how do you convert y" equation into Sturm-Liouville form For $y = y(x)$, convert the following equation 
$$ y''- 2xy' + 2vy = 0;$$ 
where $v$ is a constant, into a Sturm-Liouville form 
$$ Ly = r(x)(\lambda)y,$$  
$\lambda $ is a number, where 
$$ L := \frac{d}{dx} [ p(x) \frac{d}{dx} ] + q(x).$$ 
Here $r(x) > 0$ is a weight. The trick is to first multiply the given equation of the form 
$$ n(x)y'' + m(x)y' + q(x)y + aw(x)y = 0;$$ 
where $a$ is a number, by an integrating factor 
$$I(x) = e^{\int x m(s)-n'(s)/ n(s) ds},$$ to get a Sturm-Liouville equation of the form 
$$ (I(x)n(x)y')' + I(x)q(x)y + aw(x)I(x) = 0: $$
What are , the weight r(x), p(x), and q(x), for the converted equation?

The goal is to multiply the second and first derivative terms by $p$ such that
$$
                py''-2xpy'=(py')' = py''+p'y'.
$$
So you want $p$ such that $p'=-2xp$ or $p=e^{-x^{2}}$. You can check that the resulting equation becomes
$$
           e^{-x^{2}}y''-2xe^{-x^{2}}y'+2ve^{-x^{2}}y = 0
$$
$$
             (e^{-x^{2}}y')' + 2ve^{-x^{2}}y = 0.
$$
The normal form is
$$
             Ly = -\frac{1}{2e^{-x^{2}}}\left[e^{-x^{2}}y'\right]'=vy.
$$
From this form, one has an eigenvalue problem for $Ly$ on the weighted inner-product space with weight function $w=2e^{-x^{2}}$:
$$
             (f,g)_{w} = \int_{a}^{b}f(x)g(x)2e^{-x^{2}}dx.
$$
