I want to show $$(1-x^2)u''-xu'+9u=x^3 \ \ \ \ \ \ -1\leq x\leq 1$$ can be written in formally self-adjoint form.
My attempt:
Calculate the integrating factor, $$\exp\left(-\int \frac{x \ dx}{1-x^2}\right)\frac{1}{1-x^2}=\frac{\exp(\ln(\sqrt{1-x^2}))}{1-x^2}=\frac{1}{\sqrt{1-x^2}}.$$ Then, \begin{align} \sqrt{1-x^2}u''-\frac{x}{\sqrt{1-x^2}}u'+\frac{9}{\sqrt{1-x^2}}u&=\frac{x^3}{\sqrt{1-x^2}} \\ \left(\sqrt{1-x^2} u'\right)'+\frac{9}{\sqrt{1-x^2}}u&=\frac{x^3}{\sqrt{1-x^2}} \ \ \ \ \ \ \ \ \ \ (1)\\ -\left(\sqrt{1-x^2} u'\right)'+\left(-\frac{9}{\sqrt{1-x^2}}u\right)&=-\frac{x^3}{\sqrt{1-x^2}} \end{align} Hence this is in the form $Lu=-(pu')'+qu$, where, $$p=\sqrt{1-x^2}, q=-\frac{9}{\sqrt{1-x^2}}.$$ Is this correct? I am not sure if I am allowed to express $q$ the way I did, as i'm unsure if $q$ can be negative.
