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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?

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1 Answer 1

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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.

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