I am studying boundary value problems and I have worked out the partial derivatives and worked out my $V*$ and its constants.
Now I have a heat equation to solve using $X(x)T(t)$. Using the auxiliary equation $X''(x)+k^2X(x)=0$ and subbing in my boundary conditions I know that $A=0$.
However for working out $B$ I am stuck as the second condition is $u(1,t)=100$.
How do I solve $B \sin k=100$?
When considering linear boundary value problems, you should apply the superposition principle in order to forget about non-homogenous boundary conditions. I guess your problem is something like:
$$u_t = u_{xx}, \quad 0 < x < 1, \quad t > 0,$$
with boundary and inital conditions:
$$u(0,t) = 0, \quad u(1,t) = 100, \quad t > 0; \quad u(x,0) = f(x), \quad x \in [0,1]$$
for some known function $f(x)$. Make now the substitution $u = v+w$, such that $v$ satisfies homogenous boundary conditions and $w$ "absorbe" them, so the sum of the two problems always leads to the original one. Assume $w(x,t) = A(t) x + B(t)$ and note that, for example, $w(x,t) = 100 \, x$ is a very good solution which satisfies the non-homogenous boundary conditions. Then, the problem for $v$ becomes:
$$v_t = v_{xx} - w_t + w_{xx} = v_{xx}, \quad 0 < x < 1, \quad t > 0,$$
with boundary and inital conditions:
$$v(0,t) = 0, \quad v(1,t) = 0, \quad t > 0; \quad v(x,0) = f_0(x)-w(x,0), \quad x \in [0,1],$$ and note that everything containing information from $w$ is known. Now the problem is to determine the function $v$, which satisfies an homogneous second order PDE (heat equation) subject to homogneous boundary conditions. We can now apply separation of variables (or Sturm-Liouville theory if you prefer so) to the PDE for $v$, assuming:
$$v(x,t) = X(x)T(t), \quad T \neq 0 \neq X, $$
which yields to the set of equations that you have already come up to before, i.e.:
$$\begin{align} X'' - \lambda X & = 0,\\ T' - \lambda T & = 0, \end{align} $$ where $\lambda$ is a constant, which can be negative, zero or positive. Apply the boundary conditions for $X$, which result to be: $X(0)=X(1)=0$ to solve the problem for $X$ (you don't need to solve after for $T$). This only makes sense for negative values of $\lambda$, so $\lambda = - |\lambda| = - k^2$ and it yields:
$$X(x) = A \cos k x + B \sin kx, $$
applying the aforementioned boundary conditions, we have: $A = 0$ and:
$$B \sin k = 0,$$
which is true if $B = 0$ (not valid) or $\sin kx = 0$ which happens to be if $k$ is a natural multiple of $\pi$, so:
$$k = n \pi, \quad n = 1,2,3,\ldots \equiv \mathbb{N},$$
so the solutions for every $n$ are:
$$X_n(x) = B_n \sin k_n x = \sin n\pi x,$$
where I have set $B_n \to 1$ due to the "homogenous-ness" of the equation for $X$. This functions are called eigenfunctions of the problem and you can obtain the solution $v$ by expanding it as follows:
$$v(x,t) = \sum_{n=1}^\infty X_n(x) C_n(t),$$
and solving for the Fourier coefficients, $C_n(t)$. I'm sure you can take it from here.
I hope this might help you.
Cheers!