Question

Find the general solution to the non-homogeneous system of DE: -4 51 3t X + -4 0 x'

Answer 1

And We have given [-1 cu 57 -4 4 X + 1:39 Compare with X' - Ax + b(+) we 31 Jet 5 A: 2 -4 4 non- the 1. Homogeneow اليا Now, Lineah blo) (0 Solution of the system a 4. Je Alb (7) de we just to find et consider And Il no X(t) = At Xo te e At have So, Por that -y-d 5 > -4 urd 2 d - 16 +20 2 d +420 d² = -4 2 da ti Eigen Value Eigen veeter according Now, corresponding Imaginary Eigen (A - diI) Xi = 0 to value is اور ہا۔ = > 5 4-21 J[3) = -4

>> 4+2 t4-21) x + 5720 59 ✓ -un +(4+21) y=0 9+(4-21)y zo (4+21) and [ +4-21]yco In this suituation let choose choose ya 2 then 5(2) n= 4.210 4721 4+2i — 2" 4o-zoi 18+4 So X 2 ti ges X 1 = and X2 Now [3) - [ P = [X, Xe] = ( 3 ) adj (P) = [2 2 Illa -2

and adi P IPL 2 [-, ;'] 2) p's 9.-12. and o Now, Know AME P DIAP At - pept pt git 2 2 - e At -zit e 2 zit 2e 0 1 zit 6 20 او۔ گا۔ لادے At e e -e zit Now -eil zit LEAL zitezil =e=/ с. е 2it2it e te At ado e cand 21t

الم سی الاع+ Hence e adj e Ad leau D کرام of ett At (4) ط Now, but value of Equ (2) in (e 2- ال 2 - e -e Xot (4) e * ع ع تو رو - 21 -e te e Idza ere و ۔ e داد e . e 2 طر 2 -29 ر2۔ نو لاو resit 27/ Le Idio e -e e -e X ا -" رو ع وع 3 [[ git - 21) 3re² ع3 + e -e 3 ا 34۔ لو 3 C -e 2 2 نو 3 المو رو رو 3te 3 اعد و e لو ع نو X + لا 2 e نامے

.2it -zit zit -214 e e 3t ༢ ( e + 3 4 С + X(t) Xot elit a where Xo is intial vectar This is our general Solution A