Problem source

The functions $x(t)$ and $y(t)$ satisfy the simultaneous differential equations 
\begin{alignat*}{1}
\dfrac{\mathrm{d}x}{\mathrm{d}t}+2x-5y & =0\\
\frac{\mathrm{d}y}{\mathrm{d}t}+ax-2y & =2\cos t,
\end{alignat*}
subject to $x=0,$ $\dfrac{\mathrm{d}y}{\mathrm{d}t}=0$ at $t=0.$ 
Solve these equations for $x$ and $y$ in the case when $a=1$. 
Without solving the equations explicitly, state briefly how the form of the solutions for $x$ and $y$ if $a>1$ would differ from the form when $a=1.$ 

Solution source

Letting $\mathbf{x} =\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$ and $\mathbf{A} = \begin{pmatrix} -2 & 5 \\ -a & 2 \end{pmatrix}$ then our differential equation is $\mathbf{x}' = \mathbf{Ax} + \begin{pmatrix} 0 \\2 \cos t \end{pmatrix}$.

Looking at the eigenvalues of $\mathbf{A}$, we find:

\begin{align*}
&& \det \begin{pmatrix} -2-\lambda & 5 \\ -a & 2 -\lambda \end{pmatrix} &= (\lambda^2-4)+5a\\
&&&= \lambda^2 +5a-4
\end{align*} 

Therefore if $a = 1$, $\lambda = \pm i$.

In which case we should expect the complementary solutions to be of the form $\mathbf{x} = \begin{pmatrix} A \sin t + B \cos t \\ C \sin t + D \cos t \end{pmatrix}$. The first equation tells us that $(A-5D+B)\cos t + (-B+5C)\sin t=0$ so the complementary solution is:$\mathbf{x} = \begin{pmatrix} 5(D-C) \sin t + 5C \cos t \\ C \sin t + D \cos t \end{pmatrix}$.

Looking for a particular integral, we should expect to try something like $\mathbf{x} = \begin{pmatrix} Et\cos t+Ft\sin t\\ Gt\cos t+Ht \sin t\end{pmatrix}$ and we find