π Example 290. Solving a Simple Linear System.
Solve the system:
\begin{align*}
\frac{dx}{dt} \amp = x + y\\
\frac{dy}{dt} \amp = 4x + y
\end{align*}
Solution.
Write in matrix form \(\frac{dY}{dt} = AY\) with \(A = \begin{bmatrix}1 \amp 1 \\ 4 \amp 1\end{bmatrix}\text{.}\)
\begin{equation*}
\det \begin{bmatrix} 1-r \amp 1 \\ 4 \amp 1-r \end{bmatrix} = (1-r)^2 - 4 = r^2 - 2r - 3.
\end{equation*}
Set equal to zero: \(r^2 - 2r - 3 = 0\text{.}\) Factor:
\begin{equation*}
(r - 3)(r + 1) = 0 \quad \Rightarrow \quad r_1 = 3, \ r_2 = -1.
\end{equation*}
Find eigenvectors for each eigenvalue.
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For r = 3: Solve \((A - 3I)\vec{v} = 0\text{.}\)\begin{equation*} \begin{bmatrix} -2 \amp 1 \\ 4 \amp -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \end{equation*}From the first row: \(-2v_1 + v_2 = 0 \Rightarrow v_2 = 2v_1\text{.}\) Choose \(v_1 = 1\text{,}\) giving eigenvector \(\vec{v}_1 = \begin{bmatrix}1 \\ 2\end{bmatrix}\text{.}\)
-
For r = -1: Solve \((A + I)\vec{v} = 0\text{.}\)\begin{equation*} \begin{bmatrix} 2 \amp 1 \\ 4 \amp 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \end{equation*}From the first row: \(2v_1 + v_2 = 0 \Rightarrow v_2 = -2v_1\text{.}\) Choose \(v_1 = 1\text{,}\) giving eigenvector \(\vec{v}_2 = \begin{bmatrix}1 \\ -2\end{bmatrix}\text{.}\)
Write the general solution.
\begin{equation*}
Y(t) = C_1 \vec{v}_1 e^{3t} + C_2 \vec{v}_2 e^{-t}.
\end{equation*}
Substitute the eigenvectors:
\begin{equation*}
\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} =
C_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} e^{3t} +
C_2 \begin{bmatrix} 1 \\ -2 \end{bmatrix} e^{-t}.
\end{equation*}
Or written componentwise:
\begin{equation*}
x(t) = C_1 e^{3t} + C_2 e^{-t}, \qquad
y(t) = 2C_1 e^{3t} - 2C_2 e^{-t}.
\end{equation*}
