After multiplying by the inverse of the derivative matrix, the system is
$$
\dot z+Az=e^{-4t}\binom{2}{3b/2-1}+e^{-t}\binom{c-1}{-c/2-1},
\quad
A=\begin{pmatrix}2&-1\\-2&3\end{pmatrix}.
$$
Choose eigenvectors $(1,1)^T$ and $(-1/2,1)^T$ of eigenvalues $1$ and $4$, and write
$$
\binom xy=
\begin{pmatrix}1&-1/2\\1&1\end{pmatrix}\binom{w_1}{w_2}.
$$
Then
$$
\begin{aligned}
\dot w_1+w_1&=(b/2+1)e^{-4t}+(c/2-1)e^{-t},\\
\dot w_2+4w_2&=(b-2)e^{-4t}-ce^{-t}.
\end{aligned}
$$
The zero initial data give
$$
\boxed{\begin{aligned}
w_1&=\frac{b+2}{6}(e^{-t}-e^{-4t})+\frac{c-2}{2}te^{-t},\\
w_2&=(b-2)te^{-4t}-\frac c3(e^{-t}-e^{-4t}),\\
x&=w_1-\frac12w_2,\qquad y=w_1+w_2.
\end{aligned}}
$$
The $te^{-4t}$ and $te^{-t}$ terms are <resonance>[resonant responses]. When $b=2$ or $c=2$, respectively, the corresponding resonant forcing vanishes and so does that secular factor.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 5B / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-2/5b/solution