quantum mechanics - Degeneracy in one dimension

I'm reading this wikipedia article and I'm trying to understand the proof under "Degeneracy in One Dimension". Here's what it says:

Considering a one-dimensional quantum system in a potential $V(x)$ with degenerate states $|\psi_1\rangle$ and $|\psi_2\rangle$ corresponding to the same energy eigenvalue $E$, writing the time-independent Schrödinger equation for the system:

$$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_1}{ \partial x^2} + V\psi_1 =E\psi_1$$ $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_2}{ \partial x^2} + V\psi_2 =E\psi_2$$ Multiplying the first equation by $\psi_2$ and the second by $\psi_1$ and subtracting one from the other, we get: $$\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$$ Integrating both sides $$\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$$ In case of well-defined and normalizable wave functions, the above constant vanishes, provided both the wave functions vanish at at least one point, and we find: $\psi_1(x)=c\psi_2(x)$ where $c$ is, in general, a complex constant. So, the two degenerate wave functions are not linearly independent. For bound state eigenfunctions, the wave function approaches zero in the limit $x\to\infty$ or $x\to-\infty$, so that the above constant is zero and we have no degeneracy.

Here are the parts I don't understand:

1. 
   

   How does integrating $\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$ yield $\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$. Without having explicit formulas for $\psi_1(x)$ and $\psi_2(x)$, how do we just apparently pull them both out of the integral?
   

   
   


2. 
   
   

   I don't understand how the constant in the last equation vanishes just because $\psi_1$ and $\psi_2$ vanish at infinity.
   

   
   


3. 
   

   Even if it does vanish, how does that imply $\psi_1=c\psi_2$?

   
   

Answer

(1). $\psi_1\psi_2^"-\psi_2\psi_1^"=\frac{d}{dx}(\psi_1\psi_2'-\psi_2\psi_1')=0$.

(2). At infinity, it is zero, so the constant must be 0

(3). Integrate $\psi_1\psi_2'=\psi_2\psi_1'$, you will get that