Explanation Given: Point x = 0 is ordinary point of linear differential equation. Consider the solution as y = ∑ n = 1 ∞ c n x n . The given algebraic system is, 2 c 2 + 2 c 1 + c 0 = 0 6 c 3 + 4 c 2 + c 1 = 0 12 c 4 + 6 c 3 + c 2 − 1 3 c 1 = 0 20 c 5 + 8 c 4 + c 3 − 2 3 c 2 = 0 Theorem used: Existence of power series solutions theorem: If x = x 0 is an ordinary point of the differential equation y ″ + P ( x ) y ' + Q ( x ) y = 0 then find two linearly independent solutions in the form of power series centered at x 0 , that is, y = ∑ n = 1 ∞ c n x n . A power series solution converges at least on some interval defined by | x − x 0 | < R , where R is the distance from x 0 to the closest singular point. Calculation: The given algebraic system is, 2 c 2 + 2 c 1 + c 0 = 0 (1) 6 c 3 + 4 c 2 + c 1 = 0 (2) 12 c 4 + 6 c 3 + c 2 − 1 3 c 1 = 0 (3) 20 c 5 + 8 c 4 + c 3 − 2 3 c 2 = 0 (4) From equation (1), 2 c 2 + 2 c 1 + c 0 = 0 c 2 = − c 1 − 1 2 c 0 Put value of c 2 in equation (2), 6 c 3 + 4 c 2 + c 1 = 0 c 3 = − 2 3 c 2 − 1 6 c 1 c 3 = − 2 3 ( − c 1 − 1 2 c 0 ) − 1 6 c 1 c 3 = 1 2 c 1 + 1 3 c 0 Now, substitute value of c 2 and c 3 in equation (3), c 4 = − 6 12 c 3 − 1 12 c 2 + 1 36 c 1 c 4 = − 1 2 ( 1 2 c 1 + 1 3 c 0 ) − 1 12 ( − c 1 − 1 2 c 0 ) + 1 36 c 1 c 4 = − 5 36 c 1 − 1 8 c 0 Use value of c

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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