Question

Use the method of variation of parameters

Find the general solution to the non-homogeneous system of DE: -4 5 X + -4 4. x'

Answer 1

-4 31 x + The eigen values of the coefficient matrise is geven by, 5 - -4 4-X => A-4)(+4) +20= 0 - 16 +20 = 0 .4 ms tei is given by, for *1 = 21 the eigen vector 5 4 4-21 => 64-21) 24 +5+2=0 434 +14-23)*2 = 0 -4x,= 4-21) xq ► 2745 (2-2)X2 het K2 = 2 4 = 2-1 eigen vector V = 03:13] +zit xit V, e P1] 2 - (m. (wszt + i smat) 2 260524 + sinat ti (2 sinat-60521) 2 10324 t. zi sinat

2 es & + หn24" 224 of 24 + 1 [2 2 528 2724 the complementary Solution is กช24 -toy2k X, () - C 2 to5 244 กท24 4 (2 2 Lc524 2 แท24 Cits consider the matine, โท 24 - Los ! 2 to s2แ+ที่ท2} X : 2 in 24 2 Ln5 26 T 2724 - 2Ce62 Adj X- to 528-2กชน 2. tos2 4 24 21 2% to 328-2.124 2 (52F 2 to 5 244 24 411 2/cos2 4 4 25 26 detx= • a Sa+ดท 264 216 2 ( หท + 4 to 326) * 2 2 26 (es2-2724 Adj x x - det { - 2 Los24 2 c5244124

Herre g(1) Lore] 2in21 ws 2k-zsinat Now X-!.911) 200521 m2 -02 3d sinat 67 sinal . TC - et osat at sinal informasi - 310524 3052+ + 2 sinat - 37 sinad - 22 cosat the particular solution is, Y p = x lxycedo 200524 + Sim rrus errein 2 simat-cosat 125001 - rue / 20521 asimas [azt wszt+& inztwoszt-2t wsztsinat + 2 sim 2t-3dsin at sinaluszt+ sir 232-524 | -365*2!+ 3 vina beosz!-3 tsis'se - 2 W2460924 -3716n°2946528)+ (simg446544) -3+(60524 simo) --34+% TE- x0) = Metap general solution is, asimat-loszl 260531 simet -34 ** **)1"}( [Answer