Please explain to me only how the yellow circled part of this example problem is obtained. Please do not try to answer
Posted: Tue Jul 05, 2022 9:57 am
Please explain to me only how the yellow circled part of thisexample problem is obtained. Please do not try to answer theexample problem again. Only explain the yellow circledpart.
Example 6.10 Find a series solution for the second order initial value problem x + x = 0, x (0) = = 1, x (0) = 0. Solution Let = y so that the second order initial value problem becomes a system of two first order equations with x (0) = 1, y (0) = 0. x = Y ý = -X
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
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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