Explanation Given: The initial value problem is ( x 2 − x + 1 ) y ″ − y ′ − y = 0 ; y ( 0 ) = 0 , y ′ ( 0 ) = 1 Approach: Rewrite the given equation a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = 0 in the form y ″ + p ( x ) y ′ + q ( x ) y = 0 Here, p ( x ) = a 1 ( x ) a 2 ( x ) and q ( x ) = a 0 ( x ) a 2 ( x ) If x = 0 is an ordinary point for this equation then, the general solution of the given equation is y ( x ) = ∑ n = 0 ∞ a n x n . Calculation: The initial value problem is ( x 2 − x + 1 ) y ″ − y ′ − y = 0 ... ( 1 ) Since p ( x ) = x 2 − x + 1 and q ( x ) = − 1 are analytic at x = 0 therefore, write the general solution of the given problem as y ( x ) = ∑ n = 0 ∞ a n x n This gives y ′ ( x ) = ∑ n = 1 ∞ n a n x n − 1 y ″ ( x ) = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 Substitute these values in equation ( 1 ) ( x 2 − x + 1 ) ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − ∑ n = 1 ∞ n a n x n − 1 − ∑ n = 0 ∞ a n x n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − ∑ n = 2 ∞ n ( n − 1 ) a n x n − 1 + ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − ∑ n = 1 ∞ n a n x n − 1 − ∑ n = 0 ∞ a n x n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − ∑ n = 2 ∞ n ( n − 1 ) a n x n − 1 + ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − a 1 x 0 − ∑ n = 2 ∞ n a n x n − 1 − a 0 x 0 − a 1 x 1 − ∑ n = 2 ∞ a n x n = 0 ∑ n = 2 ∞ ( n 2 − n − 1 ) a n x n − ∑ n = 2 ∞ n 2 a n x n − 1 + ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − a 1 x − ( a 0 + a 1 ) = 0 … ( 2 ) In order to make each term in the above equation like term, put k = n in the first summation, k = n − 1 in the second summation, k = n − 2 in the third summation Then, equation ( 2 ) becomes ∑ k = 2 ∞ ( k 2 − k − 1 ) a k x k − ∑ k = 1 ∞ ( k + 1 ) 2 a k + 1 x k + ∑ k = 0

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 8.3, Problem 26E

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The power series expansion of the solution to the problem about x=0.

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Chapter 8.3, Problem 25E

Chapter 8.3, Problem 27E

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