Can Laplace Transforms Reveal Bessel Functions as Solutions?

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Crush1986
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Homework Statement


[tex]xy''+y'+xy=0, y'(0)=0, y(0)=0[/tex] Using the method of Laplace transforms, show that the solution is the Bessel function of order zero.

Homework Equations


[tex]-(d/ds)L{f(x)}[/tex]

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...

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

This gives me [tex]-2sY(s)+1+sY(s)-1=0[/tex]

=> [tex]-sY(s)=0[/tex]

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

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

I know that the inverse of that is the zero order Bessel function. How would I prove with no table though?
 
Crush1986 said:
Ok so I'm on the right track for sure. My current expression is

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

I know that the inverse of that is the zero order Bessel function. How would I prove with no table though?
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.
 
SteamKing said:
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.
Yah I'm supposed to derive it I'm sure.