I've been trying to solve:
$$ \left\{\begin{array}{c} u_{t}-u_{x x}=-u,\,\, 0<x<L, t>0 \\ u_{x}(0, t)=1 \\ u(1, t)=0 \\ u(x, 0)=x^2 cos(\pi x) + 2x + 1 \end{array}\right. $$
By separation of variables. My attempt:
$$ \begin{aligned} &u(x,t)=F(t) G(x) \longrightarrow F^{\prime}(t) G(t)=F(t) G^{\prime \prime}(x)-F(t) G(x) \\ &F^{\prime}(t)=-\lambda^{2} F(t) \rightarrow \frac{F^{\prime}(t)}{F(t)}=\frac{G^{\prime \prime}(x)}{G(x)}-1=-\lambda^{2} \\ &G^{\prime \prime}-G=-\lambda^{2} G^{\prime}+\lambda^{2} F=0 \\ &F(t)=e^{-\lambda^{2} t} \end{aligned} $$
If $\lambda = \pm 1$ $$ G^{\prime \prime}=0 \longrightarrow G(x)=a x+b \\ \left.\begin{array}{rl} \longrightarrow G(0) & =b=1 \\ \longrightarrow G^{\prime}(1) & =a+1=0 \end{array}\right\} \quad G(x)=1-x $$
If $\lambda\neq \pm 1$ $$ G(x)=A \cos \left(\sqrt{\lambda^{2}-1 x}\right)+B \operatorname{sen}\left(\sqrt{\lambda^{2}-1} x\right) \\ G^{\prime}(x)=-A \sqrt{\lambda^{2}-1} \sin \left(\sqrt{\lambda^{2}-1} x\right) +B \sqrt{\lambda^{2}-1} \cos \left(\sqrt{x^{2}-1} x\right)\\ \left.\begin{array}{rl} \longrightarrow G(0) & =A=1 \\ \longrightarrow G^{\prime}(1) & =-\sqrt{\lambda^{2}-1} \sin \left(\sqrt{\lambda^{2}-1}\right)+B \sqrt{\lambda^{2}-1} \cos \left(\sqrt{\lambda^{2}-1}\right)=0 \end{array}\right\} \\ G(x)= \cos \left(\sqrt{\lambda^{2}-1 x}\right)+\operatorname{tan(\sqrt{\lambda^{2}-1})}\operatorname{sen}\left(\sqrt{\lambda^{2}-1} x\right) $$
From here, I don't know how to keep going. Thanks a lot!
Hint.
Assuming
$$ G(x) = c_1\cos(\sqrt{\lambda^2-1}x)+c_2\sin(\sqrt{\lambda^2-1}x) $$
From the conditions on $G'(0)$ and $G(1)$ we have to determine $c_1,c_2$
$$ \left( \begin{array}{cc} \sin \left(\sqrt{\lambda ^2-1}\right) & \cos \left(\sqrt{\lambda ^2-1}\right) \\ \sqrt{\lambda ^2-1} & 0 \\ \end{array} \right) \left( \begin{array}{c} c_1 \\ c_2 \end{array} \right)=\left( \begin{array}{c} 0 \\ 0 \end{array} \right) $$
To avoid the trivial solution we choose $\lambda$ such that
$$ \det\left( \begin{array}{cc} \sin \left(\sqrt{\lambda ^2-1}\right) & \cos \left(\sqrt{\lambda ^2-1}\right) \\ \sqrt{\lambda ^2-1} & 0 \\ \end{array} \right)=-\sqrt{\lambda ^2-1} \cos \left(\sqrt{\lambda ^2-1}\right)=0 $$
or
$$ \lambda^2=\{1\}\cup\{\left(\frac{\pi}{2}(2k-1)\right)^2+1,\ \ k\in \mathbb{N}\} $$