Question

2. Solve the initial-boundary value problem 2% for 0 < x < 6, t > 0, u(0,t) = u(6,t) = 0 for t > 0, u(x,0) = x(3 - x) for 0 5736. (60 pts.)

Answer 1

given equation It - 2 Jun ulot) = u(6,t) = u (2,0) = x (B-2) ocaco, & yo forty, o for a2246 pet ulat x (8) +(H) x cas TH = 2x!!(0) TCH and xu - uxa o ti_2uT=0 from the bounderry Condition. . ulot = u(bit) ao 20 T('209) x6.0 (TCH" ů (6(H) 20-) (6) 20 loset: uro X (212 ALIB. x (0) = o x (6) = ausing B. c we get A B = 0 So we regeet uz o lase,

Case-2, MEN which gives x cas - Acth Be-12 o = ATB and 0 2 A Be-sa. this gives A=B=0 So we nepect the case MEN cose 3: uz Xco ne X (2) Aloola AB singa x (1 = 0 gives A=0 then the solution becomes X (212 B sinar X (6) = 0 = Bein Ab = 0 for non trivial solution Sin6A 20 sin 62 sinns & a no Wzma 242.. So xn (2) = Br Sin nou and the other equation T-2MT-0 - t'+ 2 y = 0 - T't hurt = 0

and the other canation T'-2ut no = T'+2. M T =0 where unCart this given Th(t) =Dne to then the general solution XmC21 Tgn (t) nout Bresin (MT) De 187 n=12 = en sin manat Wher. Br Drach nath that is more general solution apart12 Iunla solutamente I un part) = 8 en sin mange menn enz **(3-) zin Cuma) da

and e tar my sin nye to - [634- Cached - (3-24) & -sin (7/6) { ta) eos chung 33 home to & 2.63 E-coren til 2012 +-(-1)"} so 2.63 (2n-1)373 3 when 22m-1 " .. حر2ر1=m when n= an SoucoH) 2.6 [romos Sin Max e-62m-1) r4/18 3.13