Solve Bessel's equation through certain substitutions

[ILikeMath]

1. Homework Statement , relevant equation
$$x^{2}y'' + xy' + (4x^{4}-\frac{1}{4})y = 0$$

2. The attempt at a solution
I tried substituting z = x²

From this I have $$\frac{dy}{dx} = 2x \frac{dy}{dz}$$
and $$\frac{d}{dx}(2x\frac{dy}{dz}) = 2xy'' + 2y'$$

Then the original equation becomes:
$$2z^{3/2}y'' + 4zy' + (4z^{2}-\frac{1}{4})y = 0$$

where derivatives of y are now with respect to the new variable z.
This does not look like a Bessel equation and I'm not sure how to make it look like one. Did I use the wrong substitution?

I know how to solve once it's in the correct form, but could someone help me get it there please?

 

[206PiruBlood]

try $$z^{2}=4x^{4}$$

 

[ILikeMath]

If I use z² = 4x⁴ I get:
z = 2x² and dz/dx = z' = 4x

$$(\frac{dy}{dx}) = \frac{dy}{dz} \frac{dz}{dx} = 4x \frac{dy}{dz}$$

Then

$$\frac{d}{dx}(4x\frac{dy}{dz})=4\frac{dy}{dz}+\frac{d}{dz}(\frac{dy}{dz})\frac{dz}{dx}=4\frac{dy}{dz}+4x\frac{d^{2}y}{dz^{2}}$$

Plugging in:
x²(4y' + 4xy'') + x(4xy') + (4x⁴ - 1/4)y = 0
if z² = 4x⁴
(z/2)^3/2 y'' + zy' + (z²/4 - 1/16)y = 0
so it still doesn't work. The x³ term messes the whole thing up...

 

[ILikeMath]

Never mind, I got it. Thanks for your help!