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## Homework Statement

The question is

Ytt- c^2Yxx =0 on the doman 0<x< +infinity

where initia conditions are y(x,0) = e^-x^2 = f(x) , Yt(x,0) =x*e^-x^2 = g(n)

and boundary condition is y(0,t) = 0 and c = 2

## Homework Equations

D'Almbert solution 1/2(f(x+ct)+f(x-ct))+1/2c∫ g(n) dn over the limits (x+ct) to (x-ct)

## The Attempt at a Solution

So I just plugged in the f(x) and g(x) in the D'alembert solution to get

1/2(e^-(x+2t)^2+e^-(x-2t)^2)+1/4 ∫ g(n)

but since we need to make it an odd function to apply D'alembert I then did -f(x) = f(-x) and -g(x) = g(-x)

So I split the ∫ g(n) into two parts which are

∫g(n) over the limits 0 to x-2t + ∫g(n) over the limits x+2t to 0 then in order to give it oddness to the first part

I did -∫g(-n) over the limits 0 to x-2t so the limits changed to 0 to x+2t and finally I ended up with

1/4( ∫-1/2*e^-x^2 over the limits 0 to x+2t + ∫-1/2*e^-x^2 over the limits x+2t to 0)

integration of x*e^-x^2 = -1/2*e^-x^2.

As for the f(x) we know f(-x) = -f(x) so I did

1/2(1/2(e^-(x+2t)^2-e^-(x+2t)^2) to give it oddness, now that both functions have been given oddness my wave solution is

1/2(1/2(e^-(x+2t)^2-e^-(x+2t)^2)+ 1/4( ∫-1/2*e^-x^2 over the limits 0 to x+2t + ∫-1/2*e^-x^2 over the limits x+2t to 0)

When I apply the boundary condition y(0,t) = 0 I see that

the f(x) term 1/2(1/2(e^-(x+2t)^2-e^-(x+2t)^2) = 0. As for the g(n) part after applying the limits and boundary condition

I end up with

1/4(-1/2+1/2*e^-(2t)^2+ -1/2 *e^-(2t)^2 + 1/2) = 0

Can you please tell me if everything I did above was the correct way to do it. Many thanks.