Converting to Sturm-Liouville using integration factor [closed]

Question

I know this type of question has been asked before, but I cannot find information on this specific form I am supposed to be using. We are being asked to convert a second-order DE

\begin{equation*}\alpha(x)y'' + \beta(x)y′ + \gamma(x)y = \lambda y \end{equation*}

to Sturm-Liouville form using the integration factor

\begin{equation*} F(x) = \exp\left( \int \frac{[\beta(u) − \alpha'(u)]}{\alpha(u)}\right) \end{equation*}

I've been looking for at least 3 hours for information regarding this specific I.F. and I can't find anything. Can someone please point me in the right direction or provide a starting point? Thank you

Answer 1

I'm sure the "$u$" should be a $x$: $$ \exp\left(\int\frac{\beta(x)}{\alpha(x)}dx-\int\frac{\alpha'(x)}{\alpha(x)}dx\right)=\exp\left(\int\frac{\beta(x)}{\alpha(x)}dx\right)\frac{1}{\alpha(x)} $$ Then multiply your original equation $$ \alpha(x)y''(x)+\beta(x)y'(x)+\gamma(x)y(x)=\lambda y(x) $$ by the expression on the right of the first line to obtain: $$ \exp\left(\int\frac{\beta(x)}{\alpha(x)}dx\right)\frac{1}{\alpha(x)}(\alpha(x)y''(x)+\beta(x)y'(x)+\gamma(x)y(x)-\lambda y(x))=0 \\ \exp\left(\int\frac{\beta(x)}{\alpha(x)}dx\right)\left(y''(x)+\frac{\beta(x)}{\alpha(x)}y'(x)+\frac{\gamma(x)}{\alpha(x)}y(x)-\frac{\lambda}{\alpha(x)}y(x)\right)=0 $$ This gives a Sturm-Liouville form that needs to be broken across a line: $$ \frac{d}{dx}\left[\exp\left(\int\frac{\beta(x)}{\alpha(x)}dx\right)\frac{dy}{dx}\right]+\frac{\gamma(x)}{\alpha(x)}\exp\left(\int\frac{\beta(x)}{\alpha(x)}dx\right)y(x)= \\=\frac{\lambda}{\alpha(x)}\exp\left(\int\frac{\beta(x)}{\alpha(x)}dx\right)y(x). $$