I was reading this article and I'm trying to follow this author's proof. The author jumps from
$$\psi_1(x)\frac{\partial^2\psi_2(x)}{\partial x^2}-\psi_2(x)\frac{\partial^2\psi_1(x)}{\partial x^2}=0$$ to
$$\psi_1(x)\frac{\partial\psi_2(x)}{\partial x}-\psi_2(x)\frac{\partial\psi_1(x)}{\partial x}=Constant$$
I've tried integrating by parts but it just seems to get mucky fast. Can anyone help me clarify how this is such a simple procedure?
For clarity, let $\partial$ stand for $\frac{\partial}{\partial x}$ and $\partial^2$ for the iterated $\partial$. Now by the product rule run slightly in reverse
$$\psi_1 \partial^2 \psi_2 = \partial(\psi_1 \partial\psi_2) - \partial\psi_1\partial\psi_2$$
and
$$\psi_2 \partial^2 \psi_1 = \partial(\psi_2 \partial\psi_1) - \partial\psi_2\partial\psi_1$$
Finally, we have
$$\psi_1 \partial^2 \psi_2 - \psi_2 \partial^2 \psi_1 = \partial(\psi_1 \partial\psi_2) - \partial\psi_1\partial\psi_2 - (\partial(\psi_2 \partial\psi_1) - \partial\psi_2\partial\psi_1)$$
which equals
$$\partial(\psi_1 \partial\psi_2) - \partial(\psi_2\partial\psi_1) = \partial(\psi_1 \partial\psi_2-\psi_2\partial\psi_1) $$
which means that
$$\psi_1 \partial\psi_2-\psi_2\partial\psi_1 = \text{constant}$$