Can Laplace Transforms Reveal Bessel Functions as Solutions?

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The discussion focuses on solving the differential equation xy'' + y' + xy = 0 using Laplace transforms, with the expected solution being the Bessel function of order zero. The initial attempts led to an incorrect result of y = 0, prompting the realization that the Laplace transform of the term xy'' was likely mishandled. After further consideration, the user arrived at the expression y = L^-1[1/(s^2+1)^(1/2)], recognizing that its inverse is the zero order Bessel function. The conversation highlights the need to derive the Laplace transform of the Bessel function from first principles, as opposed to relying solely on reference tables. Ultimately, the user is encouraged to prove the relationship without external resources.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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