How can I find the $C^1([1,2])$ functions $y = y(x)$ that satisfy the following system of integro-differential equations?
$$\int_1^2 x^4 \, yy' \, dx = 0, \qquad \int_1^2 y \, dx = 0$$
Okay, here is the context for the question. I want to find the eigenvalues and eigenfunctions of the Sturm-Liouville problem
$$L^2(y) + \lambda y = 0, \qquad y(1) = 0, \qquad y(2) = 0$$
where $L(y) = x^2 y'$.
Set $z = xy$ and differentiate it twice to obtain
$$xz'' = x^2 y'' + 2xy' = (x^2 y')'$$
Then the Sturm-Liouville problem becomes
$$x^4 z'' + \lambda z = 0, \qquad z(1) = 0, \qquad z(2) = 0$$
Multiply times $w = z'$ and integrate from $1$ to $2$, obtaining
$$\int_1^2 x^4 ww' \, dx = -\lambda \int_1^2 zz' \, dx = 0$$
Then I need to find the functions $w = w(x)$ that satisfy
$$\int_1^2 x^4 ww' \, dx = 0, \qquad \int_1^2 w \, dx = 0$$
Which is the original question, except I have renamed $w$ to $y$.
This problem is neither from any class that I'm taking (in case this is a concern for anyone), nor from my job, so I'm under no particular urgency to get an answer. I just asked here because I thought it would be nice to get other people's thoughts on this problem.
Substitution of $z(x) = x y(x)$ was a step back... Let us write equation in the form $$ x^4 y''(x)+2 x^3 y'(x)+\lambda y(x) = 0. $$ We divide it by $x^4$ in order to obtain coefficient $1$ at $y''(x)$ $$ y''(x)+\frac{2 y'(x)}{x}+\frac{\lambda y(x)}{x^4} = 0. $$ Let us try a substitute (substitute of only this kind can transform this equation into an equation with constant coefficients) $t = c \int \sqrt{x^{-4}} dx = c \int x^{-2} dx = - cx^{-1}$, where $c$ is some real constant. Choose $c = -1$ and make the substitute $t = x^{-1}$: $$x = \frac{1}{t},\quad \frac{d y}{d x} = \frac{d y}{d t} \frac{d t}{d x}=-\frac{d y}{d t} \frac{1}{x^2} = -\frac{d y}{d t} t^2,\\ \frac{d^2 y}{d x^2} = \frac{d}{d t}\left(\frac{d y}{d x}\right) \frac{d t}{d x} = t^4 y''(t)+2 t^3 y'(t),$$ so we obtain $$ t^4 y''(t)+\lambda t^4 y(t) = 0, $$ or $$ y''(t)+\lambda y(t) = 0, $$ with conditions $y(x = 1) = y(t = 1) = 0$ and $y(x = 2) = y\left(t = \dfrac{1}{2}\right) = 0$.
Further solution leads to the system of eigenfunctions $$ y_n(t) = \sin((2 + 4 n)\pi t), $$ or $$ y_n(x) = \sin\left(\frac{(2 + 4 n)\pi}{x}\right). $$