# Bessel's Eq. of order 0 and solution help

1. Dec 12, 2005

### mkkrnfoo85

Hello all,
I'm studying for my diff/eq final, and I am having a lot of trouble understanding the answer to a question involving Bessel's equation of order 3/2. So, here's the question. please help::::::

The following question concerns Bessel's Equation:

$$x^2y''+xy'+(x^2-(\frac{3}{2})^2)y=0$$

The Bessel Function $$J_{3/2}(x)$$ is the Frobenius series solution which is finite at x = 0 Find the first three terms in its series expansion around x = 0.
Ok, so I have the definition of the Bessel Function as:

$$J_r(x)=\sum_{n=0}^{\infty} a_nx^{n+r}, a_0=1$$

where 'r' is the root of the indicial equation:

$$r(r-1)+(1)r-(\frac{3}{2})^2$$

The answer to this question says::

$$J_{3/2}=x^{3/2}(a_0+a_1x+a_2x^2+a_3x^3+...)$$

$$y=x^{3/2}(1+0x-\frac{1}{10}x^2+0x^3+...)$$

I don't know how they changed the $$a_0,a_1,a_2,...$$ values to those numbers right there.
Thanks for the help.
-Mark

Last edited: Dec 12, 2005
2. Dec 13, 2005

### incognitO

just insert the series $J_{\frac{3}{2}}$ on your O.D.E. The only way the equality will hold is if all the therms on the O.D.E. are zero. That will give you a condition over the $a_n$'s.

Last edited: Dec 13, 2005
3. Dec 13, 2005

### mkkrnfoo85

thx, Just one more question...

thanks, incognitO. That actually helped me put things together. I just had one more question:::

why does the term $$a_0$$ = 1??

Is it just part of the definition of the Bessel Function, or something else?

4. Dec 13, 2005

### Dr Transport

Definition, if memory serves me correctly, the terms alternate. You can find the solution in terms of $$a_{n+2} = f(n)a_{n}$$.

5. Dec 13, 2005

### incognitO

in series solutions with recurrence relations as the type mentioned by Dr Transport, is costumary to choose $a_0=1$, $a_1=0$ to find the first independent solution and then $a_0=0$ and $a_1=1$ to find the second independent solution. In the case of your Bessel function, this is not necesary scince $J_{-3/2}$ is also a solution. So the choice of $a_0=1$ comes from definition, wich makes Bessel functions special functions.

Last edited: Dec 13, 2005
6. Dec 15, 2005

### mkkrnfoo85

thanks.

Thanks incognitO, and Dr Transport. Really helped me with the understanding.

-Mark

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