$$\begin{align} &\frac{\partial^2u}{\partial t^2}+\frac{\partial u}{\partial t}+u = \alpha^2\frac{\partial^2u}{\partial x^2},\qquad && 0<x<L,\qquad t>0\\ &u(0,t) = u(L,t) = 0, && t>0\\[2ex] &u(x,0) = f(x), &&0<x<L\\ &\frac{\partial u}{\partial t}(x,0) = 0, && 0<x<L \end{align} $$
I need to derive the formal solution.
I used the method of separation of variables. What I did so far is:
Let $u(x,t) = X(x)T(t)$ and get : $$ \begin{cases} X'' = \lambda X\\ T''+T'+T = \alpha^2\lambda T \end{cases} $$
I found that $X(x) = Csin\frac{n\pi x}L$, but having trouble find T(t).
$$\begin{align}T''+T'+T &= \alpha^2\lambda T \\ T''+T'+(1 - \alpha^2\lambda)T &= 0\\ r^2+r +(1 - \alpha^2\lambda) & =0\\[2ex] r = -\frac12 \pm &\frac{\sqrt{1-4(1 - \alpha^2\lambda)}}2\\ r_n = r + \frac12\\[2ex] \text{and finally get:}\\ T(t) = C_1 e^{-\frac t2}cos(r_nt) &+C_2 e^{-\frac t2}sin(r_nt) \end{align} $$
Is it correct?
Does $T(t)$ has to satisfy $T'(0)=0$?