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
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets define
$$
\varphi\pars{x,t} \equiv \expo{-x + t}\,\mrm{u}\pars{x,t}\iff
\mrm{u}\pars{x,t} = \expo{x - t}\varphi\pars{x,t}
$$
$\ds{\varphi\pars{x,t}}$ satisfies
$$
\partiald{\varphi\pars{x,t}}{t} =
\partiald[2]{\varphi\pars{x,t}}{x}\,,\qquad\qquad
\left\{\begin{array}{}
\ds{\varphi\pars{0,t} = \varphi\pars{\ell,t}= 0\,,\ \forall\ t}
\\[2mm]
\ds{\varphi\pars{x,0} = \expo{-x}\,\mrm{u}\pars{x,0} = x}
\end{array}\right. 
$$

\begin{align}
\varphi\pars{x,t} & =
\sum_{k\ \in\ \Omega}A_{k}\pars{t}\sin\pars{kx}\,,\qquad
\Omega \equiv \braces{k \in \mathbb{R}\ \mid\ k = n\,{\pi \over \ell}\,,\ n = 1,2,3,\ldots}\label{1}\tag{1}
\end{align}
Expression \eqref{1}, for $\ds{\varphi\pars{x,t}}$, already satisfies the boundary conditions
$\ds{\pars{~\varphi\pars{0,t} = \varphi\pars{\ell,t} = 0\,,\ \forall\ t~}}$. Moreover,
\begin{align}
\sum_{q\ \in\ \Omega}\braces{\totald{A_{q}\pars{t}}{t} + q^{2}}\sin\pars{qx} = 0
\end{align}
Multiply both sides by
$\ds{\left.\sin\pars{kx}\right\vert_{\ k\ \in\ \Omega}}$ and integrate over $\ds{\pars{0,\ell}}$. It leads to
$$
0 = \totald{A_{k}\pars{t}}{t} + k^{2}A_{k}\pars{t}\implies
A_{k}\pars{t} = A_{k}\pars{0}\expo{-k^{2}t}
$$

$\ds{\varphi\pars{x,t}}$ is reduced to:
\begin{align}
\varphi\pars{x,t} & =
\sum_{k\ \in\ \Omega}A_{k}\pars{0}\expo{-k^{2}t}\sin\pars{kx}
\\[5mm] \implies &
x = \varphi\pars{x,0} = \sum_{k\ \in\ \Omega}A_{k}\pars{0}\sin\pars{kx}
\\[5mm] 
\implies &
\int_{0}^{\ell}x\sin\pars{kx}\,\dd x
\\[5mm] & =
\sum_{q\ \in\ \Omega}
A_{q}\pars{0}\int_{0}^{\ell}\sin\pars{kx}\sin\pars{qx}\,\dd x =
{\ell \over 2}\,A_{k}\pars{0}
\\[5mm]
\implies
A_{k}\pars{0} & =
{2 \over \ell}\int_{0}^{\ell}x\sin\pars{kx}\,\dd x \\[5mm] & =
\pars{-1}^{n + 1}\,{2 \over n\pi/\ell}\,,\qquad k = n\,{\pi \over \ell}
\end{align}

Then,
$$
\varphi\pars{x,t} =
{2\ell \over \pi}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n}\,
\sin\pars{{n\pi \over \ell}\,x}
\exp\pars{-\bracks{n\,{\pi \over \ell}}^{2}t}
$$
and
$$\bbx{%
\mrm{u}\pars{x,t} =
{2\ell \over \pi}\,\expo{x - t}
\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n}\,
\sin\pars{{n\pi \over \ell}\,x}\exp\pars{-\bracks{n\,{\pi \over \ell}}^{2}t}}
$$
A: 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).
