Question

8. Find the solution to the given systems. (20pts) 0 3 (a) X' = X X(0) = =) 1 2

Answer 1

8. a) x' 12 X -12 ; X(0) = (6) 3 5 e) -1/2 ) Let A= S 12 -1/2 Characteristie polynomial of A, 1 - Ź -> (3--)(-1-7)= 0 => a = for, LL 2 - AIR (0) (%):13 1- 9 = 0 0 = 9 Hence corresponding eigenvector () for A2 Ź 2 1 =) U= 0 (18)(u)-(8) Hence corresponding eigerwestor () ... the general solution is Ge 12+ (!) + G 6-12* (i) using initial condition, X(0) ali) +G (i) х = 3 > (3) > 3 - G G+ G 5 - Q =2 Hence the required solution is x = 3 e Bekzt (1) +2e=\24 () CS Scanned with CamScanner

b d2 dt de 5x+Y du i =-22+ 3y => 6-10%) = * = ar Characteristic equation of A, 5-2 1 = - 2 3-2 2²8 15-8ata ²+2=0 82 +17 = 0 » 2-87 +16+1 = 0 » (0-4) = -1 1 =) 2-4 = 11 > a = 4 I 4t 4 IL For i = [---][U]-[:] = (-i) ut = 0 - 24+ (-1-1)=0 These two are identical to (1-1) utle=0 =) na (1+1) u 4t Hence corresponding eigen vector [ir] - žlt) = et (cost St+isint) [ii] ti sint E-cost-sint) +i (Cost-sint) costt ett CS Scanned with CamScanner

4t Gettrost X(t)= Gent cost sint tie -cost-sint cost-sint Hence the general solution is 4tr sint cost-sini the sint YCZ) = Gent (- cos t-sint) tegeAt(cost-sint) .., cost-sint cost + Ge at CS Scanned with CamScanner