### Author Topic: Problem2  (Read 5596 times)

#### Aida Razi

• Sr. Member
•    • Posts: 62
• Karma: 15 ##### Problem2
« on: October 15, 2012, 01:34:28 AM »
Solution is attached,

#### Thomas Nutz

• Full Member
•   • Posts: 26
• Karma: 1 ##### Re: Problem2
« Reply #1 on: October 15, 2012, 12:04:21 PM »
I obtain the solution
$$u(x,t)=\frac{1}{2}((x-ct)e^{x+ct}+(x-ct)e^{x-ct})+ \int_{x-ct}^{x+ct}e^{-x}dx+\int_0^{t}\int_{x-ct}^{x+ct}e^{-x-t}dxdt$$
which becomes
$$\frac{2ce^{-x}}{1-c^2}+\frac{e^{-t}(e^{-ct}+e^{ct})}{1+c}$$
(plugging in $c=5$ doesn't really simplify it)

Did anyone get the same?
« Last Edit: October 15, 2012, 12:47:53 PM by Victor Ivrii »

#### Victor Ivrii ##### Re: Problem2
« Reply #2 on: October 15, 2012, 12:53:20 PM »
Thomas,

I fixed you LaTeX (it was a double subscript error \int_{0{}_{t} but your formula is wrong: you should integrate by $x$ (inner) from $x-c(t-s)$ to $x+c(t-s)$ and integrand is $e^{x-s}$ and then you integrate by $s$ from $0$ to $t$; also forgot $\frac{1}{2c}$ factor.

I have not checked other terms. You can see that your "solution" does not satisfy initial conditions (or equations)

#### Betty Zhu

• Newbie
• • Posts: 1
• Karma: 0 ##### Re: Problem2
« Reply #3 on: October 15, 2012, 04:03:18 PM »
Razi, I agree with you, I got the same answer....hope it is the correct one...