for $\mathbf{X}(t)$; As an example, the term in row 1, column 1 of the above matrix has an inverse Laplace transform of $\frac{4}{5}e^{7t} + \frac{1}{5}e^{2t}$.
Can someone please guide me as to where I am going wrong?
Best Answer
The solutions of this differential equation form a two-dimensional space. That space has many possible bases.
The columns of your eigenvector-based matrix $\begin{pmatrix}e^{2t}\\-4e^{2t}\end{pmatrix}$ and $\begin{pmatrix}e^{7t}\\e^{7t}\end{pmatrix}$ are one such basis.
The columns of the Laplace transform-based matrix $\begin{pmatrix}\frac45e^{7t}+\frac15e^{2t}\\ \frac45e^{7t}-\frac45e^{2t}\end{pmatrix}$ and $\begin{pmatrix}\frac15e^{7t}-\frac15e^{2t}\\ \frac15e^{7t}+\frac45e^{2t}\end{pmatrix}$ are another.
Both choices are valid; the only difference is what constants we multiply by to reach any particular solution.