3 4 12 1 x x4 2 4 a1 x 13 1 x x5 3 5 we

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e first two coefficients a0 and a1 in the power series can be determined from the initial conditions, dy (0) = a1 . dx y (0) = a0 , Then the recursion formula can be used to determine the remaining coefficients by the process of induction. Indeed it follows from (1.5) with n = 0 that a2 = − a0 1 = − a0 . 2·1 2 Similarly, it follows from (1.5) with n = 1 that a3 = − a1 1 = − a1 , 3·2 3! and with n = 2 that a4 = − a2 1 11 = a0 = a0 . 4·3 4·32 4! Continuing in this manner, we find that a2n = (−1)n a0 , (2n)! a2n+1 = 8 (−1)n a1 . (2n + 1)! Substitution into (1.4) yields y = a0 + a1 x − = a0 1 − 1 1 1 a0 x2 − a1 x3 + a0 x4 + · · · 2! 3! 4! 12 1 x + x4 − · · · 2! 4! + a1 x − 13 1 x + x5 − · · · 3! 5! . We recognize that the expressions within parentheses are power series expansions of the functions sin x and cos x, and hence we obtain the familiar expression for the solution to the equation of simple harmonic motion, y = a0 cos x + a1 sin x. The method we have described—assuming a solution to the differential eq...
View Full Document

This document was uploaded on 01/12/2014.

Ask a homework question - tutors are online