Explanation Given: Given differential equation is x 2 y ″ − x y ′ + 2 y = 0 ; x 0 = 2 . Approach: To write general solution of a differential equation, p ( x ) and q ( x ) must be analytic at a point x = a , then x = a would be an ordinary point and the given differential equation can be expressed in general solution in the form y ( x ) = ∑ n = 0 ∞ a n x n . Calculation: Consider the given problem. x 2 y ″ − x y ′ + 2 y = 0 ... ( 1 ) Because the initial value is x 0 = 2 , general form of the equation can be written as below. y ( x ) = ∑ n = 0 ∞ a n ( x − 2 ) n ... ( 2 ) Substitute t + 2 for x in equation ( 2 ) . y ( t + 2 ) = Y ( t ) Differentiate above equation with respect to t upto second derivative. d y d t = d Y d t d 2 y d t 2 = d 2 Y d t 2 Substitute above values in equation ( 1 ) . x 2 d 2 Y d t 2 − x d Y d t + 2 Y ( t ) = 0 … ( 3 ) Rewrite the equation ( 2 ) as below. Y ( t ) = ∑ n = 0 ∞ a n ( t ) n … ( 4 ) Differentiate equation ( 4 ) upto second derivative. Y ′ = ∑ n = 1 ∞ n a n ( t ) n − 1 Y ″ = ∑ n = 2 ∞ n ( n − 1 ) a n ( t ) n − 2 … ( 5 ) Substitute ∑ n = 2 ∞ n ( n − 1 ) a n ( t ) n − 2 for Y ″ , ∑ n = 1 ∞ n a n ( t ) n − 1 for Y ′ and ∑ n = 0 ∞ a n ( t ) n for Y ( t ) into equation ( 1 ) . ( t + 2 ) 2 ∑ n = 0 ∞ n ( n − 1 ) a n ( t ) n − 2 − ( t + 2 ) ∑ n = 0 ∞ n a n ( t ) n − 1 + 2 ∑ n = 0 ∞ a n ( t ) n = 0 ( t 2 + 4 t + 4 ) ∑ n = 0 ∞ n ( n − 1 ) a n ( t ) n − 2 − ( t + 2 ) ∑ n = 0 ∞ n a n ( t ) n − 1 + 2 ∑ n = 0 ∞ a n ( t ) n = 0 [ ∑ n = 0 ∞ n ( n − 1 ) a n ( t ) n + 4 ∑ n = 0 ∞ n ( n − 1 ) a n ( t ) n − 1 + 4 ∑ n = 0 ∞ n ( n − 1 ) a n ( t ) n − 2 − ∑ n = 0 ∞ n a n ( t ) n − 2 ∑ n = 0 ∞ n a n ( t ) n − 1 + 2 ∑ n = 0 ∞ a n ( t ) n ] = 0 … ( 6 ) Substitute n − 1 for k in second and n − 2 for k in third summation in equation ( 6 ) . ∑ k = 0 ∞ [ k ( k − 1 ) a k + 4 ( k + 1 ) ( k + 2 ) a k + 1 + 4 ( k + 1 ) ( k + 2 ) a k + 2 − k a k − 2 ( k + 1 ) a k + 1 + 2 a k ] = 0 ∑ k = 0 ∞ [ ( k 2 − 2 k + 2 ) a k + ( k + 1 ) a k + 1 ( 4 k − 2 ) + 4 ( k + 1 ) ( k + 2 ) a k + 2 ] t k = 0 Compare the constant term and coefficient of t k in above equation

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 8.4, Problem 10E

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First four nonzero terms in a power series expansion about x0 for the solution of given differential equation.

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Chapter 8.4, Problem 9E

Chapter 8.4, Problem 11E

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First four nonzero terms in a power series expansion about x 0 for the solution of given differential equation.

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First four nonzero terms in a power series expansion about x0 for the solution of given differential equation.

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First four nonzero terms in a power series expansion about x 0 for the solution of given differential equation.

Solution Summary: The author explains the first four nonzero terms in a power series expansion about x_0 for the solution of given differential equation.

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To find: First four nonzero terms in a power series expansion about x0 for the solution of given differential equation.

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Chapter 8 Solutions

To find:

First four nonzero terms in a power series expansion about x0 for the solution of given differential equation.