Question

Solve using the Laplace Transform the problem with border values

au x2 au at2 para 0<x<1,7 > 0 sujeto a las condiciones u(0,t) = 0, u(,0) = 0, ди u(1,t) = 0 2 sin(72) + 4 sin(372) at lt=0

Answer 1

O = OC <!, Given d²u d²u t>o. dna dt² subject to the conditions, u (0,t)=0 ut) = 0 u (no) 4(4,0) 2 Sin (TTX) + 4 Sin (31130) -5 - O 4 du dt 14-0 on both sides of o, ', we get Taking laplace transform Leona Ld²u dt² s² ū (ns) su (9,0) – 4 (0,0) = d? ū(1,8) da? or da ū (n, s) - s? ū (n. 8) = -8 (0) – 2 SindTix)-4 Siu(371x1) dre? Using ega O&0 d² ū (n,s) - 8² ū (n,x) = -2 Sin (TX) - 4 Sin (374) da? or (D²-8²) ū (n,s) OL -2 Sin (TTX) – 4 Sin (31T2) the Homogeneous form of a — O. For CF, con onsidering D2-82) ū (n%8) Theu, Auxiliary ean is m-s=0 m=ts

2 This implies, C.F = Ge toes fecom p. Ia (-2 Sin T18 - 4 Sin 317) Da82 11 (-2 Sim TT) - (4 Sin 31T2) 02-8 - 2 Siutta Sin 31Tn -172-82 4 - 97²-8² - 2 Sin Tret Slu 3tr T2+g 99²78² Hence, complete solution of o is ū(a,s)= qe se te su Sin tnt Sin The + 2 T²+8² 4 9T2+8² ** Now, from @ = 0 from ③ ulo,+)= 0 u(1,t) On taking Laplace Transform, on taking laplace transform, we Get ū(0, 3) = 0 ū (1, 8) = 0. Using K., we Get Using ū(0, 1) = at C₂+ -0 = ū(1,8) = ge+Ge$++0=0 O = a +(2=0 = q entgets of

3 Using in 9 , we get ces. с е or aces_p%) = 0 Either G=0 or es-es=0 which is not possible Hence C = 0. from 6 thus Cz=0 .. C=C= Therefore ** becomes ū (ms) = 2 Sinttat Sin 31 (112+2 4 (97²78² Taking Inverse lablace transform on both sides, we get u(n, t) = 2 Sin The (74) + Sinart R2 ]u[v+) = ? Su Ta Stu Tt + 4 I Sintt) + 4 Sin 37x 31 Slutta Sintttu Siu 3tr. Sin 37t 3TT