Converting differential equation into sturm -Liouville

Question

Convert the following equation to the form of a Sturm-Louville equation. $$3x^2y''(x)+4xy'+6y(x)+\lambda y(x)=0,x>0$$

I used this.

But the substitution was tedious and I couldn't find the sturm-Liouville form. Please help me out!

Answer 1

Your problem may be written as $$ a(x)y''(x)+b(x)y'(x)+c(x)y(x)+\lambda d(x)y(x)=0, $$ where $$ a(x)=3x^2,\; b(x)=4x,\; c(x)=6,\;\; d(x)=1. $$ Using the problem I referenced, start with the substitution $$ y=\rho(x)f(x), $$ where $$ \rho(x)=\sqrt{3x^2}\exp\left(-\int\frac{4x}{6x^2}dx\right) \\ =\sqrt{3}x\exp\left(-\frac{2}{3}\ln(x)\right) \\ =\sqrt{3}xx^{-2/3}=\sqrt{3}x^{1/3} $$ This transforms your problem in $y$ to a new problem in $f$: $$ -\frac{d}{dx}\left(3x^2\frac{df}{dx}\right)+\left(\frac{(b-a')^2}{4a}+\frac{(b-a')'}{2}-c\right)f=\lambda f(x). $$ Using $b-a'=4x-6x=-2x$ gives the following Sturm-Liouville equation in normal form: $$ -\frac{d}{dx}\left(3x^2\frac{df}{dx}\right) +\left(\frac{1}{3}-1-6\right)f=\lambda f \\ -\frac{d}{dx}\left(3x^2\frac{df}{dx}\right)-\frac{20}{3}f=\lambda f$$