Question

Use Laplace Transform to solve this problem: Determine the solution of the following initial value problem y"(t) + 3y'(t) + 2y(t) = f(t), y(0) = 0, y'(0) = 0. where f(t) = 1 when t lies between 2n and 2n + 1, and f(t) = 0 when t lies between 2n – 1 and 2n (for non-negative integers n). This is a square-wave forcing term.

Answer 1

– 1i) +34'+ 2y = 414). where f(t) = ro Li y 10) = 0, y'(0) = 0. tt (20-1,20) te (20,2041] L (Y) = Y(s), 217') = Sy(s)-910), 4(y") = 5² 715) - sy 10)-4110) taking Laplace transform both sides of eqn (i) we get Lly" + 34') + 2y) = S(+14) sº (5) – $(e) y'(4)+ 3($}(s) - (e)) + 2 )(s) = 1 - 1 Iš ,te(20, 2001] s?y15) +35415) +27(5) = 10, (5²/35+2) Y (5) = Y) = 3/542ste) 3 (5+1) (642) y (s) = 365+) (542) sosti partial fraction method we get

5(51)(572) = A or, A (5+1)(542) + B (5)(5+ 2) +((s)(+1) = 1 (A + B + C) 5 ² + ( 3A + 2B + C) 5+ (2A) = 1 equating coefficients of so sl.82 both sides we get 2 A= 1 At Bt A : C =0 3A + 2Btc=0 B+c=Ź, 2B+C = 2B+C 2 - 3 B A = =-1 c = - 1 2 - B = - - 4 + 1 = £ £ , B=-1, (= Ź . as sti 215427 yes) = taking us -state 1st2) inverge laplace transform both sides we get yit -e-t tet 2 1, tel 20, 2n+17 -o tt (20:1,207