I would like to solve $\begin{cases} u_t=4u_{xx}+\sin{(\pi x)}+\sin{(2\pi x)} & 0<x<1, t>0 \\ u(0,t)=0 \\ u(1,t)=0 \\ u(x,0)=0\end{cases}\tag*{}$ using an eigenfunction expansion.
My attempt:
By solving an eigenvalue problem $\begin{cases} \phi''(x) = \lambda \phi (x), 0<x<1 \\ \phi (0)=0 \\ \phi (1)=0\end{cases}\tag*{}$
we obtain that the solution is $\phi_n(x)=B_n \sin{n\pi x}$($B_n$:constant) and the corresponding eigenvalue is $\lambda = -n^2\pi^2$. Here $n=1,2,3,\cdots$.
So we perform the eigenfunction expansion and we have $\displaystyle u(x,t) =\sum_{n=1}^{\infty} c_n(t)\sin{n\pi x}\tag*{}$ where $c_n(t)$ is a function of $t$. This $c_n$ satisfies, by initial condition, $\displaystyle u(x,0) = 0 = \sum_{n=1}^{\infty} c_n(0)\sin{n\pi x}\tag*{}$ Therefore $c_n(0)=0$ for all $n$.
Now substitute the expansion to the PDE: $\displaystyle\sum_{n=1}^{\infty} \dot{c_n}(t)\sin{n\pi x} =\sum_{n=1}^{\infty} (-4n^2\pi^2)c_n(t)\sin{n\pi x}+\sin{(\pi x)}+\sin{(2\pi x)}\tag*{}$
This means $\dot{c_n}(t) = \begin{cases} -4n^2\pi^2 c_n(t) & n\neq 1,2 \\ (1-4\pi^2)c_1(t) & n=1 \\ (1-16\pi^2)c_2(t) & n=2 \end{cases} \tag*{}$
Solve this and we get $c_n(t) = \begin{cases} A_ne^{-4n^2\pi^2t} & n\neq 1,2 \\ A_1e^{(1-4\pi^2)t} & n=1 \\ A_2e^{(1-16\pi^2)t} & n=2 \end{cases} \tag*{}$
For some constant $A_n$.
However, by the initial condition, $A_n=c_n(0)=0$. This leads to $u(x,t)=0$. But it obviously does not satisfy the PDE.
What went wrong in the solution? I suspect there is a typo in the problem. For reference, here is the original problem:
The differential equations for $c_1(t)$ and $c_2(t)$ should be \begin{align*} \dot{c_1} & = -4\pi^2c_1 + 1 \\ \dot{c_2} & = -16\pi^2c_2 + 1 \end{align*}