Question

Solve the IVP y' + y = f(t), y(0) = 0, where f is the 27-periodic function given by f(t) -1, 0<t<T, <t<21, f(t) = f(t + 27).

Answer 1

The given initial value problem (IVP) is y'+y=flt), where it is the 21-periodic function given by flt={ 7(0) = 0 osta 1 aste f(t) = f(t+20) 2R • L{ fltl} f(t) dt l-e-225 1 le-zas {fest. (6-1) 4 + sate stilt ***1!1+[++$1" ke-zis {tle asco-le-zas Le-as]} 25 1-6-2a Scitets) {t (1-2e as te 20s)} ute-them f(-x) 4 =-let! = -1 (1-e-as). (itets)-1 --} (1-ets) (1-es te 225 & 3TS. ..) = -1 (1-ets) È Drena nas tajl=1-x+m2 uzne no The given differential equation is y' + y = f(t) Taking Laplace transform both sides L{y} + L{y} = L { f(t)} → SY(31-7 (0) + YIS) = (1-eas). Ž-5 has >> $415) -0 +43 = -f(1-es) Ž-o? ennas > (5+995) = levo) con emas given, y (0) =0

Y(8)= -1 S(5+1) (1-2 *) Žu-o?echas Taking inverse Laplace transform both sides Y(H) = ['{J (+) (sets). Eine nas] no →YLt)= L eintijas - È she has no 565+y } nzo Sist) • Let - Sist) sti B = (5] ,-- then, Alsti] = 0+1=1 --=-1 sists = f-sti [ { $(sto} = ["{+ - +1} = 1- et śctu} = [l-e-47- 4n+a)).ult-(n+1)x) L'{enas sustip] - [l-e-(tha)].4Ct-na) If I { F()} = f(t) - (htlas and by second shifting property, putting these values in , we get (then, L'{e as F(S)} = f(H-Wulta) ylt) Živ". [l-e-(4-6+)a)), ult-c1453) - E-1" [l-eckna). ult-ha) ho nzo