Consider the fourth-order linear ODE $$ \label{eq1} v^{(4)} + \frac{-C_2 - 2\alpha \phi}{C_4}v'' + \frac{4\alpha \phi'}{C_4}v' + \frac{k_1 + 3k_3\phi^2 -2\alpha \phi''}{C_4}v = 0. $$ Without getting into too much details, this ODE is obtained from a linearization of a nonlinear ODE about a solution $\phi(x) = A sech(Bx)^2$ (where the constants A, B depend on the parameters $C_2, C_4$, etc.), therefore $\phi'(x)$ is one bounded solution. My question is, are there good ways to prove it is the only bounded solution, or, if there is another one, can we prove this second solution $w$ has $w(0) = w''(0) = 0$? Are there any methods, which allow one to answer such types of questions?

We know that in the far field ($x \to \pm \infty$) the ODE looks like $$ v^{(4)} - \frac{C_2}{C_4}v'' + \frac{k_1}{C_4}v = 0, $$ and this constant coefficient equation has four solutions $e^{\pm\lambda_1 x}$, $e^{\pm\lambda_2 x}$ where $0 < \lambda_1 < \lambda_2$, Therefore, all solutions of the original equation converge exponentially fast to $0$ or $\infty$ as $x \to \infty$. Furthermore, there are at most two solutions, which converge to zero as $x \to \infty$, so there cannot possibly be more than two bounded solutions (and we already know one). Generically, there is exactly one bounded solution, but I do not know how to prove it in this particular example.

This question comes from trying to prove transversality of certain stable and unstable manifolds in a dynamical system, and I am happy to provide details, but for the sake of brevity, I am keeping the question as self-contained as possible.