Question

Find the general solution. Please and thanks.

2. x'(t) = 0 1 1 1 0 0 1 2(t) 1 1 0 1 3. x'(t) = 1) =(t) i 1 -i

Answer 1

efem fimel amaithen arjen wecilar A B = P + x E EN G for da-1 Zz is another (A-1) = eigen verston 1 1 2 for da and 8:1 1 1 1 VE):( 1 1 Zg change in reduced form by Sow operation 1 -2 1 1 1 1 1 1 R2 R- RL BB-Ri Hoo 1-2 3 o 3 2 (6190 Z1 + Z₂ + Z₂ = -2 Let 2 = Z=8 CScanned with cam ganter 2 & t-s

181:3-8 Zz t + 8 1 for of 50 Sro 어니 2 Izle Zz [] then Generd salution -2 his = G -X 1 1 1 Com e 2 t 1 + 3.94 1 CScanned with CamScanner

azat then A = at 749 1 H2 " + d3 eigen vector eigen rector corresponding to sigen value d=-1 10-1-1) 1 y Lest " 1 2 Y2 I ork) pa 93 eigen vector 1 1 9, Y Ya 1 1 YnfY+Y₂ T o found by any eq Let 9,= ; then 9,-A- yg -1 2 1 . + 8 Ya 1 for t = 1; s=2 yi -1.1 140 1 Scanned with CamScanner eigen rectan

td tot dt 172 then general solution G car- 201 has a G [ c1[] Gen G.S. = 2- O #(+) = 1 alt) d I 1 for eigen values A la-jI=0 Ord 1. 1 0-1 1 1 1 - 1 r() lea) -14.1-0).+ (1+d -d (²-1) td +2 tdth -12-31+2) Scanned with CamScanner

d = -2,-2, 2 eigen valmes finel eigen recitan corresponding eigen valios d=2 tedy, 211 22 = -2 1 1-2 03 - 271, f3y + 3O 2 -2% + ago 21 +82-272 = 0 a + 2x equo -27, 2011 ta+22=0 -402 +223 - 382 +33= 24 = x put in equ -20 + 22 +22=0 -22, --2012 2 = 22 CSscanned with CamScanner

x' (t) 1 에버 - 1 A- 11 ri 1 for eigen valises A-d2| = 0 ~ - 1-4 (1 -119.1) - Li). - 0 (2-1) (1-4) - 1-1-1-yea -2d=0 d (d-2)=0 d=og eigenvalues eigen value d=2 to eigen rectar corresponding 1 :) :) co Isto tako sto che camscaldeer

an V = -2 - 2 v, 4 Vazo ne: zuz V = ive (2.1) V, a tiva than let V₂ = 8 te, a aith V = ait then -L f V₂ 1 Lest I eigen Neetan Eigen rectan Corresponding eigen value d=2 ( A-2d09-0 13-2 eigen werden - į * )=(:) 1-2 CSscanned with CamScanner