Question

In Exercises 1-15 solve the initial-boundary value problem. In some of these exercises, Theorem 11.3.5(b) or Exercise 11.3.35 will simplify the computation of the coefficients in the Fourier sine series.

Answer 1

fen) = u(4,0) = x (429), glu) = q (u, of ooches ulnut) = x(x) T (t) be the soth Uit = 64 uun, Ocnc3, to ulo,t) = 0 su(3,4) to Let then & T" X = 64X"T 2) T" x 4 = 1 645 Case I when a=0 then X" = 0 = x= Ant B X100=0= B=0 X (3) = 0 - 3 A=0 ) A=0 x = 0 uln, t)=0 is a trivial soth. Case II + Bet ze 0-32 when = mx 0 x = 2x = x= Alte X (0) = 0 = A + B =0 2) A =-B X(3) 0 =) Ale 31 _ ( 3 ) = 0 A = 0 [e31 when 770) x = 0 u (nyt) = 0, a trivial solution. when a=-uco X" = - 72 => x = A cos ax + Bsinde X(0) = 0 = A=0 X (3) = 0 B sin 30 = 0 sinna non trivial sot tet B to then sin 30-05 sin na Can III & for 3 2) a= na B sin nta 3 6-82t 874 P" + 648T = 0 T = ct c cos (877) + sin(at) e cos Cent) - + D sin ( F )

In(at) = 1 Ancos sna. 2 Mu,t) E Ancos An cos 817 & + Basin Butt)] sin nax Bensin 18 n t ) sin sin (max) cos sunt + Basin 18 vnt nail

gen) sin nan da Y (^, =) = {(Ancas na Bjt Bn sin /h Sim non 5 nzi Ans 2 3 I f(x) sin nan da 3 0 Bn = 2 3 8 na 3 لهم لم the An = 2 2/ 1 x (ang) sin van de 2/3 o x3 sin nende - 68 x sin van de % [-x² Cos nax ( 2 ) + 2x * sin nax (1) - 6x (-40sni%) (wa) - 6 sin (nap) (mm) • Ex es nix ( 1 ) + sin max (1) tas gin) = 0 cos Inak I - 243 + 108 x 18 x + ] 3 na 4373 T + sin (TX) [12 x - 324 - 54 1994 (ait) - nan coa 18 in 2 ht - 20² + 1088 + 18x nt h²3 INT 18x² -324 -54 ht Σ n=1 sin inta) che ha mai Jeroen sin (47)