I have the problem:
$$x^2y''+(1/4)y=\lambda y, 1<x<e$$
$$y(1)=y(e)=0$$
And I'm trying to find a weighting function for its solutions (I already calculated the solutions and eigenvalues). If we plug $x=e^t$ we can see the equation becomes:
$$Y''-Y'+(1/4-\lambda)Y=0, 0<x<1$$
This seems like something I should be able to transform into Sturm-Liouville form, but since the coefficients of Y'' and $\lambda Y$ are different it doesn't seem feasible.
Any suggestions?
Thanks!
Best Answer
The equation $$a_2(x)y''(x)+a_1(x)y'(x)+a_0(x)y(x)+\lambda y=0, \quad a<x<b,$$ can be brought into the so-called self-adjoint form given by $${1\over w(x)}[(p(x)y')'+q(x)y]+\lambda y=0,$$ via $$ w(x)=\exp\left(\int {a_1(x)-a_2'(x)\over a_2(x)}\,dx\right), \quad p(x)=a_2(x)w(x), \quad q(x)=a_0(x)w(x). $$ (It's a good exercise to show why this is true.)
The $w(x)$ here is the weight function.