Question

Use the variation of parameters formula to find a general solution of the system x'(0) AX(t) + f(t), where A and f(t) are given -4 2 А. FU) 21 12 +21 Let x(t) = xy()+ X(t), where x, (t) is the general solution corresponding to the homogeneous system, and X(t) is a particular solution to the nonhomogeneous system. Find X. (t) and X.(1).

Answer 1

8 To find General sol of Oc(t) = A x(+) + f(t) A = 4 g & f(t) = [t 12+2t 9 -4 g -| 4 10 = alt) = 4 972(1) + X(£)+ + 12+at' 11 the complementasy septem x = 1 V -4 2 2-1 2 3 -4-2 2 ga 4 -1-1 = 5 Chassachecustic pobyosomiat f coefficient mato'x is (4-4) (-1-A) - 4 -(4+d) E(lta)) - 4 (472)(1+)4 : 4+ + A + d + 1²- y 12+ 51 Put chanache sustic polynomial 1750=0 =) ded+5) =0 6 mial o December 2017 January 2018

(A-IX) X =0 =) CA-TOX -4 2 9 -1 12) = .. dzo d =-5 These are two aigen values. Now we need to find corresponding Eigen vectors =O (:) Jed -484+ 2x2 24 - Ny - Uy + 2x + 2x 2 = 0 & 9 - X, =0 -2x + 1 =0 =) =) dx = Nz for arbitarany 24 =1, ne Eigen ch sunday So y = [1] ܘܲ Rg = +224 get Ng=2 vector is ee ot Week 49 - Day 344 for d=-5 (A-(-5)I)X = 04+2x2 2x+482 17:) 2017 January 2018

8 za 24 + Qxg =0 204 + 4xg=0 lc 4+ 2X2 4= -QXg for Ng 1, we have x = -2 vector is -2 Eigen vector is 1 2 e-5t Ya = 12 is complementary pot to 1 a +G-2 2 3 q 2 e-57 4 solution is 0-54 te Yes to-20-57 These y, a Ye are lunearly independent Solutions Y= a fundamental mateix we hant to find particulas Solution Yu=f 2 5 ti -de-ot 0-57 6

8 1 -2e-57 ul معا 0-57 H ť 12tat deteguminant of 10 Y is Wronskian e-5t + 40-5t 50-51 20-57 e-57 11 be By Caamers quile 4 = -ge -56 52-57 t 19+9+ e-57 57 50-57 = 10-5 Egue-54 - 4+18-57] 240-5t + 50-5t t [ [ au + 5t- 24 # 5e5t 11 + 5 5 4 Ug = se-it t 19+9+! +12+&+ et-1) = q 19057 5 5e3t

u' gutst 1 12e-st . 5 Integrating we get u=/ 24 t + 5 +420-57 Slogt 15 u = Hea 244 +5byt) 5 as So yu Yp Yp 20-14 gut + 5logt) 25 q e-st - 120-57 General sol y = Yes + Up