# Solutions of D.E - Bessel Function

1. Nov 14, 2004

### irony of truth

Hello, I hope someone can show me where I got stuck/wrong.

Verify that the Bessel function of index 0 is a solution to the differential equation xy" + y' + xy = 0.

Note that my "<= 1" DOES NOT mean less than or equal to 1 but an arrow pointing to the left... it is said to be "equation 1".

It is given that the bessel function of index 0 is:

y = (from n= 0 to infinity){(-1)^n x^(2n) / [(n!)^2 2^(2n)] } <=0

getting the 1st derivative (w/ respect to x) is

y' = (from n= 1 to infinity){(-1)^n x^(2n-1) / [(n!)(n-1)! 2^(2n - 1)] } <= 1

Can I say this?

y' = (from n= 0 to infinity){(-1)^(n+1) x^(2n+1) / [(n!)(n+1)! 2^(2n + 1)] } <= 2

going back to <= 1,

y" = (from n= 1 to infinity){(-1)^n (2n -1)x^(2n-2) / [(n!)(n-1)! 2^(2n - 1)] } <= 3 or

y" = (from n= 0 to infinity){(-1)^(n+1) (2n + 1)x^(2n) / [(n!)(n+1)! 2^(2n + 1)] } <= 4

Substituting y", y' and y with <=4, <=2 and <=0 respectively to the given differential equation... but my can't simplify it to 0. There are terms in my series that cause me not to cancel them out to become zero.. where are my mistakes?

2. Nov 15, 2004

### arildno

3. Nov 15, 2004

### irony of truth

Thank you very much.. I understood about the solutions...