Question

1/2 POINTS TU -49 12.4.002. a²u azu Solve the wave equation a204 = 4,0<x<L, t > 0 (see (1) in Section 12.4) subject to the given conditions. - ax at u(0, t) = 0, u(L, t) = 0, t> u(x, 0) = 0, y = x(L – x), 0<x<L Ot t=0 u(x, t) = 0 u(x, t) = 0 n=1 Need Help? Read It Talk to a Tutor | -71 POINTS ZILLDIFFEQ9 12.4.003. azu Solve the wave equation, a? . ax azu. 10<x<L, t> 0 (see (1) in Section 12.4), subject to the given conditions. az2' u(0, t) = 0, u(1, t) = 0, t> 0 u(x, 0) = 0, 04 = sin(x), 0<x< 11 at t=0 u(x, t) = Holn

***I know this is technically two questions, but im all out of questions after this and would be appreaciated to answer anyway:)

also please ignore the text inside the first answer, that is my attempt, but incorrect

Answer 1

Solution With detailed Explanation:

1.

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o Given problem is utt = a lan olact, t u(0,1) = 0 = 1 (L,t), t>0 U12,0) = 0 , Utt(,0) = x (L-x),0<x<L Let wat) = X(2) T (t) be the solution. then Utt = XT". Una = X"T apo using these above value is the problem, Ay is called , T" = 0 -) (say) Separation constant 7 x *x"+XX=0 T"+xa&T=0 (2) XT"=a2x"T ART - Now. From the Bes. u(0,+) = 0 = Xeo) TCH) = 0 => X(0) = 0 [": T707 Similarely Ult) = 0 = X(L) = 0 1 0 So, we get a and order diffeq (SLP) X" + x = 0, X(0) = X(L) = 0 Auxillary equľ m²+x=0 Case-I Fore x=0, m²=0,7 m=0,0 (repeated root) X (2) = C, +Car X (0) = 0 → G = 0, X(L) = 0 -> C2 = 0 . x(x)=0 in the trivial Sol. But we in nontteivial solutim. So we cannot consider this case. Case-II For x 20 x = -lu, reto. m²l =0 > m= Ile (real and distinct rcoote ) - X(x) = Cielet Celex Now, in the trivial Sol. But we carce intrested

na at L In (t) = de Cop (naat) + en sin (naat), in the Solution wheree do, en aree arbitray Constents for all NEN Now, Inta 11 ntat ( centaste o Unca,t) = Xn (2) To(t) &nen - do Cop (mant ) + en Sis (nat) Jasin (142)trend = (an Cop (oxat) + bn Sis(172!)) Sin (12); * nen (adjusting the arcbitruy Using the preenciple of superpositim, uca,t) = 3 Un (at) te [an concret) + bo Sin (193)] sin (hik). To find the Coeff. of an. and by u(a,0)=0 >> Ž an sin (192) = 0. fourier Sine Series, an- 2 so sin (113) df = 0 . ag=0:&n]

bon non We get the required solution. Sin (naat). Sinha Wast) = [ (2n-1). na with a sense that). Si (192) Answer: U (2,4)=Late ja mesin (nta!). Sin (14)/

Note: Your Answer given in the question is correct. Just compact your constant term.

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TERRILARER To find the Coeff of an and by u(a,0) = 0 = 3 an sin(na) = 0 Comparing the Coeffi of both sides we get an o Vn. aho Mecant) = ſeancen sin (ant) 4(an) bacoy (ant) | Sin(ne) Ut(a,0) = Sina >> Ž (an) by cargo Sin (ma) = Sinne = b; (a) Sinn + baka) Sin@w) + b3 (3a) Sin (@a) +... =Sinn Comparing the Coeff. we get ba=t, anbn=0 - b,=1 and bn=0 ( n= unft &nfl bn = { 0 otherwise Sol": .. the required ucat) = Sin (ot). Sin (19x). Answere uim,t) = Sin (at) Sin (1)