Proving ODE does not have bounded solutions Consider the equation
\begin{equation}
        x'' + (a+b\cos t)x = 0
    \end{equation}
and let $u, v$ be solutions such that
\begin{equation}
    u(0) = 1, u'(0) = 0; v(0) = 0, v'(0) = 1
\end{equation}
Set $F(a, b) = u(2\pi) + v'(2\pi)$ and show that if $|F(a, b)| > 2$ then no solutions remain bounded for all real $t$.
My attempt is below:
Consider $F_{a, b}(t) = u(t) + v'(t)$. We know that $|F_{a, b}(2\pi)|>2$. We can now differentiate $F_{a, b}(t)$ with respect to $t$ to obtain:
\begin{align}
    F'_{a, b}(t) &= u'(t) + v''(t)\\
    &= u'(t) - (a+b\cos(t))v(t)
\end{align}
where in the last equality we used the fact that $v$ is a solution to the given ODE. We can differentiate $F'_{a, b}(t)$ again to obtain:
\begin{align}
    F''_{a, b}(t) &= u''(t) - (a+b\cos(t))v'(t) + b\sin(t)v(t)\\
    &= -(a+b\cos(t))u(t) - (a+b\cos(t))v'(t) + b\sin(t)v(t)\\
    &= -(a+b\cos(t))(u(t)+v'(t)) + b\sin(t)v(t)\\
    &= -(a+b\cos(t))F_{a,b}(t) +  b\sin(t)v(t)
\end{align}
This looks promising because now I have $F_{a, b}(t)$ and $v(t)$ in the same equation but I'm having trouble finishing this off myself. Some hints would be greatly appreciated.
 A: The equation is linear, so
$$
\Phi(t)=\pmatrix{u(t)&v(t)\\u'(t)&v'(t)}
$$ is a fundamental matrix of the first order system for the components $(x,x')$. Note that

*

*For the full periods one can easily confirm that $\Phi(k·2\pi)=\Phi(2\pi)^k$.

*For the quantity in the claim one finds $F(a,b)=\operatorname{trace}(\Phi(2\pi))$.

*If $|F(a,b)|>2$, then one eigenvalue of $\Phi(2\pi)$ has absolute value larger than $1$.

Now one has 2 cases, either the eigenvalues are a complex conjugate pair, then the eigen-solutions corresponding to the eigen-decomposition of $\Phi(2\pi)$ are unbounded for $t\to+\infty$, and also all their linear combinations.
In the case of real eigenvalues it is possible that the smaller one is $\pm1$, which would lead to a bounded basis solution. This would have to be excluded with further arguments. If the smaller eigenvalue is inside the unit circle, then the corresponding eigen-solution is unbounded in direction $-\infty$.

As there is no first-derivative term we know that the Wronskian $W[u,v](t)=\detΦ(t)$ is constant, thus has always the value $\detΦ(0)=1$. Thus the product of the eigenvalues of $Φ(2π)$ is always $1$, which reduces the possible cases discussed above. As far as I can see the only remaining case is the one of two real eigenvalues, one with absolute value larger than $1$, and one smaller $1$.
