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1) Insert the formal power series into the differential equation, we derive an equation 2)zr-1 - 0 So we have the indicial equation and a recurrence relation

1) Insert the formal power series into the differential equation, we derive an equation 2)zr-1 - 0 So we have the indicial equation and a recurrence relation

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The second order equation 81a?y" + 81ry' + (81x? – 16) y = 0 has a regular singular point at a = 0, and has series solutions of the form 00 (1) Insert the formal power series into the differential equation, we derive an equation Cn+ n-2 So we have the indicial equation and a recurrence relation Cn_2 for n = 2, 3, . (2) From the indicial equation we can solve the indicial roots (enter your results as a comma separated list). For any one of the indicial roots, we have c = (3) Let r, be the smaller indicial root. Then r and the recurrence relation becomes Cn 2 for n = 2, 3, - . . Let co =1 From the recurrence relation, we have a solution I+1+ z+2+ T+3+ z+4+ +... Y1 = z+ (4) Let r2 be the larger indicial root Then T2 = and the recurrence relation becomes Cn= Cn 2 for n = 2, 3, -.. Let co = 1 From the recurrence relation, we have another solution. 2 = 1+ T+2+ Ta+0 +... (5) The general solution is given by y = Ay + By, with arbitrary constants A, B.
Transcribed Image Text:The second order equation 81a?y" + 81ry' + (81x? – 16) y = 0 has a regular singular point at a = 0, and has series solutions of the form 00 (1) Insert the formal power series into the differential equation, we derive an equation Cn+ n-2 So we have the indicial equation and a recurrence relation Cn_2 for n = 2, 3, . (2) From the indicial equation we can solve the indicial roots (enter your results as a comma separated list). For any one of the indicial roots, we have c = (3) Let r, be the smaller indicial root. Then r and the recurrence relation becomes Cn 2 for n = 2, 3, - . . Let co =1 From the recurrence relation, we have a solution I+1+ z+2+ T+3+ z+4+ +... Y1 = z+ (4) Let r2 be the larger indicial root Then T2 = and the recurrence relation becomes Cn= Cn 2 for n = 2, 3, -.. Let co = 1 From the recurrence relation, we have another solution. 2 = 1+ T+2+ Ta+0 +... (5) The general solution is given by y = Ay + By, with arbitrary constants A, B.
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