3. Consider the 2nd order linear ODE: (1-x²)y" - xy' + a²y = 0 (1) where a is a real constant. (a) Solve this equation about the ordinary point xo = 0 by finding a power series solution, con- vergent for x < 1 by: (i) finding the recurrence relation and (ii) determining two linearly independent solutions y₁, y2 (write out the first four terms of y₁ and y2). (b) Now assume a = 2. Show that one of the linearly independent solutions is a polynomial (i.e. the series terminates). Do the same for a = 3. Using these results, argue that if a equals a positive integer n, then one of the solutions to (1) will be a polynomial of degree n.
3. Consider the 2nd order linear ODE: (1-x²)y" - xy' + a²y = 0 (1) where a is a real constant. (a) Solve this equation about the ordinary point xo = 0 by finding a power series solution, con- vergent for x < 1 by: (i) finding the recurrence relation and (ii) determining two linearly independent solutions y₁, y2 (write out the first four terms of y₁ and y2). (b) Now assume a = 2. Show that one of the linearly independent solutions is a polynomial (i.e. the series terminates). Do the same for a = 3. Using these results, argue that if a equals a positive integer n, then one of the solutions to (1) will be a polynomial of degree n.
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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Transcribed Image Text:3. Consider the 2nd order linear ODE:
(1 — x²)y" - xy' + a²y = 0
(1)
where a is a real constant.
(a) Solve this equation about the ordinary point äî = 0 by finding a power series solution, con-
vergent for x < 1 by: (i) finding the recurrence relation and (ii) determining two linearly
independent solutions y₁, y2 (write out the first four terms of y₁ and y2).
(b) Now assume a = 2. Show that one of the linearly independent solutions is a polynomial (i.e.
the series terminates). Do the same for a = 3. Using these results, argue that if a equals a
positive integer n, then one of the solutions to (1) will be a polynomial of degree n.
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