Question

3. (5) Solve the initial-value problem: ( 20) 106 x, subject to x(0) =

Answer 1

Sol. x'=(8%) * ; *(0) = ). The first thing that we need to calculate is the eigenvalues for the matrix A = (6 %): det (A-NI) = 0 → 127 = 0 → (2-0)(6-1) = 0 = 1= 2,6 Eigenmec forze corresponding to the eigenvalue 2 is : Av; = de 5 %:)61)=2*) 29, = 2n, ³n, is arbitrary 7. 6%2 = 212 => 492=0 → M2=0 put 14 =1. So, eigenvectos corresponding to eigenvalue 2 alan len 16n2

Me - Gule 1e 0 B Eigenvectos corresponding to the eigenvalue G is : cl Bil 7 o 6 ( B2 / 1 + 1) - D = 2P) = 6B=> 48, = 0 37 P, = 0 ' 682=682 * B2 is arbitrary put B2 =1. So, eigenvector corresponding to eigenvalue 6 So, the general solution is X(+) = ceart (6) + Gect (O) / where ç, & cq are arbitrary constants. put xe (0) = (-6); we get 6) = 4e*(!) + se% ) = (&) {e=1} G=-4 & Cq = 6 - Hence, the solution for the initial-value problem is X(t) = -4e2t (6) + Seat (7)

an method : 7=66)* ; 260)=(%) (*):( 32) () (7) 2x del = 2x, - 0 dX de = 6X2 from 6; we get dx = adt 24 Integrate on both side, we get In 18y) = 2t +C ; Where cis arbitrary a constant C attc at •2+ x = e alto = e. eat = ceea where c,=ec 29 = G,ezt put 2,(0) = -4 ; we get -4 = Geo G=-4] {: 2:19 from 3 we get dx₂ = 6dt Integrate on both side, we get :

- len az = 67 + Ao where Ao is ambitioning O constant ng = est+to = eho eft = Geft where G=eto [x2= Geft ] put X (0) = 6 ; we get 6 = Geº = 9=6] {ie=1} So, 2x2 = 6e6t7 Hence, solution alt) for the given initial value problem 1-4 ezt alt) = 6e6t