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This problem is from our recitation which I do not have solutions for, and I'm stuck on the very last part where I need to satisfy the $u_t(x,0)$ initial condition.

The problem is:

$$ \left\{ \begin{split} u_{tt} &= c^2u_{xx}, \quad\qquad x>0,t>0\\ u(&0,t) = 0 \qquad\qquad\qquad t>0\\ u(&x,0) = \sin(x) = f(x)\;\quad x>0\\ u_t&(x,0) = e^{-x} = g(x)\quad\quad x>0 \end{split} \right. $$ Since we have Neumann conditions we use an even extension. Let $$ u(x,0) = \begin{cases} f(x) &\text{if } x > 0 \\ f(-x) &\text{if } x < 0 \end{cases} $$ and $$ u_t(x,0) = \begin{cases} g(x) &\text{if } x > 0 \\ g(-x) &\text{if } x < 0 \end{cases} $$ Then D'Alembert's formula gives: $$ \begin{split} u(x,t) &= \frac{1}{2}\big(f(x-ct)+f(x+ct)\big) + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds\\ &= \frac{1}{2}\big(f(x)+f(x)\big) + \frac{1}{2c}\int_{x}^{x} g(s)\,ds \\ &= f(x)=\sin(x) \end{split} $$ thus the First initial condition is satisfied. For the second one we have $$ u_t(x,t) = \frac{1}{2}\big(-cf(x-ct)+cf(x+ct)\big) + \frac{d}{dt}\bigg [ \frac{1}{2c} \int_{x-ct}^{x+ct} g(s)\,ds \bigg ] $$

My problem is I'm not sure what to do with the $$ \frac{d}{dt}\bigg [ \frac{1}{2c} \int_{x-ct}^{x+ct} g(s)\,ds \bigg ] $$ term. Should I differentiate the integral and then plug in $t = 0$ for the whole expression? How do I get the term $e^{-x}$ from this?

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FTC says $$\frac{d}{dt}\int_{a(t)}^{b(t)}f(s)ds = f(b(t))b'(t)-f(a(t))a'(t).$$ Substituting your functions, we get $$\frac{d}{dt}\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)ds = \frac{1}{2c}(g(x)c+g(x)c) = g(x),$$ as desired.

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