Question

HW help.

4. Solve the initial value problem. 2 -1 6 x(0)

Answer 1

& 4 1 6 2 ie 20 & 402 dis zlo) given initial value problem is, x 40)= (6) o 1x, los )-40) dae ( ) The characteristic equation of the coefficient matrin () is, 1- (2+6) + (276- (-1*4) = 0 -2, +602 toke. 2 A = iie X-8% + 16 = Xo »- 2.2.4t 4 = (2-4) > O IN 4,4 5o, x=3 is a double root. We charcer teristic verton a first find velue X=4 to the charecteristic corres isponding di and Let >= 4 d = 2) 22 2 :) (2) -4 (20)

from which we at once find that d, ond d must satists, 2d, +42 4dg -d, +6% = 4 d₂ lie -22, + 4 dy = 0 - di +222 = 0 On d = 222 A simple montrrival solution of this system in obviously 2,=2 de = 1., and this a shenesterintis vector to = 4 in d= 4.* r = 44 e corresponding (3) () {"" lineonly independent (&+ + p) ett, where (9 – 11)} = 4. Now, solution in of the form, X=4 and (3) * = ($) satinties This To satisfies

[() - (19) - () . - 2 1)(2) = (3) From the we at Once find sotisty that P. and be 쟈, " 22 = 1.. A triral simple solut of this system is mon. b = 1. () (1)] - 4 L二 t c. (2*)e (+1) e 3 0. in another soh The general solution. 44 (241 4) e c () bituary where ond . ae ari constant t . 지) = 2ce + 2+1]e 4t, 일(아 가 +21 At e 옥

Now initial condition 31 (0) (0) 2,(0) ( O) = 6 ) 71, 10) = -1, 7200= 6 Then. by we have. 2,00) 2c, & q = -1 210) 4, + < = 6 (29+% (9+2) -1-6 5-7 a -7. { 13 1. we have By 4+ #g(t) = -148"* +13 (2++1) é 41 Hz (t) = 47 -fe + 13 (t+1) e solution for 6 Which is the required general