Example 27.4.1. Three coupled differential equations.
Consider the following system, with the provided initial values:
\begin{align*}
\amp
\left\{\begin{array}{rcrcrcr}
y_1' \amp = \amp -9 y_1 \amp - \amp 14 y_2 \amp + \amp 14 y_3 \text{,} \\
y_2' \amp = \amp 9 y_1 \amp + \amp 16 y_2 \amp - \amp 13 y_3 \text{,} \\
y_3' \amp = \amp 2 y_1 \amp + \amp 4 y_2 \amp - \amp 1 y_3 \text{,}
\end{array}\right.
\amp
\amp
\left\{\begin{array}{rcr}
y_1(0) \amp = \amp 6 \text{,} \\
y_2(0) \amp = \amp -4 \text{,} \\
y_3(0) \amp = \amp 4 \text{.}
\end{array}\right.
\end{align*}
In matrix form, we express this as
\begin{equation*}
\ddt
\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}
= \left[\begin{array}{rrr}
-9 \amp -14 \amp 14 \\
9 \amp 16 \amp -13 \\
2 \amp 4 \amp -1
\end{array}\right]
\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}\text{.}
\end{equation*}
The characteristic polynomial of the coefficient matrix is
\begin{equation*}
\lambda^3 - 6 \lambda^2 - \lambda + 30
= (\lambda + 2)(\lambda - 3)(\lambda - 5)\text{,}
\end{equation*}
so the eigenvalues are \(\lambda_1 = -2\text{,}\) \(\lambda_2 = 3\text{,}\) \(\lambda_3 = 5\text{.}\) Computing eigenspaces by row reducing \(\lambda I - A\) for the various eigenvalues as usual, we obtain
\begin{gather*}
E_{\lambda_1}(A) =
\Span\left\{ \left[\begin{array}{r} -2 \\ 1 \\ 0 \end{array}\right] \right\} \text{,}
\qquad
E_{\lambda_2}(A) =
\Span\left\{ \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \right\} \text{,}\\
E_{\lambda_3}(A) =
\Span\left\{ \left[\begin{array}{r} -1 \\ 2 \\ 1 \end{array}\right] \right\} \text{.}
\end{gather*}
So we can use the transition matrix
\begin{equation*}
P = \left[\begin{array}{rrr}
-2 \amp 0 \amp -1 \\
1 \amp 1 \amp 2 \\
0 \amp 1 \amp 1
\end{array}\right]
\end{equation*}
and its inverse to convert between the original system and a decoupled one involving transformed functions
\begin{equation*}
\uvec{w}(t) = \inv{P} \uvec{y}(t) \text{.}
\end{equation*}
With our ordering of the columns of \(P\text{,}\) the general solution of the decoupled system is
\begin{equation*}
\uvec{w}(t) = \begin{bmatrix} c_1 e^{-2 t} \\ c_2 e^{3 t} \\ c_3 e^{5 t} \end{bmatrix} \text{,}
\end{equation*}
where \(c_1,c_2,c_3\) are arbitrary parameters. The general solution to the coupled system is then obtained by converting back via \(\uvec{y}(t) = P \uvec{w}(t)\text{,}\) yielding
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
y_1(t) \amp = \amp -2 c_1 e^{-2 t} \amp \amp \amp - \amp c_3 e^{5 t} \text{,} \\
y_2(t) \amp = \amp c_1 e^{-2 t} \amp + \amp c_2 e^{3 t} \amp + \amp 2 c_3 e^{5 t} \text{,} \\
y_3(t) \amp = \amp \amp \amp c_2 e^{3 t} \amp + \amp c_3 e^{5 t} \text{.}
\end{array}\right.
\end{equation*}
It's easier to work with the decoupled system to determine the parameter values that correspond to the initial values. Convert the vector of initial values \(\uvec{y}(0)\) to a vector of initial values for the decoupled system:
\begin{equation*}
\uvec{w}(0) = \inv{P} \uvec{y}(0) = \left[\begin{array}{r} 2 \\ 14 \\ -10 \end{array}\right] \text{.}
\end{equation*}
These three transformed initial values are precisely the values of the parameters \(c_1,c_2,c_3\) we need to create a particular solution that meets the initial values for \(y_1(t),y_2(t),y_3(t)\) provided at the beginning of this example.
Inserting these parameter values, we obtain particular solution
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
y_1(t) \amp = \amp -4 e^{-2 t} \amp \amp \amp + \amp 10 e^{5 t} \text{,} \\
y_2(t) \amp = \amp 2 e^{-2 t} \amp + \amp 14 e^{3 t} \amp - \amp 20 e^{5 t} \text{,} \\
y_3(t) \amp = \amp \amp \amp 14 e^{3 t} \amp - \amp 10 e^{5 t} \text{.}
\end{array}\right.
\end{equation*}
