Question

1.

Find the inverse Laplace transforms of the following function: 2w7 F(s) = s($2 + 2Cwns + wa) "US 25 (0<5<1)

problem 2. and 3. as follows

Answer 1

o f(3) = 2 wn 8(8²+2&wns twn) into partial Let is redue it fractions BS+C 2 wnt . 82+2&wntown 8(32+28 on 2 + wna) awn = A (3²+266a8t wat) + (BS+() = gur = A 8²+26wn As + Awn t B527 CS > gwn = (A+B) 8² + (24WnA+C) 3+ Awn Now compare, constant on both sides .. giant = Awn? A=2) Now compare 8² coefficient on both sides . - A+B=05 BE-A =)B= -2 sides Now compare & coefficient on both » 240n A+ c = 0 s c= -266n (2) = = -4¢wn :: F(8)= 2+ -28-48 wn 8²+28 Wn8+ wir → F(8)= -9 (8+ 2%wn) s +2& Waist wna

& win in the denominator Now add and subtract of the second taim - F(8) = 3/5 -2 5²28wnst&wn-san = f(B)- -2-etaren - &t {wn)² + wn²-4² want -> FCB) = (8 5.ch) - -2n South Caroline (stqwn+ (wnlinça) ² (Bt&wn + (con Ji-q2) 3+ gwn => F(X) = $-24 *5709? Cansi-fo] 1 . Lawn Sie -2- B+Ewy +[wnstreafeone Thus we reduced the given function into standard forms and now we can easily write the inverse Laplace Transform as .. f(E)= 2 u(t) - 2 e sunt cos (undirşit) 7 Cum se poate in conseq=t)

Garten Alle rental equation of stabiema tih'XoX=4 ; xolab 3 ན རོ་མེད། གས་ ༦༤༌༥ -༠ the apply haplace Transform both the ནས -5x(༤) - () + ཙམ་ཡིག (5x3) - x(༠)] tinx()=0 མ་རྣམ།(3) 1 - 5） ཡེ ་ག ༼x x（3) -༠) 1 conX 2 = ༡ རྗེ– ཉ ཊ ་ག ངེ རྟ གསོ ) + [ཁག་བཅུ བ བ ] ༤༠ ཁ་ ༠༡༼༤ ཚན་རིག འག ) - ༠ ཙམིག རི ག x(8)= a +b + 2açıuch རུ– ༠༤ ༩+ “ » ༩3) འ(s) +E+ ༢བ་ཏ+ ང་ངོད ༤, ༤་རེག ༉ + tག- འ(༤༥ ༦༦་ད+( 4༢༤༠༩) 5-+2$canst can? problem, we have - In the above already seen that 8²+28 lunit wn? - ( +s;ངས་འ) + ་འ (i-༩)

= X(8)= a(st&wn) btaçon 846607576m +(1-3) - why (6²) / braqwn) (3+{wn + wa (1-62) ( wasire?) Now, by applying Inverse replace Transform се оешале да (б. -- ) + Bota Freebi) e sunt.co (əndine= t) wrd-& ③ Given differential equation is ä + 3x +48 = 2 sint ; x =0; 1(0)=0 | 3 dx +48 = agint Реля teace Transten on wh Ardeg => 8x(8) +38*(3) + 4 x(3) = 2 () :59 (+33 44) X(B) = X)= (2-3) 68739ren resolve into Partial Nowe we have to fractions

AB+3 (8+o 2 )(8+3 82H 82 +39 +4 & MP 34 359 3 to those 3) 9- (63+)(343344}+(3+0) (304) = 2 = A 2 3A 3² +4A 3 + Bg² + 3B8+ 4B t (83+Cs+ D 32+D 1 = 2= (A+C) 83 + (JA +B +D)² + (4A +3B tc) in & (4B + D) By comparing various coefficients LHS, and Ris, we can write that A+C =0 ZA+B+0= 4A + 3B+C =0 4B+039 By solving above equations, we can write that .. wp wt wt w/ stå . : 3 + 2/3 (32+ 1) (3 +38+4) 8+ 1 8 273844 We can write (82738+4) as (3 + 2)² + 2 4

>*(3)= + $ to as meningkatan 3(71) >> X(3)= et + Ś ut will (*) (8+27(0)? . 8+ 3 . .* $ het (1945€ 14 (9) 357. (6-7)+(9)" Now by applying Inverse Leplace Transform "(t)= 3 cost ust) + 3 sint u(t) Itse * confiture ce tin tuces + 4 e i sin tult)

7) allt = colu(t) + Ź Bintluck) tite it (ft)uce) is e 3 sin tucts :

out as a formula s used a List of L{ucts}= i s-a -{eace wit} = Banyan wat [{ent sro with bang -{X2}} = sx(3)= x(0) L{ xC2y3 = 8x(9) - 3*(g) – (0) {smwtje {costot) Rest of the solving involves adjustment os various terims in order to get them to above standard forms: P2