Suppose u(x,t)=Ae^(−t)sin x solves u_t=u_xx on 0<x<π,t>0, with zero endpoint values and initial data u(x,0)=60 for 0<x≤π/2 and 40 for π/2<x<π. Find πA.
The official grading is **marks to all**. At t=0 the proposed form would require A sin x to equal a piecewise constant function, which no single A can do. A full sine series is needed. If the intended target were only its first Fourier coefficient, A₁=(2/π)[60∫₀^(π/2)sin x dx+40∫_(π/2)^πsin x dx]=200/π, giving πA₁=200. That coefficient is not an exact single-mode solution of the printed initial-value problem, so MTA is preserved.
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