Question

Solve, using Laplace Transforms: Y" + 4y = ui(t) - u3(t), y(0) = 1; y'(O) = 0

Answer 1

e s -35 e s e S -3S es e +$. Now, The giver differential equation is y"+44 4(1) - U21) - 90 -1, Y'O) = 0 Now, taking Laplace trans formie both sides, we get → L{y"} +4.L{y} = {u(t)}- L{us(t)} »s? L{y}-530-y' (0) + 4.4{y} = es > ($+4).L{y} -5= > (${+4). Y(5) = + y(s) = S.(52+4) s(S²+4) (54+4) Laplace Powerse trans for both sides is e-35 "{YLDY - LUE -LY >(ii) Now, as {*} =1 =C6 s S -35 S es e's + taking we get [s:(524) L +4 s (5²+4) 한 L L2374 $(t-3): g(3).dz sin(24-266 2. = 1. 2 sin (27) dt

= 1 o Sin(27).d7 = - * [cos(22)], - (605(@t) - 1) +(1-Cus(t)) x 2 sim 2Z & sinet 4 2 This is required Solution. Note > Ua (t)f(t-a) .: Now, but to lessed from from Chi) we get> >y(t) = U(U). Sin? (t-1) - UCL). sin?[E-3) + cos (at) - 460) = { [4-6. sin(t-1) – u{). sinº (+29)+cos/29 é as F(s) cony Where F(s) is the Laplace transform of flt) ① L{ va(t)} = FLs). G(s);} where f(t) and g (t) are the laplau inverse transform of F(s) and q (s) respectively T 다. -as e S, LI f(t-34(.dz.