Explanation Given: The equation is, ( 1 − x 2 ) y ″ − 2 x y ′ + 2 y = 0 ... ( 1 ) Approach: The linear second-order differential equation given by, ( 1 − x 2 ) y ″ − 2 x y ′ + n ( n + 1 ) y = 0 , n is a fixed parameter is called Legendre’s Equation. Set z = x − 1 , the equation is transformed as, z ( z + 2 ) d 2 y d z 2 + 2 ( z + 1 ) d y d z − n ( n + 1 ) y = 0 ... ( 2 ) The indicial equation for equation ( 2 ) is given as, r ( r − 1 ) + r = 0 The solution is given as, y 1 ( x ) = 1 + ∑ k = 1 ∞ ( − n ) k ( n + 1 ) k k ! ( 1 ) k ( 1 − x 2 ) k Calculation: The given equation is Legendre’s equation. Let z = x − 1 Therefore, the equation ( 1 ) is transformed as, z ( z + 2 ) d 2 y d z 2 + 2 ( z + 1 ) d y d z − 2 y = 0 … ( 3 ) The indicial equation is given by, r ( r − 1 ) + r = 0 r 2 = 0 r = 0 , 0 Let y ( z ) = ∑ k = 0 ∞ a k z k be the solution to equation ( 3 ) Differentiate y ( z ) as, y ′ ( z ) = ∑ k = 0 ∞ k a k z k − 1 Again differentiate y ′ ( z ) as, y ″ ( z ) = ∑ k = 0 ∞ k ( k − 1 ) a k z k − 2 Substitute ∑ k = 0 ∞ a k z k for y ( z ) , ∑ k = 0 ∞ k a k z k − 1 for y ′ ( z ) , and ∑ k = 0 ∞ k ( k − 1 ) a k z k − 2 for y ″ ( z ) in equation ( 3 ) as, z ( z + 2 ) d 2 y d z 2 + 2 ( z + 1 ) d y d z − 2 y = 0 z ( z + 2 ) ( ∑ k = 0 ∞ k ( k − 1 ) a k z k − 2 ) + 2 ( z + 1 ) ( ∑ k = 0 ∞ k a k z k − 1 ) − 2 ( ∑ k = 0 ∞ a k z k ) = 0 Solve further to get the solution as, y 1 ( x ) = 1 − ∑ k = 1 ∞ x 2 k 2 k − 1 Let y ( x ) = v ( x ) y 1 ( x ) Substitute the expression for y 1 ( x ) as, y ( x ) = v ( x ) ( 1 − ∑ k = 1 ∞ x 2 k 2 k − 1 ) Differentiate y ( x ) with respect to x . y ′ ( x ) = v ′ ( x ) ( 1 − ∑ k = 1 ∞ x 2 k 2 k − 1 ) − v ( x ) ∑ k = 1 ∞ 2 k x 2 k − 1 2 k − 1 Again differentiate y ′ ( x ) with respect to x

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 8.8, Problem 36E

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The general solution about x=0 for the given equation

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Chapter 8.8, Problem 35E

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Fundamentals of Differential Equations and Boundary Value Problems

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The general solution about x = 0 for the given equation

Solution Summary: The author explains the general solution about x=0 for the given equation. The linear second-order differential equation is called Legendre’s Equation.

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The general solution about x=0 for the given equation

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The general solution about x = 0 for the given equation

Solution Summary: The author explains the general solution about x=0 for the given equation. The linear second-order differential equation is called Legendre’s Equation.

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To find: The general solution about x=0 for the given equation

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Chapter 8 Solutions

To find:

The general solution about x=0 for the given equation