Question

Find the general solution to X1 2 X1 -2 4 32 32

Answer 1

2 Uz с -> -( 1+j) 4,-24, 50 - (1+i) 4 = 242 let ₂= c then u = 2 (1+i) -2 (1-1) +1-1 = -2(1-1) = -(1-1) = (-1 + i) c. Hence. u = e) = c (-21). vector of A. For x = 3-1, let v = ( ) be, the eigen Then AV = (3-12 → 2 athere e(40) Complex number: - 2 4 It (20 - 202 -(3-1), + 402 3.1) V? Equating both sides we have, 20,-2V₂ = (3-7), and V, + 402 = (3-1). -> (–1 + i) v, - 2 ₂ = 0 and + (1+i) 2 = 0. net de = c then from and equation & = -(1+i) c. and also ( +i) {(1+i)0} – ze ={-290-47-3e-20 (1+BC-2C = 0. Then --[***] - [2) fave care Complex number.

the solution that we get from the first eigenvalue and eigen vector is, (3+) H = 1 est (cost + isint) 7 X(t) = => X,(t) = /-e3+(cost +isint)+ iet (cost+i sint 3+ ( cost ti sint) est(-sint cost) ext(-cost-sint) est cost + il est sint u(t) + iult). cleasly. Ult), V(t) are linearly independent solution's Then we use them to form a general solution and they are both real solution. So the general solution to is 2 (t) = C, 46+oct) est(-cost- sint) ezt +(-sint+cost)) → X(t) = CI + l2 et cost 3t sint. Where C, are arbitrary constants