I have the following problem
$\left\{\begin{array}{l} u_{t t}=u_{x x}+2 u, \quad t>0, \quad 0<x<\pi \\ \left.u\right|_{t=0}=0,\left.\quad u_{t}\right|_{t=0}=0 \\ \left.u_{x}\right|_{x=0}=1,\left.u_{x}\right|_{x=\pi}=1 \end{array}\right.$
It is clear that it needs to use D'Alembert's approach:
$u(x, t)=\frac{\varphi(x+a t)+\varphi(x-a t)}{2}+\frac{1}{2 a} \int_{x-a t}^{x+a t} \psi(\alpha) d \alpha$
where
$u(x, 0)=\varphi(x), \quad u_{t}(x, 0)=\psi(x)$
In my case I have $\varphi(x) = \psi(x) = 0$. What id does mean and how to solve the problem in this case? And also, how to treat this term $2u$ in the equation?
D'Alembert's solution is hardly adapted to solve the wave equation with an additional term $2u$. Instead, let us use separation of variables. To do so, we need to substract the stationary solution $$ u^*(x) = \frac{\sin(x\sqrt{2}) - \tan\left(\frac{\pi}{\sqrt{2}}\right) \cos(x\sqrt{2})}{\sqrt{2}} $$ by introducing $v = u^* + u$. Thus, $v$ satisfies the PDE problem $v_{tt} = v_{xx} + 2 v$ with homogeneous Neumann boundary conditions, and with the initial conditions $v = u^*$ and $v_t = 0$ at $t=0$. Separation of variables $v(x,t) = X(x)T(t)$ gives $$ T'' + \lambda T = 0, \qquad X'' + (\lambda+2)X = 0 $$ where $\lambda$ is the separation constant. For $\lambda >0$, solutions are of the form \begin{aligned} T(t) &= A \cos(t\sqrt{\lambda}) + B \sin(t\sqrt{\lambda})\\ X(x) &= C \cos(x\sqrt{\lambda+2}) + D \sin(x\sqrt{\lambda+2}) \end{aligned} Hope you can take it from here.