a. [3 pts] Suppose the current Q flowing through an electrical circuit satisfies the differential equation Q" + Q'+2Q=2 cos(t) Which of the following is the general solution? 0Q = Ge 2 cos (at) + cge i sin (wit) + cos (t) + 2 sin (t) 0Q = Cle-2 t cos 2t) + cge-2 t sin wut + cos (t) + sin (t) OQ = Ge 2 cos t) + cge ? sin 2 + cos (t) + sin (t) University of Waterloo Calculus Bank/Differential Equations/ODEs/Models/Q05b (2 7 b. [3 pts] Write down the equation describing the solution's limiting behaviour as too. o IMPORTANT: Do not use the equation editor s as results may be unpredictable (if you happen to select it, select it a second time to disable it). Q (t) =Now behaviours of Q(t ) at t -> go t -7 00 Q ( ) . ce-t/ 7 Cos 2 Cop- 1 / 2 Sin + cost taint 2 Now at t - no lim Q (t ) = lim . e -t/2 ce - 1/ 2 t-700 Cos/ V 2 sin + cost taint * ) Lim C Cos 7 2 6 sin + e t / 2 . e t/ 2. + cost + sint ? Since we know that sint & cost ase bounded functions and it's value between - 181 ie ( - 1 , ] and at t-7 00 it values always be bounded and in between . [- 1 , JJSo at t-700, C, Cos / Vi t 2 is bounded t -80 , C, sin t ) is bounded t 200 cost is bounded 1 70o sint is bounded t / 2 00 / 2 And as too, e 2 e e Lim Q (t)z kim SC, cos Co Sin 10 / +1 t t700 t / 2 1/2. e + cost + sint? Bounded Value + Boundedvalue + Bounded value + Bounded Value 0 + 0 + Bounded value + Bounded value Here at t - Do lim Q (t ) z Limit does not exist t -700 Because Bounded Value between(- , ) it does not gives Exact . Value so limit doesn't exist