Question

Solve the given initial-value problem. x = (-: 1. )x. xco) - (-:) X(t) =

Answer 1

Solne Given, initial value problem. ac - 6 2 xo) 1 -1 8 12 let A = 6 .X' X -1 9 12 حقا . 22 and x= 24 Sot X2 be's Ax. In order to find the Solution we have to find eigenvalues and corresponding of Matrix A. eigenvectors Characteristics equation of Matrix A is. given by det (A-1I) = O, Where A is an eigen values of a Matrise Į and I is an identity matrix Therefore we have. det 6 -1 + g 12 -1 0 - : ] ] or det 6-7 -1 9 12-1

or (6-1) (12-1) +9 = 0 = 0, or or 72-18dtd tg 42 -18d tal = 0. 08 27 ( A - 9.1 O. de g, g. Repeated eigenvalues, let u = 내 U2 Corresponeling (2) be the eigenvectors. to eigenvalue dag where then, we have U20 a. A u= gu. (A-9 I u = 0 on g -3 1 u Uz -- Apply Rq » Rq - 1 R g -3 o Hun O U₂ -34, + gu₂=0 O, = 342 let U2 = 1 then 4 = 3

Eigenvectors. ur [] 3 1. Let U= V be the another eigenvetor Corresponding to eigenvalue dag where وفا to Then we have А 19 - gu tu. or [A-9 I] e u. 08 -3 g I u -1. 3. Apply R2 R R2 Rq - / R -3 g 부 3 1. 1.] دفا 5 I 3 O 0 Solving we get -34, + 9 V₂ = 3 08 1 - 4 + 3 u v 3 % -1 08 let V₂ = t then = 3t-1.

Cenit ut Coretzt. Date eigenvector v " - Bt - 1. ti فا =#3 + - 1 ve -1 + 3 1. 4 Therefore, Generals solution of given System. is xlt- X(t) Get [3] + C e egt 1 t 3 1. ED + Conelition aco) = Using initial have I we x 10) [2,(0) ci 3 1.. + Cq -1 X210). 1 8 30, - 다 we have d. Comparing ta ₂ and 36,+! 3x tla C2=251 26.

Date Page Tutti ing values of C and (₂ in (5) we get alt) 8e94 Ano [ 1. 3 1 +25eat {E}] tt 3 1. which is the required Solution of initial value problem (1)