Can someone tell how to solve this type of problem please?
$$\begin{cases}u_{tt}=c^2u_{xx}+e^t\sin(5t)&t>0,\,x\in(0,\pi)\\u(0,t)=0=u(\pi,t)&t>0\\u(x,0)=0&x\in(0,\pi)\\u_t(x,0)=\sin(3x)&x\in(0,\pi)\end{cases}$$
I searched on internet but all the problem I saw had the following conditions:
$u(0,t) =$ (some function)
$u(x,0) =$ (some function)
$u_t(x,0) = 0$
But in this problem the first two conditions are equal 0 for me and the third one is not. How can I solve this please?
Disclaimer: there might be mistakes in the algebra here, I haven't thoroughly checked it. The main point of this answer is to guide you on how this can be solved.
You have an inhomogeneous equation with homogeneous boundary conditions - this is the form we generally want such an equation to be in. Try separation of variables $u(x,t)=\sum_{n\ge0}X_n(x)T_n(t)$. Considering the homogeneous problem to start with, $X$ satisfies the following equation. $$X''(x)+\lambda X(x)=0\\X(0)=X(\pi)=0\\\implies X_n(x)=\sin(nx),\,\,\lambda=n^2,\,\,n\in\Bbb N$$ Therefore we have $u(x,t)=\sum_{n\ge1}T_n(t)\sin(nx)$. Now note that one of the initial conditions has a $\sin3x$ in it, and no other sine terms. So we will only keep the $n=3$ term.
Now differentiate this and compare to the inhomogeneous problem we have. $$u_{tt}-c^2u_{xx}=\left[T_3''(t)+9c^2T_3(t)\right]\sin(3x)=e^t\sin(5t)$$ Integrate with respect to $x$ from $0$ to $\pi$ to get $$\frac23\left[T_3''(t)+9c^2T_3(t)\right]=\pi e^t\sin(5t)\\T_3''(t)+9c^2T_3(t)=\frac{3\pi}2e^t\sin(5t)$$ We also have boundary conditions for $T_3$ as follows$$T_3(0)=0,T_3'(0)=1$$
This is a linear ordinary differential equation which can be solved. Once solved, combine with $\sin 3x$ and you have the solution.
Edit:
Here is the solution that you got for $T_3$:$$T_3(t)=c_1\cos(3ct)+c_2\sin(3ct)-\frac9{130}\pi e^t\sin(5t)-\frac3{65}\pi e^t\cos(5t)$$ Use the initial conditions to fix the values of $c_1$ and $c_2$. Then your final solution is $u(x,t)=y(t)\times\sin3x$.