# Solve Bessel's equation through certain substitutions

1. Oct 5, 2009

### ILikeMath

1. The problem statement, all variables and given/known data, 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?

2. Oct 5, 2009

### 206PiruBlood

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

3. Oct 5, 2009

### 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....

Last edited: Oct 5, 2009
4. Oct 6, 2009

### ILikeMath

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