Question on PDE solving using Variable separable. I am trying to solve the equation $u_t + 2u_x = u_{xx}$ , with initial conditions 
subject to the boundary conditions $u(0, t) = 0,u(l, t) = 0$, and the initial condition $u(x, t) = xexp(x)$.
I assume the solution of the form $u(x,t) =X(x)T(t).$
I get two equations $\frac{T'}{T} = k$ and $\frac{X'' - 2X'}{X} = k.$
Solving for T we get $T(t) = Aexp(kt)$ 
Solving for $X(t) = c'_1[exp(\sqrt{1+k})x] +c'_2[exp(-\sqrt{1+k})x]$
$u(x,t) = XT = exp(kt)(c_1[exp(\sqrt{1+k})x] +c_2[exp(-\sqrt{1+k})x)]$
When I proceed with using the initial conditions by simple substitution I get $c_1+c_2 = 0$, which yields $u(x,t) = 0$.
How do I proceed? Please give me hint.
Thanks

The nontrivial solutions to the $X$ eigenvalue equation $k X + 2 X' - X'' = 0$ with $X(0)=0$, $X(\ell) = 0$ are 
$$ X(x) = e^x \sin(n \pi x/\ell), \; k = -1 - n^2 \pi^2/\ell^2 $$
for positive integers $n$.
Thus you should get
$$ u(x,t) = \sum_{n=1}^\infty c_n e^{x - (1 + n^2 \pi^2/\ell^2) t} \sin(n \pi x/\ell)$$
The $c_n$ must then be determined to satisfy your initial condition (Hint: Fourier series).
