Can someone tell me if this is the right way to find the u(x,t)
$ u_t = 5u_{xx}$
$ u(0,t) = u(\pi,t) = 0$
$u(x,0) = f(x) = \left\{\begin{array}{l}-x & 0<x\leq\frac{\pi}{2} \\x+\pi & \frac{\pi}{2}<x\leq\pi \\\end{array} \right.$
Analysis:
$\beta = 5$
$L = \pi$
$u(x,t) = \sum_{n=1}^{\infty} b_n sin(\frac{n\pi x}{L})e^{\frac{-\beta n^2\pi^2t}{L^2}}$
$f(x)_{odd} = \sum_{n=1}^{\infty} b_nsin(\frac{n\pi x}{L})$
$f(x)_{odd} = f(x),0<x\leq\pi$
for t = 0:
$u(x,0) = \sum_{n=1}^{\infty} b_n sin(\frac{n\pi x}{L})e^{0}= \sum_{n=1}^{\infty} b_nsin(\frac{n\pi x}{L}) = f(x)_{odd}$