The correct option is (c) \(3∑_{n=1}^∞ \) e^-n^2 π^2 t sin(nπx)
The explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e^-p^2t)
When x=0, c=0 and when x=1, p=nπ.
When t=0, 3 sin (nπx) = \(∑_{n=1}^∞ \) bn e^-n^2 π^2 t sin(nπx)
Therefore bn =3 for all n
Hence the solution is \(3∑_{n=1}^∞ \) e^-n^2 π^2 t sin(nπx).