Question

(1 point) Solve the nonhomogeneous heat problem u, = Uxx + 5 sin(5x), 0<x<1, u(0,t) = 0, u1,t) = 0 u(x,0) = 4 sin(4x) u(x, t) = Steady State Solution lim 700 u(x, t) =

Answer 1

Given ut a uxx+ 5 sin(5x) ssin (5*) let - uxa ucoit) = 0 u(s, t)=0 < x <37 U(4,0) = =usin(4x) f(x) b ( + ) 58in (52) = $600) $(30) 5 sin (5*) we know that solution of nonhomogenay Heat equation kant uca, to 2 bne (x) t nel 2 Pn le kidnt-1) sore) kan fo hal +) f(x) da bn fro) for (X) do j -e, 2 I of bn dn un = not = n1 = for (x) sin una sin (nsc) 24 Jasin (48) 8in (28) dx cos (n+4x) & cosin(n-4) x In-+ I "[cas (n = 4)x 17 sin inte87 (nte)

bn *15 sin (n=4257 sincnt459 (9-4) n+4 -0+0] nery 8 sinna otog 3 (02-16) 32sinns &nen T(12-16) $(0) In (X) da th ī M bn 11 8 sinnst n2-16 l 2 Hana Ni5 = 5 sin 5x sinna da so so ST a sin (massinso) doc ज cos (nts)x] doc #teos (n=5 os (n=50 sinin +5Jx737 sin (n-5) a hos) De]" 듭 (nts) ml5 sin(n-5) sin(n+ 5j7 (7-5) (n+5) 0 +0J losinnst & [ = n2-25 - 50 sin (no) t nen (52-25) when nŁ 4 non 4 n4 lim 328innor case) n4 t (n2-16) J bn :

lim 32 Acos no n>at s 2n cosur 32 284 ( applying it Hospital Rule) b4 4 similarly on=le nes & AEN n=5 lim -so sin(at) n05 n2-25 = म -50 C os 57 r 2x 5 - 5 (-1) = 5 K = 1 das -up? -n2 U18, +) (ut (lit) = K uxx' (Xit) +0100) & bn eltnit sinmay + Ethione (u-ent) sikose na n=1 o bn - u 4 en als * DEN n #4 n = 4 XnEN n&5 n=5 ull,t) sin 5x +5 - 25 4 e 16t sin 40 t t-e-257 25+) u(ait) = licht &e 1st sinua +5(1-ery sin5x), 4Xo sintx + 5 (1-0) sinsa - 25 lim ta hoo -25

dim uexit) 5 too 2 5 sinise sin (5x) 5