Question

(1 point) Take the Laplace transform of the following initial value problem and solve for Y(8) = L{y(t)}; ſ1, 0<t<1 y" – 6y' - 27y= { O, 1<t y(0) = 0, y'(0) = 0 Y(8) = (1-e^(-s)(s(s^2-6s-27)) Now find the inverse transform: y(t) = (Notation: write uſt-c) for the Heaviside step function uct) with step at t = c.) Note: 1 | 1 s(8 – 9)(8 + 3) 36 6 10 + s $+37108 8-9

Answer 1

y cul 974 2 0 " - 64' - 274 - no 1st 1 9(0)=0 y 1000 Taking Laplace transform on bothe side 224") – 62294 – 27 L3g = sestadt & o = ² 2 2 4 3- sy (0) - 9' (0) - Ch.29 - (0)] - 27 L3 ar 2 2245 [ 5-65-27 SZ & opo 47751 sises-29 [- ] S

124) I s (57-65-27) S(57-65-27) 1 y(s) = S (5²65-27) 5(5²65-27 Mow taking invese da place 4 rll & .) - 65-97 2.27s toiset tar 1273 36(5+3) 18/S-9) 108 (59) 8-65-27 41)+13 36 t It t 10 - ult-1)/ 3 (t-1) - e t -3 (t-1) 12 C

, -3t of Jy (t) = - I ult) + its e + Toot - 4LE-D ( te glt-1) -377-7 et Rate you et Plose ! Alor comments have problem.