Explanation Given: The differential equation is x y ″ + ( x − 6 ) y ' − 3 y = 0 . Calculation: The differential equation is given as follows: x y ″ + ( x − 6 ) y ' − 3 y = 0 ...... ( 1 ) Suppose that y = ∑ n = 0 ∞ c n x n + r . Differentiate the above equation. y ' = ∑ n = 0 ∞ ( n + r ) c n x n + r − 1 Again differentiate the above equation. y ″ = ∑ n = 2 ∞ ( n + r ) ( n + r − 1 ) c n x n + r − 2 Substitute the value of y ′ and y ″ equation (1) in above expression, x y ″ + ( x − 6 ) y ' − 3 y = 0 x ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) c n x n + r − 2 + ( x − 6 ) ∑ n = 0 ∞ ( n + r ) c n x n + r − 1 − 3 ∑ n = 0 ∞ c n x n + r = 0 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) c n x n + r − 1 + ∑ n = 0 ∞ ( n + r ) c n x n + r − 6 ∑ n = 0 ∞ ( n + r ) c n x n + r − 1 − 3 ∑ n = 0 ∞ c n x n + r = 0 Further simplify the above equation. [ ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) c n x n + r − 1 + ∑ n = 0 ∞ ( n + r ) c n x n + r − 6 ∑ n = 1 ∞ ( n + r ) c n x n + r − 1 − 3 ∑ n = 0 ∞ c n x n + r ] = 0 arcsin θ ∑ n = 0 ∞ ( n + r ) ( n + r − 7 ) c n x n + r − 1 ︸ n = k + ∑ n = 1 ∞ ( n + r − 3 ) c n x n + r ︸ n = k − 1 = 0 ∑ k = 0 ∞ ( k + r ) ( k + r − 7 ) c k x k + r − 1 + ∑ k = 1 ∞ ( k + r − 4 ) c k − 1 x k + r − 1 = 0 r ( r − 7 ) c 0 x r − 1 + x k + r − 1 { ∑ k = 1 ∞ ( k + r ) ( k + r − 7 ) c k + ∑ k = 1 ∞ ( k + r − 4 ) c k − 1 } = 0 Compare the coefficients. r ( r − 7 ) = 0 ( k + r ) ( k + r − 7 ) c k + ( k + r − 4 ) c k − 1 = 0 Two values of r are obtained as 0 , 7 . For r = 7 , the recurrence relation is obtained as follows: ( k + 7 ) ( k + 7 − 7 ) c k + ( k + 7 − 4 ) c k − 1 = 0 c k = − ( k + 3 ) ( k ) ( k + 7 ) c k − 1 , k = 0 , 1 , 2 , 3 …

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781337515573

Author: ZILL

Publisher: Cengage

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Chapter 6.3, Problem 31E

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The number of solutions of the differential equation xy″+(x−6)y'−3y=0 with the use of recurrence relation by the method of Frobenius.

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Chapter 6.3, Problem 30E

Chapter 6.3, Problem 32E

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The number of solutions of the differential equation x y ″ + ( x − 6 ) y ' − 3 y = 0 with the use of recurrence relation by the method of Frobenius.

Solution Summary: The author compares the number of solutions of the differential equation xy′′+(x-6)y '-3y=0 with the use of recurr

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The number of solutions of the differential equation xy″+(x−6)y'−3y=0 with the use of recurrence relation by the method of Frobenius.

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