π Example 283. Solving a Partially Coupled System.
Solve the system:
\begin{align*}
\frac{dx}{dt} \amp = -x \qquad x(0) = 1\\
\frac{dy}{dt} \amp = -2y + x \qquad y(0) = 0
\end{align*}
Solution.
Step 1: Solve the independent equation for \(x\text{.}\) This is a familiar exponential decay:
\begin{equation*}
\frac{dx}{dt} = -x \quad \Rightarrow \quad x(t) = e^{-t}.
\end{equation*}
\begin{equation*}
\frac{dy}{dt} = -2y + e^{-t}.
\end{equation*}
This is a linear first-order equation. Use an integrating factor to solve:
\begin{equation*}
y' + 2y = e^{-t}, \qquad \mu(t) = e^{2t}.
\end{equation*}
Multiply by \(e^{2t}\) and rewrite the left side as a derivative:
\begin{equation*}
\frac{d}{dt}\left(e^{2t} y \right) = e^{t}.
\end{equation*}
Integrate:
\begin{equation*}
e^{2t} y = \int e^{t} dt = e^{t} + C.
\end{equation*}
Solve for \(y(t)\text{:}\)
\begin{equation*}
y(t) = e^{-t} + C e^{-2t}.
\end{equation*}
Use the initial condition \(y(0) = 0\text{:}\)
\begin{equation*}
0 = 1 + C \quad \Rightarrow \quad C = -1.
\end{equation*}
So the solution is:
\begin{equation*}
x(t) = e^{-t}, \qquad y(t) = e^{-t} - e^{-2t}.
\end{equation*}
\(x(t)\) decays smoothly, while \(y(t)\) rises briefly (pulled by \(x\)) before eventually decaying as well. This βone-way influenceβ is the hallmark of a partially coupled system.
