Question

In Exercises 1-15 solve the initial-boundary value problem. In some of these exercises, Theorem 11.3.5(b) or Exercise 11.3.35 will simplify the computation of the coefficients in the Fourier sine series.

Answer 1

Doar <3 tyo 64 uan ulo,t) - U13,t)=0 Unio)=0, we solve Method ols Veniable Ulust)= X(u) qet) this by it (1,0)= x(429) sepenation putting this in o XTH 64 Xunt Xxx X Tet 649 -A2 a) X(X) = 4 cos da tqsinan act) = 3 o 8 att Cy sinsat urnit) = (( salute sin du) ( CG Cos 8xt + Cysin & x) ulo,t)=0 G=0 Ulast) = (2 sin An) C G c) 8xt + Cysingat) u(3,t) so -) Sin 3420 A = 1117 Unlu,t) = OFTA ( An Cojen I t + Bn sin snitt & UCHA) = S(Sin IT (An cws 8 ntt, OT ... Sin OTX 3°+ By SingT€) n=1 acnol=0 2) Anzo Jucnit) - Ž Bn sin O ile sin sort 3 n=1 (40) = 4(x2-9) 19 Ž Bosman sin = x(x2-9) )

is a hat Range fohnen sine series where 20T 3. Bo g(x²-a) sin nite Sin amen du - Br 1 UNT 2 / 4 d [-me as minus 2 + 342 sin af ht 6 sin m 33 2912 + Gu contik 27 n33 91 MITTY -9(-se co nhu 3 + Sinne acelet YOT [ 80 Ti +436 co) OTT + n33 143 20TY Co) 2013 (-19 2ohy ula,t) = 42 43 4-1 sin ni sin 9-1 20YTY 3

I use method of separation of variable to solve this problem