Convert with variable change the form Sturm- Liouville to $y^{\prime\prime}+(\lambda \rho_{1}(x)+q_{1}(x))y=0 $

Question

I have the problem of Sturm Liouville in $[a,b]:$ $$ \left(p(x) y^{\prime}\right)^{\prime}+(\lambda \rho(x)-q(x))y=0, $$ developing the expression I got: $$ p(x)y^{\prime\prime}+p(x)^{\prime}.y^{\prime}+(\lambda \rho(x)-q(x))y=0 $$ by changing the variable I must arrive at the expression: $$ y^{\prime\prime}+(\lambda \rho_{1}(x)+q_{1}(x))y=0 $$

the problem is not about reducing but making a change of variable,any help will be appreciated.

Answer 1

Let $y=h(x)u$, do your product rules and collect all your terms to get an ODE of the form \begin{align} f_1(x)u''+f_2(x)u'+f_3(x)u=0. \end{align} Set $f_2(x)$ to zero. This will give you an easy ODE to solve for $h(x)$ in terms of $p(x)$ which will simplify it to the form you desire.