Changing Independent Variable in the Bessel Equation

[Potatochip911]

Homework Statement

Given the bessel equation $$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} -(1-x)y=0$$ show that when changing the variable to $u = 2\sqrt{x}$ the equation becomes $$u^2\frac{d^2y}{du^2}+u\frac{dy}{du}+(u^2-4)y = 0$$

Homework Equations

The Attempt at a Solution

$u=2\sqrt{x}$, $du/dx=\frac{1}{\sqrt{x}}=\frac{2}{u}$, $x = u^2/4$ and $x^2=u^4/16$, since $u=u(x)$ we have $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{2}{u}\frac{dy}{du}$$ and obtain the operator $\frac{d}{dx} = \frac{2}{u}\frac{d}{du}$ then $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\left(\frac{2}{u}\frac{d}{du}\right)\left(\frac{2}{u}\frac{dy}{du}\right)=\frac{4}{u^2}\frac{d^2y}{du^2}$$ plugging into the original now we obtain $$\frac{u^4}{16}\frac{4}{u^2}\frac{d^2y}{du^2}+\frac{u^2}{4}\frac{2}{u}\frac{dy}{du}-\left(1-\frac{u^2}{4}\right)y = \frac{u^2}{4}\frac{d^2y}{du^2}+\frac{u}{2}\frac{dy}{du}+\left(\frac{u^2}{4}-1\right)y=0$$ multiplying by 4 we obtain $$u^2\frac{d^2y}{du^2}+2u\frac{dy}{du}+(u^2-4)y=0$$ where I somehow have an additional factor of 2 on the $dy/du$ term despite getting the correct coefficients for the first and last term.

 

[Ssnow]

Hi, attention $\left(\frac{2}{u}\frac{d}{du}\right)\left(\frac{2}{u}\frac{dy}{du}\right)\not=\frac{4}{u^2}\frac{d^2y}{du^2}$, you must apply the product rule for derivation! ...