Bessel's Eq. of order 0 and solution help

[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

 

[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.

 

[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?

 

[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}.

 

[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 science J_{-3/2} is also a solution. So the choice of a_0=1 comes from definition, which makes Bessel functions special functions.

 

[mkkrnfoo85]

thanks.

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

-Mark