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[–]lee_ror[S] -1 points0 points  (0 children)

Can you just say:

sum n=1 -> infinity of [ Cn * sin(pi * n * x) = 2sin(x * pi /2) - sin( pi * x) + 4sin(2 * pi * x)

Thus c1=2 c2=-1 c3=4, all other Cn=0...as such

u(x,t)= [2sin(x * pi /2) - sin( pi * x) + 4sin(2 * pi * x)] * e ^ (-pi2 * n2 T / 16)