Question

(1 point) Solve the boundary value problem by using the Laplace transform 22 w ²w + sin(6ax) sin(16t) = 0 < x < 1, t> 0 дх2 dt2 w(0,t) = 0, w(1,t) = 0, t> 0, w(x,0) = 0, dw -(x,0) = 0, 0 < x < 1. dt First take the Laplace transform of the partial differential equation. Let W be the Laplace transform of w. Then W satisfies the ordinary differential equation W" = subject to W(0) = and W(1) = Solve for W: W = Take the inverse Laplace transform, we then have the solution w(x, t) =

Answer 1

2t2 0 KALI, tyo 2 x2 (2) at Given that the boundary value problem is 22w + sin (67729 sin (164)= 22w W(0,t=0, Wet, t) =0, to W(30)=0, 2w (O) = 0, 0<nci Taking Laplace toams form of both sides of the partial differential equation (1), we have L { var tey + L{sin (6t72)sio (164)3 =k{ LEW ()3 + sio (6tTM)L{ Sih (169} $12{W(6193-swho WE (10) 220 272 2 d2 da2

dw dx2+ da2 ola 16 Sin (671x) s?W - SXO - O, by equs 6). st (16)2 d²W -sw= 16 sin (671x) S+162 (028W=- 16 Sin (6112) 541162 where DE linear differential equn. Therefore egun (A) is a now-homogeneous and order Agains from equation (2), we get Lwe (ore} = 0 and L4 W (116)}=0 → W () = 0 and w(1)=0. :: W satófies the ordinary differential equn W"-s'w= 16 Sin (671997 (59+16). 1651n (671) S8+16) subject to W(0)-10 and W (1) = 0 Now we selve for w of the equn (A). The general arlution of equn (4) is W = Wet we where well is the solution of (325) W=0. and wp (1) is the particular solution Let we (7) We (1) = emx be any sohn of (875) W=0. .. we wave (mis emno. mx 520, (nem to » W"=3 wa of 6+). ² mats.