Question

Solve the following initial value problem using the method of Laplace transform. y" + 2y' +10y = f(t); y(0)= 1, y'(0) = 0, where, f(0) = 10, Ost<10, 20, 10<t.

Answer 1

Sofi- P Conceptit Let L {f(t)} = F(s) Then, Juse: Let'it)} = sfcs) - f 10) L {f(t)} = 5² F (S) - sf (o)-fo) to ostdo 3 tilo ; y loa) 120 y 10) = 0 Take lapan on both sides & let teo} Lęy its} = Yes) Lly") + 2 Ly') + 1o Legs - L{f(t)} in ={ s² Yes - syco) -y '0)] + 2[sYis3 - yoos] + lo Y Ls) ł Let It)} nost - 10 -st 3² Pis -5 +2(3 Yes) -1) + 10% = lodt 1 • 200t → Yesu (sº +25 +1] -3 -2 = (0 + 20 10

Yeo) [s? 42s +1} - (5+2) = 10 cele 1] + 20 (o-e ] -105 -10 Vis [stas to] (s+2) = 10 - 10+ lue 5 S e s » Yes) [ 5² 425 +10] = 10(lte los) + (8+2) S Y (3)= lolltelos) & (542) (51 (5423410) (2+25+10) You = 10 + loes + tll + 1 (s) (x + 15 +3² (s) (+13 (3) [s+ 1)² + 37] [(5+1) +3] Let's calculate Laplace invence of tim e = A + Bsto 3 st25+10 (3 [stast 16 is stastio ! e As RASTION & CSABS on (s) C3²+25+1 = 3 [A (B] + [244 ] 410A (3) C 5 7 22 st lo]. compare coeffisents! - 10A = 10 Atbi0 2 A + C D ad & A= B=-A=4 ce-2A = -2 d. 8 +2540

(5+2) I sti x s coop - base] (st1²422 432 Stll 2422 txl sint T inst Result (o

Il tem : SC 3725110) Apply Laplace inverse rule ft { fies)} = f ct) than I e a F(s)} = H (t-a) f (tra) where Hits in heavesicle stepfanctim . Mere Fos) = 10 & a=lo (5) 65425110) 1 10 3 is calculated in term (S103²125110) wwind us= 1 & è cost - è sinst akt[etos Fiss} = H (1-10) / (+18) -(t-10) =HE (t-10) SM (3 (t-1 I-e cos(3 (t-10) e ) lost 1o-t i-e cos (31-30) - (3 + LCD = MC +-10) [1- en los ( 36-30) - et sm (34-30)] L/Resun@ (T) = et [eoszty - Rent @ / Nai, w term = (st1 ) (5+1) 73²

Les - To ten (54) 432 E l ) = certyf sin3t - Reant ② causine to results , . & D atlyco) h*(9) +"'CD) +2"(T)+7%) y = {c* cost ] Hct -to poceo scx 20) - " sin Cat=20)]+ e cos U ant et cost U t é sinst ا y = i +H(t-10) i-e eus(3t-30) - ا ا ا ا ا ا نا نی ن ی