Explanation Given: The initial value problem is ( x + 2 ) y ″ + 3 y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 1 . Theorem used: Frobeniu’s Theorem: For a differential equation a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = 0 , there exists at least one solution of the form as follows: y = ( x − x 0 ) r ∑ n = 0 ∞ c n ( x − x 0 ) n = ∑ n = 0 ∞ c n ( x − x 0 ) n + r Calculation: The series solution for a differential equation is given as follows: y = ∑ n = 0 ∞ c n x n . Differentiate the above equation. y ′ = ∑ n = 1 ∞ n c n x n − 1 Again differentiate the above equation. y ″ = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 Substitute the values of y , y ′ and y ″ in ( x + 2 ) y ″ + 3 y = 0 . ( x + 2 ) y ″ + 3 y = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 . x ︸ k = n − 2 + 2 ∑ n = 1 ∞ n ( n − 1 ) c n x n − 2 ︸ k = n − 2 + 3 ∑ n = 0 ∞ c n x n ︸ k = n ∑ n = 2 ∞ n ( n − 1 ) c n x n − 1 ︸ k = n − 1 + 2 ∑ n = 1 ∞ n ( n − 1 ) c n x n − 2 ︸ k = n − 2 + 3 ∑ n = 0 ∞ c n x n ︸ k = n = ∑ k = 0 ∞ ( k + 1 ) k c k + 1 x k + 2 ∑ k = 0 ∞ ( k + 2 ) ( k + 1 ) c k + 2 x k + 3 ∑ k = 0 ∞ c k x k Further solve the above expression. ( x + 2 ) y ″ + 3 y = 4 c 2 + 3 c 0 + ∑ k = 1 ∞ [ ( k + 1 ) k c k + 1 + 2 ( k + 2 ) ( k + 1 ) c k + 2 + 3 c k ] x k = 0 For the above equation to be zero, it has to satisfy two equations as shown below. 4 c 2 + 3 c 0 and ( k + 1 ) k c k + 1 + 2 ( k + 2 ) ( k + 1 ) c k + 2 + 3 c k = 0 The implications obtained from the above result is as follows: c 2 = − 3 4 c 0 2 ( k + 2 ) ( k + 1 ) c k + 2 = − ( k + 1 ) k c k + 1 − 3 c k c k + 2 = − k c k + 1 2 ( k + 2 ) − 3 c k 2 ( k + 2 ) ( k + 1 ) for k = 1 , 2 , 3

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

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ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 6, Problem 16RE

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The solution of the initial value problem (x+2)y″+3y=0 , y(0)=0 , y′(0)=1.

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Chapter 6, Problem 15RE

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The solution of the initial value problem ( x + 2 ) y ″ + 3 y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 1 .

Solution Summary: The author explains Frobeniu's Theorem: For a differential equation a_2(x)y′′+3y=0

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The solution of the initial value problem (x+2)y″+3y=0 , y(0)=0 , y′(0)=1.

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