1. ## d'alembert solution Help

Hi I have been trying to figure out how this method works.

Searched the forum but got no results. Have attached an example Im trying to do. If someone could show me it would be great.

Or even show me a similar example worked out.

Edit: I am trying another question
u_tt = u_xx
u( x, 0) = 0 = f(x) ?
u_t( x,0) = xe^-(x^2) = g(x)?

So as f(x) = 0 can I just figure out the ingegral of g(x) with limits (x+ct) and (x-tc)

The answer is supposed to be 11 but I just get

1/4 [ -e^(x+t)^2 + e^-(x-t)^2]

I use the formula and as f(x) = 0

2. Originally Posted by Niall101
Hi I have been trying to figure out how this method works.

Searched the forum but got no results. Have attached an example Im trying to do. If someone could show me it would be great.

Or even show me a similar example worked out.

Edit: I am trying another question
u_tt = u_xx
u( x, 0) = 0 = f(x) ?
u_t( x,0) = xe^-(x^2) = g(x)?

So as f(x) = 0 can I just figure out the ingegral of g(x) with limits (x+ct) and (x-tc)

The answer is supposed to be 11 but I just get

1/4 [ -e^(x+t)^2 + e^-(x-t)^2]

I use the formula and as f(x) = 0
The answer can't be 11! If the answer was $u(x,t) = 11$, then $u_t(x,t) = 0$ always and so $u_t(x,0) = 0$ and not $u_t(x,0) = xe^{-x^2} \ne 0$ for all $x$. Your answer (almost) is correct

$
u(x,t) = \frac{1}{4} \left( e^{-(x-t)^2} - e^{-(x+t)^2}\right)
$

It satisfies both the PDE and the initial conditions.

3. Hey Thanks for that. Someone on the college forum said they got 11 for the answer which threw me way off.