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Find the solution of the initial-boundary value problem

$$ \begin{aligned} & u_t=3 u_{x x}, \quad 0<x<4, t>0 \\ & u(0, t)=0, u(4, t)=0 \\ & u(x, 0)=\sin 2 \pi x \cdot \cos 3 \pi x . \end{aligned} $$

By using solution formula $$ u(x, t)=\sum_{n=1}^{\infty} A_n e^{-\left(\frac{n \pi}{4}\right)^2 3 t} \sin \frac{n \pi x}{4} $$ $\sin u(x, 0)=\sin 2 \pi x \cdot \cos 3 \pi x$ , by using trigonometric identities: $$ \begin{aligned} & u(x, 0)=\frac{1}{2}(\sin 5 \pi x-\sin \pi x) \\ & =-\frac{1}{2} \sin \pi x+\frac{1}{2} \sin 5 \pi x \\ & A_4=-\frac{1}{2} \quad A_{20}=\frac{1}{2} \\ & u(x, t)=-\frac{1}{2} e^{-3 \pi^2 t} \sin \pi x+\frac{1}{2} e^{-75 \pi^2 t} \sin \pi x \end{aligned} $$

What coefficient should I use? $A_1, A_5$ or above and why? I was wondering whether my process is correct, can anyone help me to do a simple check?

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    $\begingroup$ You should use $A_4$ and $A_{20}$. The $n$ in $A_n$ must be the same $n$ in $e^{-\left(\frac{n \pi}{4}\right)^2 3 t} \sin \frac{n \pi x}{4}$. (By the way, there is a typo in your solution: the last term should be $\frac{1}{2} e^{-75 \pi^2 t} \sin\color{red}{5} \pi x$.) $\endgroup$
    – Gonçalo
    Nov 5, 2023 at 6:13

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