Question

9. Solve the wave problem: 0 < x < T, t> 0; Utt: t2 0; u(T, t) = 0, u(0, t) = 0, 0 SST. u(x,0) = sin(10r), u(x, 0) = sin(4æ) + 2 sin(6x), Answer: sin(10r) sin(10t). 10 sin(4r) cos(4t) + 2 sin(6x) cos(6t) + u(x, t) =

Answer 1

Given wave equatich is Une = "At J o Lx < A> ulo, t) = 0 U(TT, t)=0 t o ulk, ol = Sin (4x) +2 Sin (6x) 4. (4,0) = sin(10x), о! к Т taking taplace transform of ea.0, we get LY Uxx} = L { uses - du (xs) = { shu (x,s) - su (4,0) - 4 g (1, 0) } using initial conditions du - Yo?u - sl sin (4%) +2 Sin (6)] - Sin (10x)} du 2 = shu s sin (ar) - 25 Sin (bre) - Sin (10x) I u 2ū = .-5 n = -s Sin (4) - 25 Sin (64) - Sin (lok) = (²-3²) ū= -s Sin (4) - 25 Sin (66) - Sin (lok) where Ded dx Auxiliary equation will be m² s2 =0 m=ts Complementary C.F. = Gesx + (2 est function will be , where a, (2 are arbitrary Constants. arbite Mao particular will be P. I-s sin (44) - 25 Sin (6x) - Sin (on)) = - 1 S sin (40) 25 Sin (6+) (6²52) I sin (lox) E s sin (Ar) - 25 sin (64) (-4² - 32) (-6²-52) - Sin (lon) (-10²_52) s sin (Ar) P.I. - 25 Sin (66) S (5² +42) - + Sin (lon) (5²+63 (52 +10²) will be Thus general solution of eq. & ū (4,5) = Cifo +P. I. I ū(4,5) = Get thesk + s sin (A) + 25 sin(66) (2+4) (+ 6) Sin (lons (213

Maw Given that ulost=0 taking haplace transform , we get L{460, +1} = L{a ū(os) = 0 - © also given that U (TT, A ) = 0 taking Laplace transform , we get LYU(TIA) = Llob = ū (IT, S) = 0 Maw Putting n=0 in eq. ③, we get u (0,5) = C, e +(2e + 25 sin(al s Sin (o) (5²+42 final (32 +62) (3²+102) = 0 = 9 + + o toto from eq. and that : we know Sin (o)=0 G+(2=0 Now Putting wat TS ū (TT, S) = Ciett's + lets + in cq. 3 we get ssin (ATT) (3 +421 25 Sin (67) (3²+62) Sin (107) (5²+102) 0 = Gete to toto from eq. 5 and we know that in general Sin (nuo n=1, 2, 3, ... Ge-Its tetts = 0 - 0 eq. 6 ets - eq. we get Ce TTS - C₂ e TTS =0 lets - eTS) = 0 -ITS Can not be zero = 2 = 0 How Putting value of (2 in eq. 6), we get G +0=0 Geo] Now Putting values of G, ( in eq. ③ we get ū (4,5) = 0 to + S Sin (A) + 25 Sin (64) (5² +42 (52+ (²) Sin (loks (8² +102) 7 ū (4,5) = o s sin (44) (5² +4²) 25 Sin (Gr (5²+62 . (824925 + 25 Sin (64) (s2 +10²)

0 = Gets + Cets to toto from eq. 5 and - we know that in general Sin (nilo, n = 1, 2, 3, ... I Gets & Gets o we get eq. 6 ets - eq. CETTS _ Ca e TTS = 0 Glets _ pTS) = 0 e Can not be zero How Putting value of (2 in eq. 6, we get G +0=0 G=0 New Putting values of G, (a in cq. ③ we get ū (4,5) = 0 to +5 Sin (44) + 25 Sin (64) Sin (lok) (52 +42 (s2 + (²) (5²+102) I I t ū (7,5) = S sin (44) 25 Sin (64) Sin (10x) (5² +4²) (52+62 - (s2 +10²) taking inverse Laplace transform '{ cu caps) = sin (4). Il { 32 + 1 2 + 2 sin (6x) [ { 32 + sin crow). [1544} In general we know that I'{ / 4 227 = Cos(at). and I'{st + a2} = Sin (at) + Sin (lon). I - ulugt) = Sin (41). Cos(at) + 2 Sin (64). Cos(6) + Sin (10x). Sin (lot) 10 Hence solution to the given wave equation is ulu, x) = sin(4x). Cos( 44 ) +2 Sin (6x). Cos (CH) + Sin (lox)Sin Clot) To