1- 4x²y" + (4x² + 1) y = 0 a. Show that the equation has a regular singular point at x = 0. b. Determine indicial equation, the recurrence relation, and the roots of the indicial equation. c. Find the first three terms of series solution corresponding to a larger root. d. If the roots are unequal and do not differ by an integer, find the first three terms of series solution corresponding to the second root.
1- 4x²y" + (4x² + 1) y = 0 a. Show that the equation has a regular singular point at x = 0. b. Determine indicial equation, the recurrence relation, and the roots of the indicial equation. c. Find the first three terms of series solution corresponding to a larger root. d. If the roots are unequal and do not differ by an integer, find the first three terms of series solution corresponding to the second root.
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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Question
![1-4x²y" +(4x² + 1) y = 0
ม
a. Show that the equation has a regular singular point at x = 0.
b. Determine indicial equation, the recurrence relation, and the roots of the indicial
equation.
c. Find the first three terms of series solution corresponding to a larger root.
d. If the roots are unequal and do not differ by an integer, find the first three terms
of series solution corresponding to the second root.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20c5837f-8a07-4fb0-b110-a8d2b8a51a74%2F1c887f46-728b-4dda-a968-21ada70f948a%2Fm48sq6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1-4x²y" +(4x² + 1) y = 0
ม
a. Show that the equation has a regular singular point at x = 0.
b. Determine indicial equation, the recurrence relation, and the roots of the indicial
equation.
c. Find the first three terms of series solution corresponding to a larger root.
d. If the roots are unequal and do not differ by an integer, find the first three terms
of series solution corresponding to the second root.
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