Question

Having difficulties with 3 problems Solving initial value problems with Laplace Transforms 1) ?′′−4?′−5?=3?^3, ?(0)=3, ?′(0)=3 2) ?′′−6? +9?=4−?(?−7), ?(0)=2, ?′(0)=0 and 3) Find the inverse Laplace Transform of (4s + 2)/(s^2 + 6s + 34)

Answer 1

1. The initial value problem is y" -49-59'= 3t3, y(0)=3,y/{09- 9 Taking Laplace toansform on both sides of the given differen- tial equation, we have Lyrip - 4L{t} -5-{ly = { * {by - 39 0 -10 - 4styles - (01 - 514 y try=34 (3-4s-5 G () } - 3S - 3 + 4x3 = 3433 -4s-5) Y (9) - 3S - 9 +1 where y(s) = L{y10} ★ YS - - 2 S24s-5 (S-45-5) ..Y(S) 의 B-55+) (3-5 (+) S4 (s-5 (+)) Taking_inverse Laplace froms form on both sides, we get yt=4_3s S4 3S 31 118 S45-5 34 SS 18 + Sion) - ls-soi) + tends ar) Fisfant - Fro-Sent ste.indo) TES-2 - Classe)(on) 은 +135/en + {o-Scry face ens 2S S-5) 3.5t, 3 18 와 나 et 2S-4+4 (3-5)(s) 2S-4 3435) SD + -월5를 타 59 (7) + 1 S-5 + 대8 을래 St. -9(+) -25 3 रे 18 + 놉 s (s-5) (S+)

3 1 3 56 e t 2 2 :. 419= 31{s:5}+{4}+i+Is I S+ 18 + I'{34 84(8-5) (s+) (5-5) (5+) get test 84(8-5) (69 (1) By method of partial fraction, we have برادران کی بات بار = ze 18 18 A s B + + E S-5 E st! 54 52 + S3 + 34(5-5) (5+) → 18 - A3/S-5) [+])+B5°/-5) (+) + s (8 5) (+) +D (5-5) (s+) +ES+ (5+) + F57/5-5) substituting s=0,-1, 5, we get and 182 EX5x6 > D= - 12 >F =-31 18=-5D 18=-6F 5 → 18= 6 E X54 = 3=EX 54 > E=3 625 Again 18=(55_484_553) +B (84-453-1554)+C (83_45455) +D(595-48) +(85+s4)+F(35_554). 718 = A+E +P)35 +(-4A+B+E -5F) $*+(-5A-4 B + e)s +(-53-4e+D)5+ (-5C-4D)s – 50 Equating the various powers of s, we have A+E+P=0, -4A+B+E-56=0,-56-40=0, -5B – 4C+D20, => 50 =-4D » C = -40 id=-E-F =-3 +3 625 A- 1872 625 5 ?C= 72 25

- 18 1) :.5B=D-4C 2-18 -AX 724 5 25 288 5 25 -18X5-288 25 - 378 25. 12) 2.B= 378 125 S3) 음대해 대학 3대해 1872 625 18722 25 125 18 13 + 5 625 378 125 25 12 -35 STS 2 t 36 125 25 5 t 18 5 + 3 425 3) et 1872 625

From equation (1), we have y(t) = zēt test 1872 3283 + 36 (2 - 2 3 4 625 19 est_zet 625 1872 625 125 25 378 +36 + 2 2 +² +628 estét This is the required solution of the given 1NP. 5 2. The initial value problem is y" – 6y+9y = 4-6(t-7), y(022, 3'()= 0 . Taking Laplace transform on both sides, we get L{y" (t)} - 6L{y (2} + 9 L{ (t)}=4[{1} - L{{ (t-7}} = $2{914)) -sy(0) – y'(0) -6544993 +5909494{y10} = § - ē75 > (365+9)Y(9) –25 +6x2 = 4-675 =(5-3)*19) 4 (5-3)2 S (5-3)2 Taking inverse Laplace troussform on both sides, "(t) = -12 [ 16.3 2} + 21'f (6333 +24576-57 =-1229171478} +2e8°c438133+24:13-333- Likes --12te3+ +28*[*}}+6e8t24s+3+7451ngel-Ikea a2e3t_12te3 4 46 te 3t [K 28-12 + $ ets >Y(9) = -12 25 Z t (8-392 ets (8-32 -[& e75 L (5-3 A -75 where F(S) = 3-3)2

By method of partial fraction, we have sty - A + + BS (6-9) + cs. putting s =0,3, Wic Nawu A29A , AF A9-9 BS 42 3C cs 쌍m, A=A (36S+9)+\3235)+Cs > 4 A+B) s+ (-6A -SB+C) s + 4 Equating the coefficients of various powers ofs, we get KEB20, - GA-Btceo) A C. :.B=-A 4 0 !,B= B=- + As = { - + - - FC + -송- est 3t e s2 'P()= (5-3)2 S2 *$b=CYF9}= } Hey = etc :: ff (4) = test

Hence 9 where unit and u a heariside from equation (1), we have "It) = 263t_bte3t+ 7 1 - e34-3test. I{EAS FC)} 19 test+9 - E'4278.FO] by and shifting theorem, we ware 9 (t) = q est la test+1 _ f ( t ) u (t-7) f (t) = test u(t-t is a step function. .. y(t) = 4 + 14e3t - Latest_$t-U14-1) This is the required solution the given ivp. 45+2 365+ 3 4 :: F(S) = 6+3)+25 Taking Loverse Laplace bansform on both sides, of 3 let F (S) = 4842 ..f@ ={FO} = P4 18345 > f(t) = 23 4 (5-3) +2 3452 }, by 1st shifting them. 45-10 S452 saestu 1345 54.52 - Hestes (56) - 283 tsin(54) ..5%) = 4€ btcos(52) – 2éb tsis (51) 2&3+ [4