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Section 3.2 Method 1: Solving \(\mathbf{x}' = A \mathbf{x} + \mathbf{g}(t)\)

Let's say \(A\) is diagonalizable. I'll also only focus on the case where \(A\) is a \(2 \times 2\) matrix.

Technically, a matrix \(A\) is diagonalizable when there exists an invertible matrix \(V\) and diagonal matrix \(D\) such that \(A = VDV^{-1}\text{.}\)

This is a bit of an outlandish definition. But, know that \(A\) is diagonalizable if \(A\) has two unique eigenvalues.

Let's say the unique eigenvalues of \(A\) are \(\lambda_1\) and \(\lambda_2\text{,}\) and that the corresponding eigenvectors are \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\) Then,

\begin{align*} A \amp = V D V^{-1}, \end{align*}

where

\begin{align*} V = \begin{bmatrix} \mathbf{v}_1 \amp \mathbf{v}_2 \end{bmatrix} \qquad \text{and} \qquad D = \begin{bmatrix} \lambda_1 \amp 0 \\ 0 \amp \lambda_2 \end{bmatrix}. \end{align*}

(\(V\) is the matrix whose first column is \(\mathbf{v}_1\) and whose second column is \(\mathbf{v}_2\text{.}\))

Since \(A\) is diagonalizable, we can write \(A = VDV^{-1}\) where

\begin{align*} V = \begin{bmatrix} \mathbf{v}_1 \amp \mathbf{v}_2 \end{bmatrix} \qquad \text{and} \qquad D = \begin{bmatrix} \lambda_1 \amp 0 \\ 0 \amp \lambda_2 \end{bmatrix}. \end{align*}

Now, solving \(\mathbf{x}' = A \mathbf{x} + \mathbf{g}(t)\) follows from a few steps.

  1. Set \(\mathbf{x} = V \mathbf{y}\) to obtain

    \begin{gather*} V\mathbf{y}' = A V\mathbf{y} + \mathbf{g}(t) \end{gather*}

  2. Multiply by \(V^{-1}\) on both sides of the new ODE to get

    \begin{gather*} \mathbf{y} = V^{-1}AV \mathbf{y} + V^{-1} \mathbf{g}(t). \end{gather*}

  3. Since \(A = VDV^{-1}\text{,}\) we have \(V^{-1}AV = D\text{.}\) Also, set \(\mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix}\) and \(\mathbf{h}(t) = V^{-1} \mathbf{g}(t) = \begin{bmatrix} h_1(t) \\ h_2(t) \end{bmatrix}\text{.}\)

  4. These give us the linear system

    \begin{align*} y_1'(t) \amp = \lambda_1 y_1(t) + h_1(t),\\ y_2'(t) \amp = \lambda_2 y_2(t) + h_2(t). \end{align*}

  5. Both of these can be solved using integrating factors. Once we have \(\mathbf{y}(t)\text{,}\) remember to get the general solution in \(\mathbf{x} = V \mathbf{y}\text{.}\)

That's pretty much all you need to know to solve \(\mathbf{x}' = A \mathbf{x} + \mathbf{g}(t)\text{.}\) Of course, it's really important to remember that \(A\) must be diagonalizable.