Question

1 point) Solve the nonhomogeneous heat problem

ut=uxx+4sin(2x),  0<x<π,ut=uxx+4sin(2x),  0<x<π,

u(0,t)=0,  u(π,t)=0u(0,t)=0,  u(π,t)=0

u(x,0)=5sin(5x)u(x,0)=5sin(5x)

  u(x,t)=u(x,t)=

Steady State Solution limt→∞u(x,t)=limt→∞u(x,t)=

(1 point) Solve the nonhomogeneous heat problem Ut = Uxx + 4 sin(2x), 0< x < , u(0,1) = 0, tu(T, t) = 0 u(x,0) = 5 sin(52) u(a,t) Steady State Solution limt u(x, t) = Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining.

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Answer 1

neous heat equation 17 solve the nonhome men homogeneous problem ut= unnt 4 sin (2x) ola<M, u(0,t)= 0, u(M,t) – 0 U(WD) = 5 sin(5x) ucut) = ? Steady state solution lim upwit) When Non-homogenous nous Heat equation of standard from ut uam + d(uit) Boundary condition u (0.t) 20, ult, t)=0 Initial condition (* 01 - f("). thus the solution of this heat equation is txo & Unn Okn<h

ucat) e nel 2 Dn Sin (MAL) į (sin (not) s ant ang (S-1) + e And 21 where Da = 2 J sin (mr) fragdn L and &a(t) = 2 J &(nit) sin(n) dn. (mit) compa saring this beat equation Td = 1) 4 sin (2x) And boundary condition 12-H) (: u(ht) = u(",t) = 0) Initial condition ucn, of a feng 5 sin 63) therefore the solution ucuit) on sin (***) E [ sin compra) eram (5-1) 4x(] 221, LIM then urnit) I Dn sin (m) e + sin veran) Je bent yu(wit). sin(ne) [ [sincang Jen (sit) 4, (5rds t na t + n21 since he nel 9.fds t Dn -+ n1 he 2 where Dn= de since sin ( hits) fex) du L=M therefore On = 4 h sin (man) flog de

Н alf sin (nn) flug da the given ful - 5 sin (52) thus Dn = NE sin(ne). 5 sin (50) da н 7 Dn = 10 Н s sin(5x) sin(nn).de H use the Result Sin (mn) Sin (nn) da I men Ildo mon M therefore Dna Din = sin (64) sin(nu) dn for n=5 then N DS 10 J sin(x) sin (suyda use Result Sie DS = Eld 3) D5 = 5 Dn=0, for n75 otherwise, And Sinee L. and a(nit) An(t) = 2 2 l acuit) sin ( 2 ) da 4 sin(27) s 4 sin(21) sin (no dn sin (2x) sin(ne) da. M this ault = 2 ale pole 8 M 7 Out 13 ol

use the Result For n=2 then M # d (t sin (2x) sin(x) da >) 42(t) = 2 / 4 2) 102 (t) = 4 ] otherwis Onlt) is therefore the required solution. Equation ③ uenit) (3)*(5-4) (25 sin(5x) [sin(2) zero. t est) + & sids 0 sinee Ds = 5 otherwis an is zero and 92=4 otherwis dn(t) = 0 t 4(5-4 -25t + sin(21) J 4. - .25€ -) 45 gt est therefore u(wit) 5 sin (5x] e -) (mt) = 5 sin(sw) @ 25sin (22). 4eht status dos u(wit) = 5 sine sng + 4 sin(2x) cenas = ucnit) a 5 sin(58) 2256 + 4 sin(2x) - 3847 (241) 2 menit) 5 sin (54) e est + sin(21 (21] (1- 1-e4t). (it) - bin [sin (suj ēst + sin(2x)(1-2 lima ucm, t) = 5 sin(su) limē + sin (2) (i-é) 21 Ans) therefore the lim ucnit) eest + sinten) i (1-4) to 5 sin (5) ēs

= lim u tyd (wit) 5. Sin (599.0 + sin(2x) (1-0) ( sinee e @ =o) therefore the Answer is lim ucnit) sin (2) (Ans).