Steps to solve semi-infinite IBVP $$ 
        \begin{matrix}
        u_t=ku_{xx}\\
        u(0,t)=0,&amp; t&gt;0 \\
        u(x,0)=xe^{-ax}, &amp; t&gt;0 \\
        \end{matrix}
$$

I'm not really sure how to go about this problem. Any help would be appreciated. Thanks!
Of course use separation of variables:

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=kX''(x)T(t)$

$\dfrac{T'(t)}{kT(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-ks^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-kts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&amp;\text{when}~s\neq0\\c_1x+c_2&amp;\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-kts^2}\cos xs~ds$

$u(0,t)=0$ :

$\int_0^\infty C_2(s)e^{-kts^2}~ds=0$

$C_2(s)=0$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds$

$u(x,0)=xe^{-ax}$ :

$\int_0^\infty C_1(s)\sin xs~ds=xe^{-ax}$

$C_1(s)=\dfrac{2}{\pi}\int_0^\infty xe^{-ax}\sin sx~dx=\dfrac{4as}{\pi(s^2+a^2)^2}$

$\therefore u(x,t)=\int_0^\infty\dfrac{4ase^{-kts^2}\sin xs}{\pi(s^2+a^2)^2}ds$
Step 1

Define $U_0$ to be an odd extension of $u_0(x)=xe^{-ax}$ to the whole real line i.e. $U_0(x)=u_0(x)$ for $x&gt;0$ and $U_0(x)=-u_0(-x)$ for $x&lt;0$.  

Step 2

Solve the Cauchy problem for the heat equation

\begin{align*}
U_t&amp;=kU_{xx} \quad t&gt;0, x\in\mathbb{R}\\
U(0,x)&amp;=U_0(x) \quad x\in\mathbb{R}
\end{align*}

Step 3

Observe that $U(t,0)=0$ and thus the restriction of $U$ to positive half-line solves your problem.