Question

02 1 2. X'= 1 1 -2 X 2 2 -1 1 2 0 3. X' = X+ 1 e tant 2

number 3 use variation of parameters for homogeneous linear systems

Answer 1

Page No.: ار 0 4 مد X =) x'= AX 4 4 - = ه و وسه 1 To find eixen Value cha (A) 4 4 - A A-de 3 1 و \ وس - -d +46 (3-به-2ص) : (هيهات = (صاف da = tita EO + LL طعه دهخدا هتم كعلالتها 8 1 ہ ۵ سل ه ہ & 5 +لا و Niz ترو -- 133 1 من دا بعدي (تيل ولن ا- وہ & -- م - - و 8

T ४५ - x 1 HA नि nz 4. ----- 31 1-1- ; (A-da I)....o 2-| O 2 10 C -2 2-2 00 LE . 0 . a) ४५ - 20 + । Hong Vig thea Om/ HAK-4 generel solition Cositat) - Sinat) Casciat) +Sinktat) कt) =_e ACAat) + Sinaty che))ena cak(54) = Sinkrt). GALat) + in.at 8 Solution पर fundamented matrinin ज Kit) =_ct) . c

1 2 X + les seves tout AX + git Ao fi :] gitla | et tont - A-11 The S= }*-**+2=0 i = 1+i die Hi . do = 1+1 for 1,1-1 IA-2,71V=0 :11%) uc 10) o W, - di'lg = 0 if Ugali Vi (30) Now Let Hilti- Liit Wilt) - (0) 10] Wilt et

=et [ Cost - Sint) i sine ( ) 2 Sint een tiet/ 2 Cost Cost - Sint Mi1t) = V+ iv 2 lost al sind 2 sint Cost el - sint Generel Solution for Homogenous Part GU+ GV Xu Homogenows Wt) + & sint Cost Now the Matrin ci get sint et cost getlost -etsint Use Parameter Method: fine Variation of X = ? met sint 1X/ et lost 2 x2 C etlost det sint - etsint -9804 -det lost get sint retlost I sintet lost i a 1 I lost et sint et Now *3 = 1 I sintet Costet a last et -sintet e tant

Sint -Sint tant Now Sint - In tontt somnade xija Cust tent -Sint tant Sint s sint at s-sint tant de {xt e Elle 4) и -Sint tant - Cost Now Particular Solutions y = xf x'g at get lost - Cost get sint et lost et sine X Sint - InlI tout + Beelty) Hyr - et sint Cost + get cust sind - det Cost Inl I tant + sert) et cos't - etsin't + et sint Incl tout + Seclti) lo det lost In Cl tanlt) + See it) met + et sint dn (ltant)+ Sertti) / General Solution: X = Xult + Uplt! X- 2 Sint Cast (-gelost done (tontt Sect) (et tetsint Iultant +Sert -sint