# Show that normable eigenstates are not degenerated in 1D [duplicate]

I'm given a particle in a piecewise continous potential $V$ and am supposed to show that normable eigenstates of the Hamilton Operator are not degenerated. My approach was to find two eigenfunctions $\psi_1(x)$ and $\psi_2(x)$ that have the same eigenvalue and the show that they are linear dependent. Plugging into the Schrödinger equation led me to $\psi_1(x) \psi''_2(x) - \psi_2(x) \psi''_1(x) =0$. For linear indepence this equation would only be true for $\psi''_1$ and $\psi''_2$ both being zero, so I thought it's somehow possible to show that they are not zero to show the linear dependence. But I'm stuck so far. The only further assumption I have is that the first differential of the functions is limited for $|x| \to \infty$.
\begin{equation} 0 = \psi_1 \psi_2'' - \psi_2 \psi_1'' = (\psi_1 \psi_2' - \psi_2 \psi_1' ) ' \end{equation} meaning the () in the right term must be a constant : \begin{equation} \psi_1 \psi_2' - \psi_2 \psi_1' = c \end{equation} Now you can evaluate that constant with your condition $| x| \rightarrow + \infty$ , $\psi_1 \rightarrow 0$ and $\psi_2 \rightarrow 0$ (since they're bound states). Thus $c = 0$ and \begin{equation} \psi_2 \psi_1' = \psi_1 \psi_2' \\ \Leftrightarrow \frac{\psi_1'}{\psi_1}=\frac{\psi_2'}{\psi_2} \\ \Leftrightarrow \frac{d}{dx} ( \ln \psi_1 - \ln \psi_2)=0 \\ \Rightarrow \ln \psi = \ln \psi_2 + \ln c' \end{equation} with $c'$ a constant. Thus \begin{equation} \psi_1(x) = c' \psi_2(x) \end{equation}