Explanation The given differential equation is, 2 x y ″ + 5 y ′ + x y = 0 (1) Consider the solution as y = ∑ n = 0 ∞ c n x n + r . Differentiating y with respect to x gives, y ′ = ∑ n = 0 ∞ ( n + r ) c n x n + r − 1 y ″ = ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) c n x n + r − 2 Use the value of y , y ′ , y ″ in equation (1). 2 x y ″ + 5 y ′ + x y = 0 2 x ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) c n x n + r − 2 + 5 ∑ n = 0 ∞ ( n + r ) c n x n + r − 1 + x ∑ n = 0 ∞ c n x n + r = 0 ∑ n = 0 ∞ 2 ( n + r ) ( n + r − 1 ) c n x n + r − 1 + ∑ n = 0 ∞ 5 ( n + r ) c n x n + r − 1 + ∑ n = 0 ∞ c n x n + r + 1 = 0 ∑ n = 0 ∞ [ 2 ( n + r ) ( n + r − 1 ) + 5 ( n + r ) ] c n x n + r − 1 + ∑ n = 0 ∞ c n x n + r + 1 = 0 Substitute n = k in first summation and n = k − 2 in the second summation, and then simplify ∑ k = 0 ∞ [ 2 ( k + r ) ( k + r − 1 ) + 5 ( k + r ) ] c k x k + r − 1 + ∑ k = 2 ∞ c k − 2 x k + r − 1 = 0 [ 2 ( r ) ( r − 1 ) + 5 r ] c 0 x r − 1 + [ 2 ( 1 + r ) ( r ) + 5 ( 1 + r ) ] c 1 x r + ∑ k = 2 ∞ [ 2 ( k + r ) ( k + r − 1 ) + 5 ( k + r ) ] c k x k + r − 1 + ∑ k = 2 ∞ c k − 2 x k + r − 1 = 0 [ 2 r 2 + 3 r ] c 0 x r − 1 + [ 2 r 2 + 7 r + 5 ] c 1 x r + ∑ k = 2 ∞ [ 2 ( k + r ) ( k + r − 1 ) c k + 5 ( k + r ) c k + c k − 2 ] x k + r − 1 = 0 Thus, it gives 2 r 2 + 3 r = 0 2 r 2 + 7 r + 5 = 0 And 2 ( k + r ) ( k + r − 1 ) c k + 5 ( k + r ) c k + c k − 2 = 0

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

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ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 6.3, Problem 16E

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To show: That the indicial roots do not differ by an integer and then find the linearly independent series solution about $x=0$ .

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That the indicial roots do not differ by an integer and then find the linearly independent series solution about x = 0 .

Solution Summary: The author shows that the indicial roots do not differ by an integer and then finds the linearly independent series solution about x=0.

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To show: That the indicial roots do not differ by an integer and then find the linearly independent series solution about $x=0$ .

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Chapter 6 Solutions

That the indicial roots do not differ by an integer and then find the linearly independent series solution about x = 0 .

Solution Summary: The author shows that the indicial roots do not differ by an integer and then finds the linearly independent series solution about x=0.

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To show: That the indicial roots do not differ by an integer and then find the linearly independent series solution about x=0<m:math><m:mrow><m:mi>x</m:mi><m:mo>=</m:mo><m:mn>0</m:mn></m:mrow></m:math>.

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Chapter 6 Solutions

To show: That the indicial roots do not differ by an integer and then find the linearly independent series solution about $x=0$ .