Question

Problem D Solve the following initial value problems using the Laplace Transform. To receive full credit, every time you use LAPLACE TRANSFORM FORMULA indicate which one you used 1. y' – 3y = te3t, y(0) = 1 2. y" – 4y = eat, y(0) = 0, y'(0) = 1 3. y' + y = H(t – 5), y(0) = 2

Answer 1

Soln we know that .nl we get Yoo-ooonyn tros . 3 y=test consider the given equation ☺ using Lapeace transform on both sides, we get [89-383= L[+€3€] + 1 [8"CO] - L[34(e)] = Lifesty Since [af (*)] = al[fc&IJ, we get L[¢*(+)3 - 31[8(*)] = [[tell] Llyn Ces] = sº'l [y(pi) - 5 L [y'rass = s'i [] (s)] - syco) > L [yicas] - SL [y(+)] - yoo) > Llylcu] = SL [8(*)] (as y co] =( Again, we know that so we get 11 sk and by first shifting property.ee L[ eat fre)] = F(S-a) , where F(S) = l [fr] L[re3] = (5-3)2 substitute these nalves in we get (Sl[ce)3-1) 3 L[g(t) -3)² SLCY(*)] -1 -3 L[YCP)] = 1 L[en] = n! L[1] = 5 shti HI we get 3 1 = (5-3) (5-3)?

=> 1 + = 다 5-32 S-3)3 grel (5-3) L[ YOU)) = LEglts] 5-3 Taking inverse Laplace transfoom on both sides, (-L8%]) i [sts J+ I [8–33] y lt) I [543] + ["[01-378 from ea? o, we get i [33] - 4234 from L[ ear] = sta [sa] = we get i [ 5 ] = 834 put these in @ ge and - eae ee yot). est 30 tte 96) (&tije 3e cans) 5 6 NO-2 Given that y"-4y=ept 5 Taking Laplace transfoom on both sides, weget L[y"-49]=L[ ett using Linearity propesty , we get L[8"(t)]- 4L[y(a) Leeft using Laplace fransfoom of degirative formula, LEYMCHI] = SM Liucas] - g"-"yco) - SM-2 ylco) yn trol LEY"(ti S` L %C4%] - Syro) - yico) = sº LCC+] -1. as y coj = 0 By formula L[eat] = tai ylco) = 1 L[eptJ = sa put these in ③ we get (s? L[%cº)] - 1) -4 LCYC*)] - = (s? -4) L[4CP)J-1 = sta we get 1 we get

It sta S-2+1 s-2 S S-2 + s-2 5+2 (59-4) L[ YCP)] = LTHC+)3 = (S-1) (S-2) ?(5+2) By partial fraration decomposition, decomposition, we get L[acer] = 3116 3116 (5-2)² G using formula Leat J = 3-a, L[e 2] = site and L[epe] =J_z L[y(4) J = Lo L[0743 + + +[BeptJ-Lo L[E??] => L [HCPJ = L [ Leeze + to 4e3f-1c e ze] Taking Inverse Laplace transform on both sides, ): we get so we get 3 16 १६ 28 4 tept 16 Noos given that y'y = HCA-5) Taking Laplace transform on both sides, LEylt y] = L[ HC-5)] using Linearinly property of Laplace & samsfoom, L[a'(x)]+ LTCA)] = LEH(-5)] using Laplace transform of derivative formula and LLH (4- a)] = we get S LEY(+)] - yco))+ L [ycks] es => € +1) L[gle)] - ? (as y cos=2) (5+1) L [ylt)J = 2+ s 5S SSH) ö as s. 11 ess ㅋ » L[ure)] = 2 1 olyan 7 sti 10

since sus th sti we get ,C5+1) -55 e srsti sti -5S -ss e ST str and e ee L[(4-5) since L[e af]=sta we get LE e eJ=st by second shifting property L[ f(-a) HC*-ay] = " FCS), get H(4-5)] = ē5s 645) substitute these in @ we ged L[J(d)]= [a HC-5)] Taking inverse Laplace ansform on both sides, we get yot) = pé* + + (1-e ) H (-5) $ + HCP-5) - ē (4-5) CANS)