Question

(1 point) Solve the nonhomogeneous heat problem u; = Uxx + 4 sin(5x), 0 < x < t, u(0, t) = 0, u(1, t) = 0 u(x,0) = 2 sin(2x) u(x, t) = Steady State Solution limt700 u(x, t) =

Answer 1

०८-८T 9 Now solution: we have nou-homogeneous heat quation ų = Uxx + 4 singx u Ot) = .. u [T, #)=0, VQ,0) = 2 sin ax taking Laplace transform, we get L {u} Luxx} + 4 pins x L sł ü, ) = da u where dace $ (ū = 2 L {u out)} (u (4,0) = sinex ż sin 2x & sin 5x 4 sin sse + su el² t chx² 8. da u Jū 2 sinaa 4 sluss dare m² Sco m- t Ts Auxilliary equation Complementary function 5x Cof Ge tee Particulas integral P. I, 2 sinaa 4 sin 5x] 11 D? S S 2_1; (sinax) 4 (sinsx) 11 .2 Das a sinaa s. (6²-s) = ta sin so (-25-6). 4 sin 5x sC8725) P. I. a sinax 8+4 General solution is

mi ū (x, t)= Cofio + Pi I. 2 sinax + + 4 sin 5x s 88 +25) 2 2 from and Tgx - 8x ū (x, t) = G. e. +Cge. s+4 u 0,t) => ū (0,t) = 0. putting in o = + C to G + ₂ •; UCIT, t)=0 = ū (T, t) = 0 ū (1,t) = 0 putting in E 4 su 51 st4 8+25) as (sinnt zo,nen non ç e we get C2 =0 putting in e IT се LITS + C e + 2 singit t ce - tic, e (4 Gs 2 sin ax 4 sin 5x + 8+4 8+4 8 8+25) as ū (uit) = (8+35) taking inverse Laplace transform le (agt) = : asinan LILL 4} + 4 sin5X LES 2 sinax. ott, + 4sinsa.Lt i(t) = 2247 +4 slnsx 254) UQ,t) = 20 sinat + $(t-eest) sînsx Palabras Stown? 25 S+25 spriat 25t :(! 25 25

it teiking Now aking - 4t lim uv, t) = lim [ ze "sin 2* +31-2354) sisx] two it os sinsk ze sine x +4.01- € ) 25 :lea .25 + 4 Clo) sinsa 4 sin 5x = 25