→ We know that, the solution of heat equation ∂2u ∂t2 = c2 ∂2u ∂x2 is given by ux,t = ∑n=1∞ Bn sin nπxL en2π2ktL2 →Using ux,0 = sin nπL, we get ux,0 = ∑n=1∞Bn sin nπx L ⇒sin nπL = B1 sin nπ L + B2 sin 2nπ L + B3 sin 3nπ L + ... ⇒B1 = 1, B2 = 0, + B3 = 0, ... So, the solution is ux,t = sin πxL e π2kt L2