Question

Solve the following differential equation with given initial conditions using the Laplace transform. y" + 5y' + 6y = ut - 1) - 5(t - 2) with y(0) -2 and y'(0) = 5. 1 AB I

Answer 1

Q. 1. Answers Given y" try toy u(t-1)-8 (+-2) f yo) = -2 y 'los-5 Taking Laplace on both the sides. Layile +5L{y's to Lf yg = 18 ut-ny #2 {S (4-2) let LGG = 48). then we know that If y s = 8²yB1-sy61-46) Hyly = 8y61-46) Luf-9) 5 = all these in above expression & get. 82Y8) 840) -Yol +5 (ryss-840) + 648) to pas [ St-9)&= pas using we е. -2.8 e > how wing y=-2 & y los = 5

For sim Sinflicting let's write you sy oo s?y +s 5 +5( SY +2 +6y 25 s +3+58 +6y +38-5+10 es - e25 -> (3P+58 +6 Y . (@s+5) s - y = e 728 Es+5) s(s+2) (3+3) (312) (83) ${$+2)(+3) now. Breaking into partial fractions tet — А & + B &+2 t.c s(s+2)(8+3) st3 ㅕ 1 A (8+2) (SA) +B( 8 )(3+3) + 6(1)(st2 put 8=0 then = 6A = A= } put s=-2, then 1-2B B - 2 pyt &=-3, then = 36

now Ls (3+2) (1+2) = tē s 41 5 6 % s 2 st2 3 8+3 6 s 27 st. St3 where Note:- Iseas.pl = ft-a).ulta) fres = 1 1 1 . Ut-1)] - 110 ēpt"). u (t-1)] + leiti. uf- Similarly [I at el let (8+2) (x+3) А Itz AB st3 A (ast) + B (8+2) put s--2, then (= A put &=-3, then je B = -)

Thow 16+2363+3) str 843 st2 ees By ett-2). 44-2 St2 843 e3lt-2). ust-2) e finally, let 24+5 (4+2)(8+3) - A &+2 4B St3 => 28+5 = A (8+3) + B + 2) put s = then t= = put 8--3, then B = B 28 +5 7 + Estatt873) st2 st3

AO Stron Gist ਨ+] +2) +3) | stol olt test () o y = es + 4 (813 ) ਨੂੰ+91)9 (343 ) how, taking inverse Laplace L - Y & = ੫੬ (18) = } ੧੯॥ ਅੱਜਦੀ छ) - AU+) (3) &3) +3) 55f2 (3) 10+2 (373) + + 873

now uhrg evu get पण =148) मो.-) K 2 हस्.िuff- हरविले. परिण t-u-D -हम- Answer.