# Green function

1.
The zero order bessel function J0(0+) = 1, J'0(0+) = 0. and J0(t) satisfies teh differential equation

2. ty''(t) + y'(t) + ty(t) = 0, t>0
Prove thata the Laplace transform of J0(t) is 1/(s + 1)^1/2

## The Attempt at a Solution

Could anyone help on this. How to I go about with the question. Tried using a general solution that is J(t) = C1e^r1t + C2e^r2t..But im not sure if that right... I wasn't able to prove...Thx

## The Attempt at a Solution

Integral
Staff Emeritus
Gold Member
Why not start by finding the Laplace transform of your DE?

thx for replying...But how do I find the Laplace transform of the differential equation? Do i convert it to general solution J(t) = C1e^r1t + C2e^r2t. Is that right? Because i tried using that general equation and found two values of t and substituted in the general equation and solve simultaneously..but still wasn't able to prove...

I've found websites and even read books. It all says the same thing about bessel equation having a general solution. http://eqworld.ipmnet.ru/en/solutions/ode/ode0213.pdf
But what i dont get is that the in the general solution what is my v in my DE. I dont see any relationship. I am so confused..

HallsofIvy
Homework Helper
If you are working with Bessel functions and Green's functions (you titled this "Green function" but I don't see anything in the problem that has anything to do with Green's function) then you should have learned long ago that functions of the form $C_1e^{r_1t}+ C_2e^{r_2t}$ satisfy linear equations with constant coefficients. Bessel's equation, the equation you are given, obviously does NOT have integer coefficients. You are not expected to "find" a solution, you are expected to find the Laplace transform of the differential equation.

Do you know how to find the Laplace transform of something like f(x) dy/dx in terms of the Laplace transform of y?

I am not quite sure but is the Lapalce transform of f(x) dy/dx is sY(s) - Yo and the laplace transform of d2y/dx2 is s2Y(s) - sY0 - Y1.

If I am wrong. Please kindly correct me and if you could kindly give me some hints to go about with the question. Thx

ty''(t) + y'(t) + ty(t) = 0
Divide the equation by t gives

y''(t) + 1/ty'(t) + ty(t) = 0

Using Laplace Transform it gives

s2*Y - s*y0 - y1 + 1/t*(S*Y - y0) + t*Y = 0

(s2 + 1/t*s + t)*Y - s - 1/t = 0

so now Y = (s = 1/t)/(s2 + 1/t*s + t)

but i still cant prove the equation.....:(

HallsofIvy
Homework Helper
The Laplace transform of tf(t) is NOT t times the Laplace transform of f(t)!

oh there was an error

ty''(t) + y'(t) + ty(t) = 0
Divide the equation by t gives

y''(t) + 1/t*y'(t) + y(t) = 0

Using Laplace Transform it gives

s2*Y - s*y0 - y1 + 1/t*(s*Y - y0) + Y = 0

(s2 + 1/t*s + 1)*Y - s - 1/t = 0

so now Y = (s + 1/t)/(s2 + 1/t*s + 1)

I have tried completing the square but it doesnt help.....Is my Y correct?