Heat equation and Fourier series

Suppose g(x) = -x (when 0 < x <= pi/2), x - pi (when pi/2 <= x < pi).

The solution to 8(d^2u / dx^2) = du/dt with the conditions
u(0,t) = u(pi,t) = 0 (when 0 < t < infinity)
u(x,0) = g(x) (when 0 < x < pi)

is u(x,t) = sum from 0 to infinity of : b_k(t) * sin(x(2k+1)).

What is b_k(t)?

I've been stuck on this one for a while. Help please!

Summarizing the separation of variables process:

Letting u(x, t) = X(x) T(t), we obtain
8X'' T = XT' ==> X''/X = T '/(8T) = -λ^2 for some constant -λ^2 (for nontrivial solutions).

So, we obtain
X'' + λ^2 X = 0
T' - 8λ^2 T = 0, with BC's X(0) = X(π) = 0.

Solving X'' + λ^2 X = 0 with BC's X(0) = X(π) = 0 yields
X_n = A_n sin(nx) for n = 1, 2, ... (with λ = n).

Solving T' - 8λ²T = T' - 8n²T = 0 yields T_n = C_n e^(-8n²t).

Hence, u_n(x, t) = b_n e^(-8n²t) sin(nx) where b_n = A_n * C_n.

Since the PDE is linear, we have
u(x, t) = Σ(n = 1 to ∞) b_n e^(-8n²t) sin(nx).
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Now, we find the values of b_n by using the remaining (initial) condition.
u(x, 0) = g(x) = Σ(n = 1 to ∞) b_n sin(nx).

Hence, b_n equals the Fourier-sine coefficients of g(x) on [0, π].

That is,
b_n = (2/π) ∫(x = 0 to π) g(x) sin(nx) dx
......= (2/π) [∫(x = 0 to π/2) -x sin(nx) dx + ∫(x = π/2 to π) (x - π) sin(nx) dx.

(Evaluate these with integration by parts.)

I hope this helps!