Question

b. x'j(t) = xi(t) – 2x2(t); X'z(t) = 2x1(t) + x2(t); x1(0) = 0, x2(0) = 4 2

Answer 1

Ⓡ X(t)= x1(+)-2xy (t) i 400) = 0 (t) = 2.2;(+) + X2 () · XCO)= 4. Matrix form of the given system is. x' (t)= A XV X(0)=b (x (1) 7 27 (9] -2 where x x (t) I' A:( 1 2 190) First we have to find the eigenvalues of matrix. A: TANI/= (1-1)2+4=8 (1-4) (1-2)2 = -4 (1-1) = 2 i & d = 1+2 i Hence di 1+2i, d2= 1-2i How, we get the eigenvector for the first eigenvector iedr=1f2 we solve the e following syetem (1-1-21) 2 ni to] 2 (1-1-22) ni n2 2 2 using second 2n - 2h i = 0 ni lhe Hence first eigenvector is ni n2 (4-1) So, the solution corresponding to this eigenvalue and eigenwester is act) eto (t) ) H] 1 1 (1+2 it zet (ati 1

= et е et (Cos (24) + i sin(24) ) [ 1 ] Seuleris formula} I cos (24) - sin (24) {multiply cosines of Cos (24) tisin (at). 'sines into the vectors - sin (2+) tiet Cos (2) [ Cos (2) sin (at) uct & luct) so the general solution of the system xCEJ = Get Coco 262 Now apply initial condition & find constants. (8)=-40)=1[i] + [6] 12=0 fra=4 sin (24) x(t)- 4. et (- Code :) x, =-4. et sin (2 t) 4 x2(the 46t cos(20) Any J + Ez et content Hence