Consider the the initial-boundary value problem $u_t=u_{xx}$ where $u(x,0)=f(x)$ and $u_x(0,t)-u(0,t)=0$ for $x>0$ and $t>0$ and $u$ remains bounded. Solve this problem by observing that the function $v=u_x-u$ satisfies the problem $v_t=v_{xx}$ where $v(x,0)=f'(x)-f(x)$ and $v(0,t)=0$ for $x>0$ and $t>0$. Let $f(x)=e^{-x}$.
So I took the Laplace transform and got $V_{xx}-sV=2e^{-x}$. I tried to guess a solution $V=Ae^{-x}$ and got $A= 2/(1-s)$ and when I inverse Laplace that I got $v=-2e^{t-x}$ but that doesn't solve the initial conditions.