I thought I had these problems down, but I found one for which I am having trouble.

$\displaystyle u_t = u_{xx} - u,\quad 0<x<\pi,\; t>0$

$\displaystyle u_x(0,t) = 1,\; u_x(\pi,t) = -1,\; t>0$

$\displaystyle u(x,0) = 0,\; 0<x<\pi$

**My attempt:**

Let $\displaystyle u(x,t) = v(x,t) + w(x,t)$ where $\displaystyle w(x,t)$ satisfies the BCs

$\displaystyle w_x(0,t) = 1,\; w_x(\pi,t) = -1,\; t>0$.

So, I picked $\displaystyle w(x,t) = \sin x$.

Now the new BCs and IC for $\displaystyle v(x,t)$ are

$\displaystyle u_t = u_{xx} - u,\quad 0<x<\pi,\; t>0$

$\displaystyle u_x(0,t) = 0,\; u_x(\pi,t) = 0,\; t>0$

$\displaystyle u(x,0) = -\sin x,\; 0<x<\pi$.

Now assume $\displaystyle v(x,t) = X(x)T(t)$ then

$\displaystyle \dfrac{T'(t)}{T(t)} = \dfrac{X''(x)}{X(x)} - 1 = -\lambda$

$\displaystyle X''(x) = \left(-\lambda + 1\right) X(x)$

*Case I:* $\displaystyle \lambda > 1$

Let $\displaystyle \omega = \sqrt{\lambda - 1}$

$\displaystyle X(x) = A_1 \cos \omega x + A_2 \sin \omega x$

$\displaystyle X'(x) = -A_1 \omega \sin \omega x + A_2 \omega \cos \omega x$

But this doesn't seem to help me solve for $\displaystyle \omega$. Normally I would solve for one constant and then solve for \omega, plug this into $\displaystyle v(x,t)$ then use the IC to solve for the last constant.