Use separation of variables to solve the following BVP with mixed boundary conditions; $$ \begin{cases} u_t = u_{xx} &x\in (0,l), t>0 \\ u(0,t) = T_0 & t>0 \\ u_x(l,t) = 0 & t>0 \\ u(x, 0) = T_1 &x\in(0,l) \end{cases} $$ where $T_0$ and $T_1$ are constants. In particular solve the above for $T_0 = 1, T_1=5$.
I tried to do it and got stuck. In particular, the condition $T_0 \neq T_1$ threw me of. This problem is in a section on how to solve boundary value problems by separation of variables and using Fourier series.
Separation of variables problems such as this require homogeneous endpoint conditions, which is usually achieved by a substitution in a case like this. $$ v(x,t)=u(x,t)-T_0. $$ The new equation for $v$ is $$ v_t = v_{xx} \\ v(0,t)=u(0,t)-T_0=0 \\ v_{x}(l,t)=u_{x}(l,t)=0 \\ v(x,0)=u(x,0)-T_0=T_1-T_0. $$ Now when you separate $v(x,t)=X(x)T(t)$, the ODE in $x$ becomes $$ \frac{X''}{X}=\lambda,\;\; X(0)=0,\; X'(l)=0. $$ The eigenfunctions are $$ \sin(\pi x/2l), \sin(3\pi x/2l),\sin(5\pi x/2l),\cdots. $$ with corresponding eigenvalues $(2n+1)^2\pi^2/4l^2$. These are automatically orthogonal, and you have to expand $v(x,t)=\sum_{n=1}^{\infty}C_nX_n(x)T_n(t)$ in such a way that $v(x,0)=T_1-T_0$, which leads to a Fourier series problem in the orthogonal eigenfunctions $X_n$, and gives the constants $C_n$.