Find the eigenvalues ​and eigenfunctions of the Sturm-Liouville problem $x^4 y'' + k^2 y = 0$

Question

Find the eigenvalues ​​and eigenfunctions of the Sturm-Liouville problem

$x^4 y'' + k^2 y = 0$
$y(\alpha) = y(\beta) = 0$, $\hspace{0.4cm}0 < \alpha < \beta$

For this problem I need to find the eigenvalues ​​and the eigenfunctions. Originally the problem has variable coefficients, so I think I can transform the problem to the form $v'' + k^2 v = 0$. I think this is possible by doing a change of variable, which one? I'm familiar with problems that have the form $y'' + \lambda y = 0$. In these cases, to find the eigenvalues ​​and eigenfunctions is simply to consider the cases where $\lambda <0$ or $\lambda = 0$ or $\lambda> 0$. However, I have had difficulty finding the eigenvalues ​​and eigenfunctions for this problem posed here. How can I do this? I appreciate any help

Answer 1

Check that $$y''+(k^2/x^4)y=0$$ has two real linearly independent solutions as $y=x\sin(k/x), x \cos(k/x)$. So its general solytion is given as $$y(x)=x(C_1 \sin(k/x)+C_2\cos(k/x)).$$ The conditions $y(a)=0=y(b)$ give $C_1\sin(k/a)+C_2 \cos(k/b)=0,$ $C_1 \sin(k/b)+ C_2 \cos(k/b))=0$. By comparing $C_1/C_2$ from these equations, we get the eigenvalues as $$\tan(k/a)-\tan(k/b)=0 \implies \sin [k(b-a)/(ab)] =\implies k_n=\frac{n\pi ab}{b-a},(b>a), n=1,2,3,4,..$$

Answer 2

$$x^4 y'' + k^2 y = 0$$ $$x^2 y'' + \dfrac {k^2}{x^2} y = 0$$ Z Ahmed's answer is correct substitute $w=\dfrac 1x$: $$y'=-w^2\dfrac {dy}{dw}$$ $$y''=2w^3 \dfrac {dy}{dw}+w^4 \dfrac {d^2y}{dw^2}$$ The DE becomes: $$wy''+2y'+k^2wy=0$$ $$(wy)''+k^2wy=0$$ Which is the form you wanted: $$v''+k^2v=0$$ Where $v=wy=\dfrac yx$.