Given the heat equation $$u_t=ku_{xx}$$
With the following Boundary Conditions $$U(0,t)=U(1,t)=100$$ And the following Initial Condition $$U(x,0)=0$$
Using the variable separation method $$U(x,t)=\sum_{n=1}^\infty b_nsin(nx\pi)e^{-kn^2\pi^2t}$$ Evaluating IC U(x,0)=0 $$0=\sum_{n=1}^\infty b_nsin(nx\pi)$$ In this case if i try to calculate $b_n$ i get $b_n=0$, and all the serie is equal to zero. How do i get $b_n$? Maybe i should solve it as an non-homogenous case?
As suggested by @mattos, let $v(x,t)=u(x,t)-100$. Then you have a problem that is set up to be solved by separation of variables: $$ v_{t}(x,t)=kv_{xx}(x,t) \\ v(0,t)=u(0,t)-100 = 0,\;\; v(1,t)=u(1,t)-100 = 0. $$ Setting $v(x,t)=X(x)T(t)$ leads to separated solutions where $$ \frac{T'(t)}{T(t)}=\lambda,\;\; \lambda=k\frac{X''(x)}{X(x)},\; X(0)=0=X(1). $$ The solutions are $X_n(x)=\sin(n\pi x)$, which gives $\lambda_n=-n^2\pi^2 k$ and $T_n(t)=e^{-n^2\pi^2k t}$ for $n=1,2,3,\cdots$. The general solution is $$ u(x,t)=100+\sum_{n=1}^{\infty}C_ne^{-n^2\pi^2 kt}\sin(n\pi x), $$ where the constants $C_n$ are determined by $$ 0=u(x,0)=100+\sum_{n=1}^{\infty}C_n\sin(n\pi x),\\ -100=\sum_{n=1}^{\infty}C_n\sin(n\pi x), \\ \int_0^{1}(-100)\sin(n\pi x)dx=C_n\int_0^{1}\sin^2(n\pi x)dx $$