# Homework Help: Help With a Laplace Transform

1. Nov 28, 2015

### Crush1986

1. The problem statement, all variables and given/known data
$$xy''+y'+xy=0, y'(0)=0, y(0)=0$$ Using the method of Laplace transforms, show that the solution is the Bessel function of order zero.

2. Relevant equations
$$-(d/ds)L{f(x)}$$

3. The attempt at a solution
The only thing I got out of this when trying to solve it was y=0. Obviously not the intended answer. In the problem I'm told that the answer is the Bessel function of 0 order.

I'm pretty sure the parts I'm messing up is the Laplace transform of xy''. I haven't tried to take the Laplace transform of anything like that before and I'm sure it is where I'm messing up. I tried to use the relevant equation up there and I got...

$$-(d/ds)[(s^2Y(s)-s)]+sY(s)-1+-(d/ds)[Y(s)]=0$$

This gives me $$-2sY(s)+1+sY(s)-1=0$$

=> $$-sY(s)=0$$

which yeah, just gives me a dumb answer. Again I'm pretty sure it's with my xy'' term. Maybe even more :(

Thanks anyone for help.

2. Nov 28, 2015

### Crush1986

Oh... as soon as I posted this I thought... Y(s) is a function of s... I need to take that into account when doing derivatives... Dang it... as always it's right after I post I have a decent idea...

Last edited: Nov 28, 2015
3. Nov 28, 2015

### Crush1986

Ok so I'm on the right track for sure. My current expression is

$$y = L^-1[1/(s^2+1)^.5]$$

I know that the inverse of that is the zero order Bessel function. How would I prove with no table though?

4. Nov 28, 2015

### SteamKing

Staff Emeritus
Well fortunately, there are tables of the Laplace transforms of all kinds of special functions, including Bessel functions. One reference which has such a table is Abramowitz & Stegun, Handbook of Mathematical Functions, p. 1029:

http://people.math.sfu.ca/~cbm/aands/

If you don't trust such a reference, I suppose you could always derive the Laplace transform of your Bessel function from the definition and compare it with your earlier result.

5. Nov 28, 2015

### Crush1986

Yah I'm supposed to derive it I'm sure.