Discussion
Let \(u(x,t)\) be the solution of the initial-value problem
\[ \begin{aligned} &\frac{\partial^{2} u}{\partial t^{2}} - 9\frac{\partial^{2} u}{\partial x^{2}} = 0, \quad x \in \mathbb{R}, \quad t \gt 0, \\ &u(x,0) = e^{x}, \quad \frac{\partial u}{\partial t}(x,0) = \sin x. \end{aligned} \]
Then, the value of \(u\left(\frac{\pi}{2}, \frac{\pi}{6}\right)\) is
Answer: C
The value is \(\dfrac{1}{2}\left(e^{\pi} + \dfrac{5}{3}\right)\).
d'Alembert's formula with wave speed c = 3:
\[ \begin{aligned} u &= \tfrac{1}{2}[f(x + 3t) + f(x - 3t)] \\ &\quad + \tfrac{1}{6}\int_{x - 3t}^{x + 3t}\sin s\,ds \end{aligned} \]
At \(x = \pi/2\), \(t = \pi/6\), the end points are \(\pi\) and 0:
\[ \begin{aligned} u &= \tfrac{1}{2}(e^{\pi} + 1) + \tfrac{1}{6}(-\cos\pi + \cos 0) \\ &= \tfrac{1}{2}(e^{\pi} + 1) + \tfrac{1}{3} = \tfrac{1}{2}\left(e^{\pi} + \tfrac{5}{3}\right) \end{aligned} \]
The 1/3 options drop the \(e^{0} = 1\) term, and the minus signs reverse the integral of the initial velocity.
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