Discussion
The diffusion equation
\(\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial u}{\partial t},\)
\(u = u(x, t),\ u(0, t) = 0 = u(\pi, t),\ u(x, 0) = \cos x \sin 5x\)
admits the solution
Answer: A
The solution is \(\frac{e^{-36t}}{2}\left[\sin 6x + e^{20t}\sin 4x\right]\).
Write the initial data as a sine series using \(\cos A\sin B = \frac{1}{2}[\sin(A + B) - \sin(A - B)]\):
\[ \cos x\sin 5x = \tfrac{1}{2}(\sin 6x + \sin 4x). \]
Each mode \(\sin nx\) satisfies the boundary conditions and decays as \(e^{-n^{2}t}\) under \(u_{xx} = u_{t}\). So
\[ \begin{aligned} u &= \tfrac{1}{2}\left(e^{-36t}\sin 6x + e^{-16t}\sin 4x\right) \\ &= \tfrac{e^{-36t}}{2}\left(\sin 6x + e^{20t}\sin 4x\right) \end{aligned} \]
The option with \(\sin 4x\) carrying \(e^{-36t}\) attaches the decay rates to the wrong modes. The others use \(\sin 3x, \sin 5x\) or \(\sin 5x, \sin x\), which are not the modes of \(\cos x\sin 5x\).
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