Int. Bessel Functions and Equation

[Aquinox]

Homework Statement

Prove that [tex]J_{n}[/tex], [tex]Y_{n}[/tex] satisfy
[tex]x^{2}*y''(x)+x*y'(x)+(x^{2}-n^{2})*y(x)=0[/tex]

where [tex]n\inZ[/tex] and [tex]x\in(R_{>0}[/tex]

Homework Equations

The standard definitions of the bessel integrals as given here:
http://en.wikipedia.org/wiki/Bessel_Functions

The Attempt at a Solution

I've tried to work on the first kind (J) by differentiating under the integral but never got to an equality. I'm pretty sure i can't integrate these (b/c of the sine in the cosine) but with differentiation I'm repeatedly stuck.
I'm pretty sure I'm missing something quite easy for the first step - and that's what I'd like to ask of you.

Thanks in advance

 

[mighty2000]

!



Thank you for your forum post. Proving that J_{n} and Y_{n} satisfy the given equation is a straightforward application of the definitions of Bessel functions. First, we can rewrite the equation as follows:

x^{2}*y''(x)+x*y'(x)+(x^{2}-n^{2})*y(x)=0
x^{2}*(d^{2}/dx^{2})y(x)+x*(d/dx)y(x)+(x^{2}-n^{2})*y(x)=0

Next, we can substitute in the definition of J_{n} and Y_{n} as given on the Wikipedia page:

J_{n}(x) = (1/pi)*∫_{0}^{pi} sin(n*t-x*sin(t)) dt
Y_{n}(x) = (1/pi)*∫_{0}^{pi} cos(n*t-x*sin(t)) dt

We can then use the chain rule to differentiate these expressions with respect to x:

(d/dx)J_{n}(x) = (1/pi)*∫_{0}^{pi} cos(n*t-x*sin(t)) * (-sin(t)) dt
(d/dx)Y_{n}(x) = (1/pi)*∫_{0}^{pi} (-sin(n*t-x*sin(t))) * (-sin(t)) dt

Using the product rule, we can then differentiate these expressions again with respect to x:

(d^{2}/dx^{2})J_{n}(x) = (1/pi)*∫_{0}^{pi} cos(n*t-x*sin(t)) * (-sin(t)) * (-sin(t)) + cos(n*t-x*sin(t)) * (-cos(t)) * (-sin(t)) dt
= (1/pi)*∫_{0}^{pi} cos(n*t-x*sin(t)) * (sin^{2}(t) + cos^{2}(t)) dt
= (1/pi)*∫_{0}^{pi} cos(n*t-x*sin(t)) dt
= J_{n}(x)

Similarly, we find that (d^{2}/dx^{2})Y_{n}(x) = Y_{n}(x). Substituting these expressions into the original equation, we get:

x^{2}*J_{n}(x) + x*J_{n}(x) + (x^{2