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Section 7.5 Applications I: Differential Equations

Definition 7.5.1.

A linear system of differential equations is a system of \(n\) coupled first-order linear differential equations of the form:
\begin{align*} x_1'(t) \amp = a_{11}x_1(t) + a_{12}x_2(t) + \cdots + a_{1n}x_n(t)\\ x_2'(t) \amp = a_{21}x_1(t) + a_{22}x_2(t) + \cdots + a_{2n}x_n(t)\\ \amp \vdots \\ x_n'(t) \amp = a_{n1}x_1(t) + a_{n2}x_2(t) + \cdots + a_{nn}x_n(t) \end{align*}
where \(x_1(t), x_2(t), \ldots, x_n(t)\) are unknown functions of the independent variable \(t\text{,}\) and \(a_{ij}\) are constant real coefficients.
This system can be written compactly in matrix form as:
\begin{equation*} \mathbf{x}'(t) = A\mathbf{x}(t) \end{equation*}
where the coefficient matrix \(A\) and the vector function \(\mathbf{x}(t)\) are defined as:
\begin{equation*} A = \begin{bmatrix} a_{11} \amp a_{12} \amp \cdots \amp a_{1n} \\ a_{21} \amp a_{22} \amp \cdots \amp a_{2n} \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ a_{n1} \amp a_{n2} \amp \cdots \amp a_{nn} \end{bmatrix}, \quad \mathbf{x}(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \\ \vdots \\ x_n(t) \end{bmatrix}, \quad \mathbf{x}'(t) = \begin{bmatrix} x_1'(t) \\ x_2'(t) \\ \vdots \\ x_n'(t) \end{bmatrix} \end{equation*}
The system is homogeneous (no external forcing terms) and has constant coefficients (the matrix \(A\) does not depend on \(t\)). Each entry \(a_{ij}\) represents the coefficient of function \(x_j(t)\) in the \(i\)-th differential equation.

Example 7.5.2. Simple Decoupled System.

Solve the system of differential equations:
\begin{align*} x_1'(t) \amp = 2x_1(t)\\ x_2'(t) \amp = 3x_2(t) \end{align*}
Solution.
It follows from calculus class that \(x_1(t)=c_1 e^{2t}\) and \(x_2(t)=c_2 e^{3t}\text{,}\) so the general solution to this system is:
\begin{gather*} \mathbf{x}(t) = \begin{bmatrix} c_1 e^{2t} \\ c_2 e^{3t} \end{bmatrix} \end{gather*}

Example 7.5.3. Coupled System of Differential Equations.

Solve the system of differential equations:
\begin{align*} y_1'(t) \amp = 2y_1(t) - 12y_2(t)\\ y_2'(t) \amp = y_1(t) - 5y_2(t) \end{align*}
Solution.
We can write this system in matrix form as \(\mathbf{y}'(t) = A\mathbf{y}(t)\) where
\begin{equation*} A = \begin{bmatrix} 2 \amp -12 \\ 1 \amp -5 \end{bmatrix} \end{equation*}
In Example 7.2.3, we computed that
\begin{equation*} A=PDP^{-1}, \text{ where } P = \begin{bmatrix} 4 \amp 3 \\ 1 \amp 1 \end{bmatrix} \text{ and } D = \begin{bmatrix} -2 \amp 0 \\ 0 \amp 5 \end{bmatrix} \end{equation*}
Thus we have
\begin{align*} \mathbf{y}'(t)\amp = PDP^{-1}\mathbf{y}(t)\\ P^{-1}\mathbf{y}'(t)\amp = DP^{-1}\mathbf{y}(t)\\ \left(P^{-1}y(t)\right)'\amp = DP^{-1}\mathbf{y}(t)\text{.} \end{align*}
Let \(\mathbf{z}(t) = P^{-1}\mathbf{y}(t)\text{.}\) Then:
\begin{align*} \mathbf{z}'(t) \amp = D\mathbf{z}(t) \end{align*}
\begin{equation*} \begin{bmatrix} z_1(t) \\ z_2(t) \end{bmatrix} = \begin{bmatrix} c_1e^{-2t} \\ c_2e^{5t} \end{bmatrix} \end{equation*}
Therefore \(\mathbf{y}(t) = P\mathbf{z}(t) = \begin{bmatrix} 4 \amp 3 \\ 1 \amp 1 \end{bmatrix} \begin{bmatrix} c_1e^{-2t} \\ c_2e^{5t} \end{bmatrix}=\begin{bmatrix} 4c_1e^{-2t} + 3c_2e^{5t} \\ c_1e^{-2t} + c_2e^{5t} \end{bmatrix}.\)
Note that \(\mathbf{y}(t) = c_1e^{-2t}\begin{bmatrix} 4 \\ 1 \end{bmatrix} + c_2e^{5t}\begin{bmatrix} 3 \\ 1 \end{bmatrix}.\)

Example 7.5.5. Coupled System of Differential Equations.

Solve the system of differential equations:
\begin{equation*} \left\{\begin{aligned} y_1'(t) \amp = 4y_1(t) - y_2(t) + 6y_3(t)\\ y_2'(t) \amp = 2y_1(t) + y_2(t) + 6y_3(t)\\ y_3'(t) \amp = 2y_1(t) - y_2(t) + 8y_3(t) \end{aligned}\right. \end{equation*}