→we know that the solution of heat equation ∂u∂t = k ∂2u∂x2 is given by ux,t = ∑n=1∞ Bn sinnπxL e- n2π2ktL2 Using ux,0 = sin nπL, we get ux,0 = ∑n=1∞ Bn sinnπxL ⇒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