Question

Solve the given initial-value problem. x=(-1-2)x+(). xco) - -() X(t) =

Answer 1

soln :- x'= ( 113 -2 J**(5) XO) to -6 7 1 3 3 4 Charceteastic en is -1 -2 1: 4-7 (x+1) (A-4) +6 EO 123*+2-0 (x-2)-1) - Eigen rector at x=2 3 (3)(1)-() -60 ) )=60 32,72X2 =0 Ill- X2=3 x = -2 -2 eigen vector 3 Now eigen vector at dal 2 (33) X2 O x + Reco leb x2 = 1 a X=- RA (1) X2 Hence Xe = Ge get (1J+oct (3) علم و CS scanned with CamScanner

et debut Now to determine Xp X et 3 e't X4 gelt geat - 082 t re et -t -ze ( -let eat eat xt (5) ) = 6 -t -ze pet det eat 16) t -t 15eloe 5e2t 5e2t !) -250t 10 e27 x05 jet -t (-25e at logat = -25 Jet at 10 fet de -t 25e Now 50 2t XP = x fx ( 3 ) at More t こ et -20 2t et 3 eat 5e27 CS Scanned with CamScanner

-2540 1 25 +15 Xp = -35 40 Hence X() Xet Xp t -1 2+/-2 X(t) = de 3 But given XO) .6 7 -6 - G 22 35 teo e° f 7 362 (uo -G-202-35 (%) = 0 Cat 36 +40 -C-262 =29 G +36 --33 C2 =-4 &l= -21 Hence +/- XA) = -21 el 2+/-2 35 -4e می 40 Cained with.CamScanner