Question

Solve the initial-boundary value problem. Theorem 11.3.5 (b) will simplify the computa- tion of the coefficients in the Fourier sine series. 64uz 0<x<3, t>0, u(0,t) = 0, u(3, t) = 0, t > 0, u(x,0) = (22 – 9), ut(2,0) = 0, 0<x<3. Utt

Answer 1

Sin M3 z a let Now from , 3 or 64 T(+) = A= equal Now from & , *6 = A To (t) = G. Cas 8ml t + H sin fullt 3 3 (+(HF) Xn (2) Tn (t) F, Sa Tu Can Go Shop 221 Si Xx(x) = For Sin in when su = ~M? lit = 64 Uxx - u Ce, t) = x (x) TCE) =) u 3 = nT, EZ M = 1 X() T" (+) = 64 X6) T (t) T"(t) x (1) X(x) = a (say) - how (Since xt are independent varieable from So both sides of ③ are equal if they are T"(t) 1 gnal to some Constant). T(t) 64 T"(t) GAT T(+) X(4) So the auxi S1² s Cox-1 when a=0 then x (x) arhich gines the sole X (2) = 4x+B Now Hence U60, 2) = 0 = x() T (t) =0 -) X(0)=0 Un (a,t) U (3,+) =0 =) x(9) 7(H) 30 =) (7) -- (:7(4) 40) nx (lan +4, Sabing dov X (0) = → B=0 1. u (x, t) = 2 Ux (x, t) u=0 which gives the tonal Sol? X(a) = Cose-2 cohen p= 1² (470) h=1 The X" (n) = M² x (x) where An & Bm are arleinen Entart. So the Ausiliary eq is in #² 50 =) m =IM =) An SxC7–9) ai nga da Now X(0) = = a +420 [x(2-3) taip (***) x(3) -0 =) ee+BIW=0 (31 g) 2 T R=B=0 She -J6x Sinh So here we also get trial sole X (1/20 + 24 Gende Con-3 6 372 (- 18x "") 324 Hence the auxiliary cat is m = - 7m-Iim ta Sa uz (26,0) = 0 Bn=0. Hence the sol? is x60 = E LOB Mr + FS24a x(0) = E=0 x(3) 20 22 FnM3 =0 7 51 M3 -o(.F70, for nu bakal 513 ) 2 (A Cos Ballit + B En Su D 2 ) sin How it (xx,0) Tx J Ans n=1 3 Hence the sole is My) = RMx +DEM* 2 3 3 3 2 3 Gos apa 0 2 3 AT la second 707) e-D-o 3. . 3 6x. Cashita er (Muro) Then sa X (n) = - M X (1) 3 3 NT