# Homework Help: Bessel Function Solution

1. Jun 8, 2010

### dats13

I am trying to solve this equation in terms of Bessel functions.

xy"-y'+(4x^3)y=0

I am sure how to do this. The first thing that comes to mind is to solve for a series solution. This solution can then be compared to the bessel function and from that I can determine the first solution and thus get the second linearly independent solution.

Would this be the correct approach or is there any other way to solve for the general solution in terms of Bessel functions?

Any advice is greatly appreciated.

2. Jun 8, 2010

### LCKurtz

That doesn't look like Bessel's equation but in any case x = 0 is a regular singular point so you would look for a solution

$$y = x^r\sum_{n=0}^\infty a_nx^n$$

3. Jun 9, 2010

### dats13

I figured out how to solve this. By comparing the given ODE to the following:

$$x^{2}\frac{d^{2}y}{d^{2}x}+x(z+2bx^{r})\frac{dy}{dx}+[c+dx^{2s}-b(1-a-r)x^{r}+b^{2}x^{2r}]y=0$$

I can then determine a, b, c, d, r and s. From this, the solution is given by:

$$y(x)=x^{\frac{1-a}{2}}e^{-\frac{b}{r}x^{r}}Z_{p}(\frac{\sqrt{d}}{s}x^{s})$$

where

$$p=\frac{1}{s}\sqrt{(\frac{1-a}{2})^{2}-c}$$

4. Jun 9, 2010

### Ben Niehoff

Hmm...

I find it a bit odd that the question asks you to solve this in terms of Bessel functions, because it has a simple solution in terms of trigonometric functions. Try transforming the independent variable to $u = x^2$.

5. Jun 9, 2010

### dats13

Ben, I agree with what you're saying.

Determining a, b, c, d, r and s, gives me the solution

$$y(x)=xZ_{\frac{1}{2}}(x^{2})$$

Now since p=1/2

$$Z_{\frac{1}{2}}(x^{2})=c_{1}J_{\frac{1}{2}}(x^{2})+c_{2}J_{-\frac{1}{2}}(x^{2})$$

Thus the solution is

$$y(x)=x[c_{1}J_{\frac{1}{2}}(x^{2})+c_{2}J_{-\frac{1}{2}}(x^{2})]=C_{1}cos(x^{2})+C_{2}sin(x^{2})$$

6. Jun 9, 2010

### vela

Staff Emeritus
What is the z in the coefficient of y'? Is that a typo?

7. Jun 9, 2010

### dats13

It is a typo. It is suppose to be "a" not "z". Thanks for pointing that out.

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