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By putting X [:] = the problem can be rewritten as 1 *-[-]* X, X (0) = - * = = −1 0 Since all entries are constants, the matrix A is analytic at t = 0. Thus the sum can yield the constant matrix ∞ ΣAk (t-to)k k=0 0 1 [4] -1 0
if and only if A1, A2, A3,. Now from AXO = A³X0 = - we notice the pattern are all the zero matrix. Hence, Ao Xn -1 0 = = 1 n 0 [] n! = 0 = 1 1 -1 LINHA -1 0 Xn-1 n ] Xo = 1 n! A²X0 A"Xo. A¹X0 = = - A so that 1 -1 0 (1) [BH] = 0 [18-8] 0 =
we notice the pattern so that and A21 2k+1X0 - - ] = (−1)k+1 Hence we have X2k+1 = A ²k+¹X0 (2k + 1)! X2k k = 0, 1,... = 1 -A²k Xo (2k)! = A2kX0 (2k + 1)! (2k)! (−1)k 0 - 1²+ i]; (−1)k+1 9 k = 0, 1, k = 0, 1, .. k = 0, 1, 2
x(t) = 2x - Σ X2t2k k=0 k=0 k=0 1 (−1)k Σ[+] Σπ[ζει (2k)! (2k + 1)! (−1)k+1 || = = = α X2k+1t2k+1 (-1) k (2k)! t2k t2k 00 k=0 (-1)k+1 (2k + 1)! k=0 k=0 [] []Σ α (-1) (2k)! -t2k (-1)k (2k + 1)!` t2k+1 -t²k+1 t2k+1