Question

Aft) Use the method of Laplace transforms and the accompanying proof results to solve the initial value problem. y" +9y' + 18y = f(t): y(0) = 0. y' (O) = 0 - 1 1 - Here, f(t) is the periodic function defined in the graph to the right. ! 1 1 1 1 1 -- 1 Click the icon to review the results of a proof. 1 i 16 Square wave Choose the correct answer to the initial value problem below. 100 O A. y(t) = 73 (-1*(1-2-31 - 4n)?uct - 4n) n=0 OB. y(t) = 18(1 - e -3ty2 OC. y(t) = 1 (1 + e -4t)t(t + 3)(t + 6) 1 OD. y(t) = (1-6-3{t-Bn)2u(t - 8n) 18 n=0

Answer 1

The given periodic function is fit= o<t<4 **** 4<t<8 flt+8)=flt) 12 its period is T = 8 ..L {flt)} se est fetelt 0 l-e-85 {"estido + f estadt legs + 0 letle] l-e-85 le-os Ś [1-e4s] (145) (lte as) Site 45) • The given initial value problem is yu+9y'+18y = f(t), y (0)=0, y'(o)=0 Taking Laplace transform both sides: L{Y"} to L{4'} + 18 L{f} = L { f(t)} [s? Yes! - sy 10)-770)] +9 [SY(s)-740) + 18 YCS) = (545) -5.0-o] +9 [5483)–0] +18Y(5)I (5²+gs+18) Y(s) selte -45) siltet silteus) → Y(S)= ! te 45 s(52+9S+18) → YIS) ! te 45 1 S(5+3)(3+6) Taking inverse Laplace transforms both sides glu) = ['! (ite-as sis+3)+6) ( 1 )

• het 1 - + B S(5+3) (5+6) 5+6 Then, Szo (0+3) (0+6) Ś 5=-3 (-3) (340) = CE s=-6 18 A-[(+3)ists ) (sesto] (561+, (-6) (-673) ..c'{ $(+][476] - [lites - te (0) - 5.2734 + to (1-2e3tte st) = + (1-2-3+)? teus = (1+e-usht -Custe: 45= 18-45) +45434 {(-1)". (645) 31543*+0;} tie 66 18 1 al-e-45 no {(1) e-tns ho :[{itets stres) Fts)} = 2 Lorean SU+y)+ جراح hio its (1-1-317-44))? ult-44) -By second shifting properly, & LI { F (S)) = f(t) then [feas f(s)} = f(ta.ult-a) putting this value in , we get » A. y(t) = - Žr (1-e ~3 (t-4h)? ult-4n) 18 n=0