A metal rod of length a cm has initial temperature function f(x) = 2 sin 3x...

E 4 -(4) Question: The given one dimensional heat equation in du die CU 33 where oca<T and t>o the given boundary conditions are 40, t) = 0 (2) 4175t) = 0 and the initial condition is L119,0) = = I Sin 31 Solution Using the method of separation of variables, let be the ulat) Solution of equation (1): Whele Х function of a only function of t only Nowa differenticting equation (3) pestically with respect to a and t, we have du X'T Ja spa x"T (6) and XT' at (7) ХТ and T = au

4T 4T Now from equations (1), (6) and We have XT' = 4X"T " I' 48) X Equation (8) will be true when each side is equal to a constant. .! LA x" T. =k 92 х Whele k is constant Now Consider the following Cc303 ; Case-I: when k=1 si flory equation (9), we have X" f10) 4 T (1) flom (i) anel (lit), we have te lan 1 x1 = yox ( x - x x) = 0 dar - MPX = 0 (DP-XX zo where I do

3) zo Equation (11) is a lineal differential equation with constant coefficients. .: Muarellicly equction of the differential equation given by (01) is WP-je = 0 = (m-1) imt») M=-))) . Complimentey function (CoFe) of equation (II) is given by Cie + 6 Algo the light fand side of equation (11) is Belo, telefone we frave x= cela toeta the) is the solcetion of equation (11) • Again from ting and (iii) of equation (10), We have Cito lo edo T هم 4T > T T 472 dT at ५२. It 40 dt _113) T VINCAT

TE 04x4 _ I ext 4X4 have infegecting, we have lory 449€ +Whey G Deary T-, hay G 42PE > bool D) 4xt, Dog(9) boga - Dogb e ifeloga I, they G ey T = Ge We have T= Ge Hly Now from equations (5), (10) and 114), we 41, t) = Gice+Goliyemt where Ci , Q and G cele calitfaly constants. butting 3150 in equation (15), we there 410, t) = GI [email protected])e4884 ,by ra? fato) elit so CitG= - Gto) Q =-C, - putting G=-C, in equation (15), we have 4119, t) = Gislede-edad e4irt -0, +10)

CiG 717) CS) Agarin putting aT in equation (16), we have 4417, ) = (etter 0414 .: 414,5) = GGTEXT-AT) 24824 f18) lohich cannot be equal to 3el0 as 4 118, t) = 0, by equation (3) end we Cito, Gto, e em to and 04X²2 = 0 Hence we have the solution 419/t) = G( c edut Gery phat is not consistent Coflegfonding to the case when k=) have O C Case I when k from equation (9), we have X" 419) 110) s from lil and (ii), we have Х!) x=0

dix ZO (6) = ci infegrating, we have dx do e again interrating, we have X = Gato (20) Again frons, cli") eind ci) of equation (19), we have T' ЧТ dT It T= G, constant. - T= G fall) Flow equations (5), (NO) and (21), we have ulait) Gol G30+) H22) Where Ci, a and G ale cistitfaly constants. putting 030 in equation (22), we have 410,t) 10+G) E by lal} GG G=0 ,{.: G603 butting G=0 in equation (22), we have 469, t) = aga -(23)

a 03 which cannot 4179, t) = GGT) (9) be equal to zelo 4115,t) = 0 by equation (3) Cito, G 70 and To we have the solution 4194, t) = G(41976) is not consistont Coffelfonding to the case when kaoo Case-III. When k=- trom, equation 19), we have T! - T24) x 4T Flow (i) and (iii) X we have -Q الا 0 > dix -jax X" 419x 47x dos 102 to ) x fast where I = Equation (25) is a lineal differential equation with constant coefficients.

(8) of М: - i Ausliellialy equation of the equation (ast is given by hy? EN 30 m2 -4? = xp gid) {** =-13 m=t12 of 2x xtiß, where x=0 and Bad .. Complimently function (CoFol of equation (25) is given by Coro (c, Cosßxtosin Bol ofl C; C83)*+Sister), 8:250 CoFe = (C, costat G Suxa) -lab Algo the right frend side of equation (2016) i's zero, Herefore we have x= COB 138 + G-tie)}) -(27) is the solution of equation (25). Again from Ill and (ii) of equation (24), we have T' T dT -410 T at 00 B=) لا -- 4T -412 11 T

al dT I. -4 rdt integrating, we have Dong T= -4894 + logo dong T-Bog G = -48 = log () =-44t,: Joglu) 1 = Nega - begget if loga =9, they T G T = GELAP we have ee -(28) From equations (5), (67) and two), we have 419, t) = G/6, Casti + 6 Shit1944' -(229) butting 0 =0) 4110, ) = G1 4. Caso + 6-80) ēl * = 0; {byty) G/C, to) C, GONE G = 0 butting G=0 in equations (29), we have Llat) = G 10+G9j. d) 8484 ... Ulast) = GG sinna ē exit 64.५२ Congos) Sinoo €30

(lo) بول رہ QG SMAT exit - putting a = I, we have LITT, t) = GG SMAT 84 XE = (3)? a to Gto Sin ATT =0 (and 84xt to a sinan Sinnt n = 1,2,3 = nit, n = 1,2,3, jun, n = 1,2,3) -: frons equation (30), we have Lla, t) = GG Sina et vit, pod=? 1 = 1,2,3 det C G = bn, n=1, 2, 3 . by Sinna e unit , n=1, 0,3 mm Now the most general solution of the givey equation (1) is given by 419t) putting tao, 419,0) a businna eo by do 14 a sinas by sinna o we have ula, t) = ln sinna e un't 431) = n=1 E bn sinne ethet We have = 2 Sinza E n 겨

(11) have +63 din 3a e 4 x 37t + by Sinha 8 4x474 2 singa bi sina toba sinaa +62 81434 + by Sinhat o sing to Singat a sin39 to Sinha + - bisingit ba Singat b3 sin 3.0 +by-Sin 47 to equating the Coblestoyding ternys, we bizo, b2=0, b3 =2, by = 0, bs=0 b6=0, b=0,- - we have b3=2 and bi=b2=by =bs Flow equation U19, t) = bi sivil e4x8t + bo linaa t tota Singa toto {"sbg=ą and bi= bazby = beme ::. 419, t) = Q Singa, ē36t is the Solution to the given one dimensional heat equation to nal (31), we have by sinna e 4net exert -489€