Question

In Problem of the solution as t- Oo 16,solve the given initial value problem. Describe the behavior 2 16. х — 5 х, х(0) 4

Answer 1

slution X = (-2 ) 1-5 4 Let n = ) 2 1-5 4/22/- (-50g +472/ » Nů = -22, the and a = -5 01 +4012 dy = -24 the due = -5h, the > Dry = -21, the, Dn2 = ~57;t the (D = 2 = (0+2987 – Y =0, 594 +0-47%= For eliminating , multifly first equation by (0-4) and adding with second equation, we get (D+2)(D-4) 94 +594 = 0 622D-875 =0 > D2-20-5) My so = [(0-3) (0+1)]" = Auxiliary equation is (M-3) (M+S) =0 = M = 3,-) i slution is ny(t) = Gent + Gent t = Gest+Get dny = 39 t - Get (differentiate ) West to Scanned with CamScanner

Putting value of ns (t) and dry in any = -21, the , we get my as z (t) = dry +226 = 39 34 - Get + zgest t2G et 374) = 5934 + çet. Then, solution is xGt) = ( 2 (t)) ( H) ( Gest+Get 264) = تاگو) . (and a -1 nito = 2 As to Scanned with CamScanner

(:es and E0 = %= % As ko, nito. Use initial condition > N; (0) = 1 and 2(o) = 3 1 = Gtel and 3=5966 (1 =1) G= ² subtract -2=-49 1, C2 = 12 Therefore final solution is (act) = + vet i As t n(t) oo, 0. 5 estret Scanned with CamScanner