End Behavior of Nonlinear ODE System - Transfer Function I'm studying for my PhD qualifying exam and have run into a problem that I cannot figure out how to solve. I'm sure there is a relatively simple trick that I'm forgetting. Can you help me solve this?
Problem Statement
Let $u(t)$ and $v(t)$ be two real functions that obey the differential relations:
$$ \begin{align} \dot{u} &= v - au\\ \dot{v} &= -u + \sin(bt) \end{align} $$
where $a$ and $b$ are positive constants and dot denotes the derivative with respect to $t$. Find the value of $b$ such that the largest value of $v$ is 5 at very large $t$ (i.e., as $t \to \infty$)
My thoughts so far:
It seems to me that transforming this system into the frequency domain will be the best way to find the answer. If we assume zero initial conditions $u(0) = v(0) = 0$, then we can solve for the Laplace transform of $v$:
$$ V(s) = \frac{b(s+a)}{(s^2 + b^2)(s^2 + as + 1)} $$
Attempt 1: I tried splitting this into pieces via partial fractions; things get very messy very quickly so I abandoned that path.
Attempt 2: Final value theorem (FVT) seemed to be the next answer, i.e.
$$ \underset{t \to \infty}{\lim} v(t) = \underset{s \to \infty}{\lim} s V(s) $$
However, FVT only applies if the roots of the denominator of $s V(s)$ all have negative real parts. Clearly, $V(s)$ has poles at $\pm bi$, so that condition is not satisfied...
Thoughts?
 A: We have:
$$
\begin{cases}
\text{u}'\left(t\right)=\text{v}\left(t\right)-\text{a}\cdot\text{u}\left(t\right)\\
\text{v}'\left(t\right)=\sin\left(\text{b}t\right)-\text{u}\left(t\right)
\end{cases}\tag1
$$
Using, Laplace transform of both sides we get:
$$
\begin{cases}
\text{s}\cdot\text{U}\left(\text{s}\right)-\text{u}\left(0\right)=\text{V}\left(\text{s}\right)-\text{a}\cdot\text{U}\left(\text{s}\right)\\
\\
\text{s}\cdot\text{V}\left(\text{s}\right)-\text{v}\left(0\right)=\frac{\text{b}}{\text{b}^2+\text{s}^2}-\text{U}\left(\text{s}\right)
\end{cases}\tag2
$$
So, we get:
$$\text{s}\cdot\text{U}\left(\text{s}\right)-\text{u}\left(0\right)=\text{V}\left(\text{s}\right)-\text{a}\cdot\text{U}\left(\text{s}\right)\space\Longleftrightarrow\space\text{U}\left(\text{s}\right)=\frac{\text{V}\left(\text{s}\right)+\text{u}\left(0\right)}{\text{s}+\text{a}}\tag3$$
So:
$$\text{s}\cdot\text{V}\left(\text{s}\right)-\text{v}\left(0\right)=\frac{\text{b}}{\text{b}^2+\text{s}^2}-\frac{\text{V}\left(\text{s}\right)+\text{u}\left(0\right)}{\text{s}+\text{a}}\tag4$$
So, for $\text{V}\left(\text{s}\right)$ we get:
$$\text{V}\left(\text{s}\right)=\frac{\frac{\text{b}}{\text{b}^2+\text{s}^2}-\frac{\text{u}\left(0\right)}{\text{s}+\text{a}}+\text{v}\left(0\right)}{\text{s}+\frac{1}{\text{a}+\text{s}}}\tag5$$
To test if we can use the final value theorem, look if the poles of $\text{s}\cdot\text{V}\left(\text{s}\right)$ lie in the left half-plane:
$$\text{s}\cdot\text{V}\left(\text{s}\right)=\text{s}\cdot\frac{\frac{\text{b}}{\text{b}^2+\text{s}^2}-\frac{\text{u}\left(0\right)}{\text{s}+\text{a}}+\text{v}\left(0\right)}{\text{s}+\frac{1}{\text{a}+\text{s}}}\tag6$$

Using that $\text{U}\left(0\right)=\text{v}\left(0\right)=0$, we get:
$$\text{V}\left(\text{s}\right)=\frac{\frac{\text{b}}{\text{b}^2+\text{s}^2}}{\text{s}+\frac{1}{\text{a}+\text{s}}}\tag7$$
Using the convolution theorem:
$$\text{v}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\frac{\text{b}}{\text{b}^2+\text{s}^2}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}+\frac{1}{\text{a}+\text{s}}}\right]_{\left(t\right)}=$$
$$\sin\left(\text{b}t\right)\space*\space e^{-\frac{\text{a}t}{2}}\left\{\cosh\left(\frac{t\sqrt{\text{a}^2-4}}{2}\right)+\frac{\text{a}\sinh\left(\frac{t\sqrt{\text{a}^2-4}}{2}\right)}{\sqrt{\text{a}^2-4}}\right\}\tag8$$
A: What about solving the equations exactly?
Let $x = \begin{bmatrix}u\\v\end{bmatrix}$, $A=\begin{bmatrix}-a&1\\0&-1\end{bmatrix}$, and $S = \begin{bmatrix}0\\\sin bt\end{bmatrix}$. Then your ODE is just $\dot{x} = Ax + S$.
Next, diagonalize: we have that $A = PDP^{-1}$ where $P = \begin{bmatrix}-1/(1-a)&1\\1&0\end{bmatrix}$ and $D=\mathrm{diag}(-1, a)$. Let $y = P^{-1}x$. Then the ODE becomes $\dot{y} = Dy + P^{-1}S$. Since the ODE is now separated, it can be solved with $y = y_0 + e^{Dt}\int_0^t e^{-Dt'}P^{-1}S(t')dt'$.
Then chug through some integrals, convert back to $u$, use some trig identities, and answer should appear...
