Can Laplace Transforms Reveal Bessel Functions as Solutions?

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Homework Help Overview

The problem involves a second-order differential equation of the form xy'' + y' + xy = 0 with initial conditions y'(0) = 0 and y(0) = 0. The goal is to demonstrate that the solution corresponds to the Bessel function of order zero using Laplace transforms.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the challenges of applying the Laplace transform to the term xy'' and express uncertainty about the resulting expressions. There is a recognition of the need to account for Y(s) as a function of s when taking derivatives. Some participants explore the inverse Laplace transform and its relation to Bessel functions, questioning how to prove the connection without reference tables.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and insights. Some have identified potential errors in their calculations and are reconsidering their approaches. There is a mention of external references for Laplace transforms, but the focus remains on deriving the necessary relationships independently.

Contextual Notes

Participants note the importance of deriving the Laplace transform of the Bessel function from first principles, as reliance on tables is discouraged in this context.

Crush1986
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Homework Statement


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.

Homework Equations


-(d/ds)L{f(x)}

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.
 
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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...
 
Last edited:
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?
 
Crush1986 said:
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?
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.
 

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