Explanation Given: The Chebyshev equation is given as, ( 1 − x 2 ) y ″ − x y ′ + a 2 y = 0 Here a is a constant. Approach: Let p and q be analytic at x = x 0 , and suppose that their power series expansion is valid for | x − x 0 | < R . Then the general solution to the differential equation given by, y ″ + p ( x ) y ′ + q ( x ) y = 0 . Can be represented as power series centered at x = x 0 with radius of convergence at least R . The coefficients in this series solution can be determined in terms of a 0 and a 1 by directly substituting y ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n . The resulting solution is of the form y ( x ) = a 0 y 1 ( x ) + a 1 y 2 ( x ) . Here y 1 and y 2 are linearly independent solutions. Calculation: The differential equation is, ( 1 − x 2 ) y ″ − x y ′ + a 2 y = 0 … ( 1 ) Assume y ( x ) = ∑ n = 0 ∞ a n x n be the solution to equation ( 1 ) . Differentiate y ( x ) with respect to x as, y ′ ( x ) = ∑ n = 0 ∞ n a n x n − 1 y ″ ( x ) = ∑ n = 0 ∞ n ( n − 1 ) a n x n − 2 Substitute the value of y ( x ) , y ′ ( x ) and y ″ ( x ) in equation ( 1 ) as, ( 1 − x 2 ) y ″ − x y ′ + a 2 y = 0 ( ( 1 − x 2 ) ( ∑ n = 0 ∞ n ( n + 1 ) a n x n − 2 ) − x ( ∑ n = 0 ∞ n a n x n − 2 ) + a 2 ( ∑ n = 0 ∞ a n x n ) ) = 0 Replacing n with k + 2 in the first summation and replace n by k in the remaining summation as, ∑ k = 0 ∞ ( k + 2 ) ( k + 1 ) a k + 2 x k − ∑ k = 0 ∞ k ( k − 1 ) a k x k − ∑ k = 0 ∞ k a k x k + a 2 ( ∑ k = 0 ∞ a k x k ) = 0 Combine the summations as, ∑ k = 0 ∞ [ ( k + 2 ) ( k + 1 ) a k + 2 − k ( k − 1 ) a k − k a k + a 2 a k ] x k = 0 Equate the coefficients equal to zero as, ( k + 2 ) ( k + 1 ) a k + 2 − k ( k − 1 ) a k − k a k + a 2 a k = 0 The recurrence relation obtained is, a k + 2 = k 2 − a 2 ( k + 1 ) ( k + 2 ) a k , k = 0 , 1 , 2.3 , … … ( 2 ) Substitute 0 for k in equation ( 2 ) as, a 0 + 2 = 0 2 − a 2 ( 0 + 1 ) ( 0 + 2 ) a 2 a 2 = − a 2 2 a 0 Substitute 1 for k in equation ( 2 ) as, To determine (b) To show: That the solution obtained in part (a) satisfies the expression, [ 1 − x 2 T N ′ ] ′ + N 1 − x 2 T N = 0 To determine (c) To prove: That the integral ∫ − 1 1 T N ( x ) T M ( x ) 1 − x 2 d x = 0 .

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 11.7, Problem 16AP

To determine

(a)

To show:

That if a=N, then the given equation has a polynomial solution of degree N.

To determine

(b)

To show:

That the solution obtained in part (a) satisfies the expression,

[1−x2TN′]′+N1−x2TN=0

To determine

(c)

To prove:

That the integral ∫−11TN(x)TM(x)1−x2dx=0.

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Chapter 11.7, Problem 15AP

Chapter 11.7, Problem 17AP

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Chapter 11 Solutions

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(a) To show: That if a = N , then the given equation has a polynomial solution of degree N .

Solution Summary: The author explains that the given equation has a polynomial solution of degree N and is given as, T_N(x)=1+x-

Concept explainers

Videos

(a)

To show:

That if a=N, then the given equation has a polynomial solution of degree N.

(b)

To show:

That the solution obtained in part (a) satisfies the expression,

[1−x2TN′]′+N1−x2TN=0

(c)

To prove:

That the integral ∫−11TN(x)TM(x)1−x2dx=0.

Want to see the full answer?

Chapter 11 Solutions

(a) To show: That if a = N , then the given equation has a polynomial solution of degree N .

Solution Summary: The author explains that the given equation has a polynomial solution of degree N and is given as, T_N(x)=1+x-

Concept explainers

Videos

(a) To show: That if a=N, then the given equation has a polynomial solution of degree N.

(b) To show: That the solution obtained in part (a) satisfies the expression, [1−x2TN′]′+N1−x2TN=0

(c) To prove: That the integral ∫−11TN(x)TM(x)1−x2dx=0.

Want to see the full answer?

Chapter 11 Solutions

(a)

To show:

That if a=N, then the given equation has a polynomial solution of degree N.

(b)

To show:

That the solution obtained in part (a) satisfies the expression,

[1−x2TN′]′+N1−x2TN=0

(c)

To prove:

That the integral ∫−11TN(x)TM(x)1−x2dx=0.