$u_t(t,x)=u_{xx}(t,x)-u(t,x)$ and $x\in(0,1), t>0$
with boundary and initial conditions
$u(t,0)=0, u(t,1)=e-e^{-1}, u(0,x)=f(x)$.
I tried using auxiliary function $v(t,x)=u(t,x)-(e-e^{-1})x$ but failed to hold the first equation.
Without auxiliary function;
Let $u(t,x)=T(t)X(x)$ then by the first equation,
$T'X=TX''-TX$,
${T'\over T}={X'' \over X}-1:=-\lambda$,
Let $\mu := \lambda-1 $, then $X''-\mu X=0$,
$ X = c_1{e}^{\sqrt{\mu}x}+c_2{e}^{-\sqrt{\mu}x}$
But with boundary condition $u(t,1)=T(t)X(1)=e-e^{-1}$, I have no idea how to proceed.
Help me, Thank you!
Hint
Observe that the function $$ w(x) = e^x - e^{-x} $$ satisfies $$ w''(x) = e^x - e^{-x} = w(x) $$ or that $$ w''(x) - w(x) = 0 $$
Furthermore, $w(0) = 0$ and $w(1) = e - e^{-1}$.
So if you decompose $u(t,x) = v(t,x) + w(x)$ you have that $v(t,x)$ solves
$$ v_t = v_{xx} - v $$
with boundary data
$$ v(t,0) = v(t,1) = 0, \qquad v(0,x) = f(x) - w(x) $$
For $v$ you can solve the equation using, e.g., Fourier series.