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Section 3.2 Method 1: Solving x′=Ax+g(t)

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Let's say A is diagonalizable. I'll also only focus on the case where A is a 2×2 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 \(\mathbf{v}_2\text{.}\))

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Since A is diagonalizable, we can write A=VDV−1 where

V=[v1v2]andD=[λ100λ2].

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Now, solving x′=Ax+g(t) follows from a few steps.

  1. Set x=Vy to obtain

    Vy′=AVy+g(t)

  2. Multiply by V−1 on both sides of the new ODE to get

    y=V−1AVy+V−1g(t).

  3. Since A=VDV−1, we have V−1AV=D. Also, set y(t)=[y1(t)y2(t)] and h(t)=V−1g(t)=[h1(t)h2(t)].

  4. These give us the linear system

    y1′(t)=λ1y1(t)+h1(t),y2′(t)=λ2y2(t)+h2(t).

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

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