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 $\phi(x) = A \operatorname{sech}(Bx)^2$ (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.