Question

24. The solution $u(x, t)$ of the heat transfer problem \begin{equation} \begin{cases} 2 \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t} & 0 < x < \pi, t > 0 \\ u(0, t) = 0 & u(\pi, t) = 0 \\ u(x, 0) = \sin(x) + \frac{1}{3} \sin(3x) & 0 < x < \pi \end{cases} \end{equation} is \begin{align} \text{A. } & u(x, t) = \sin(x) e^{-2t} + \frac{1}{3} \sin(3x) e^{-18t} \\ \text{B. } & u(x, t) = \sin(x) e^{-2t} + \frac{1}{3} \sin(2x) e^{-8t} \\ \text{C. } & u(x, t) = \sin(x) e^{-t} + \frac{1}{3} \sin(3x) e^{-9t} \\ \text{D. } & u(x, t) = \sin(x) e^{-2t} + \sin(3x) e^{-18t} \\ \text{E. } & u(x, t) = \sin(x) e^{-2t} + \sin(2x) e^{-8t} + \sin(3x) e^{-18t} \end{align}

          24. The solution $u(x, t)$ of the heat transfer problem \begin{equation*} \begin{cases} 2 \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t} & 0 < x < \pi, t > 0 \\ u(0, t) = 0 & u(\pi, t) = 0 \\ u(x, 0) = \sin(x) + \frac{1}{3} \sin(3x) & 0 < x < \pi \end{cases} \end{equation*} is \begin{align*} \text{A. } & u(x, t) = \sin(x) e^{-2t} + \frac{1}{3} \sin(3x) e^{-18t} \\ \text{B. } & u(x, t) = \sin(x) e^{-2t} + \frac{1}{3} \sin(2x) e^{-8t} \\ \text{C. } & u(x, t) = \sin(x) e^{-t} + \frac{1}{3} \sin(3x) e^{-9t} \\ \text{D. } & u(x, t) = \sin(x) e^{-2t} + \sin(3x) e^{-18t} \\ \text{E. } & u(x, t) = \sin(x) e^{-2t} + \sin(2x) e^{-8t} + \sin(3x) e^{-18t} \end{align*}

24. The solution u(x, t) of the heat transfer problem
2 (∂^2 u)/(∂ x^2) = (∂ u)/(∂ t) 0 < x < π, t > 0
u(0, t) = 0 u(π, t) = 0
u(x, 0) = sin(x) + (1)/(3)sin(3x) 0 < x < π
is
A. u(x, t) = sin(x) e^-2t + (1)/(3)sin(3x) e^-18t
B. u(x, t) = sin(x) e^-2t + (1)/(3)sin(2x) e^-8t
C. u(x, t) = sin(x) e^-t + (1)/(3)sin(3x) e^-9t
D. u(x, t) = sin(x) e^-2t + sin(3x) e^-18t
E. u(x, t) = sin(x) e^-2t + sin(2x) e^-8t + sin(3x) e^-18t

Added by Mary T.

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