Convert in a Sturm-Liouville form $$y''+R(x)y'+(Q(x)+\lambda P(x))y=0\tag1$$
My attempt:
The form of Sturm-Liouville is:
$$\frac{d}{dx}[r(x) \frac{dy}{dx}]+(q(x)+\lambda p(x))y=0\tag2$$
For obtain this, we need multiply for $\mu(x)$ the ODE $(1)$. This result in:
$$\mu y''+\mu R(x)y'+(\mu Q(x)+\lambda \mu P(x))y=0 \tag3$$
I need rewrite the first term of $(3)$.
Here, i'm stuck. I don't have a clear idea of how rewrite and then proceed for solve the exercise.
Can someone help me?
If we expand out equation $(2)$, we get $$ r(x) y'' + r'(x) y' + (q(x) + \lambda p(x)) y = 0 \ \ \ \ \ (2a).$$
So the objective is to choose the functions $\mu(x)$, $r(x)$, $q(x)$ and $p(x)$ so that your equation $(3)$ matches my equation $(2a)$.
Comparing the individual terms, the requirement is that $$ \mu(x) = r(x), \ \ \ \ \ \mu(x)R(x) = r'(x), \ \ \ \ \ \mu(x)Q(x) = q(x), \ \ \ \ \ \mu(x)P(x) = p(x).$$ The first of these equations is straightforward: it tells us that $\mu(x)$ and $r(x)$ must be the same function.
Given that $\mu(x) = r(x)$, the second equation says that $$ \mu(x) R(x) = \mu'(x),$$ and this is satisfied by
$$ \mu(x) = \exp \left(\int dx \ R(x) \right).$$
I'll leave you to figure out what $q(x)$ and $p(x)$ and finish off.