Explanation Given: The initial value problem is y ″ + x y ′ + y = 0 , y ( 1 ) = − 6 , y ′ ( 1 ) = 3 . Concept: If y is a function of initial value problem then the solution of the given function can be written in the form y = ∑ n = 0 ∞ C n ( x − 1 ) n . Calculation: Consider x − 1 = t to make calculation easier, Differentiate both sides, d x = d t So y ′ can be written as follows: y ′ = d y d x or y ′ = d y d t Similarly, y ″ = d 2 y d x 2 or y ″ = d 2 y d t 2 So the initial value problem can be written as follows: y ″ + ( t + 1 ) y ′ + y = 0 , y ( 1 ) = − 6 , y ′ ( 1 ) = 3 According to the initial value theorem, the solution is written as shown below. y = ∑ n = 0 ∞ c n ( x − 1 ) n Where x − 1 = t then it becomes y = ∑ n = 0 ∞ c n t n . Now differentiate the above function. y ′ = ∑ n = 1 ∞ n c n t n − 1 Further differentiate the above function. y ″ = ∑ n = 2 ∞ n ( n − 1 ) c n t n − 2 Substitute the value of y ′ and y ″ in equation (1). y ″ + ( t + 1 ) y ′ + y = 0 ∑ n = 2 ∞ n ( n − 1 ) c n t n − 2 + ( t + 1 ) ∑ n = 1 ∞ ` n c n t n − 1 + ∑ n = 0 ∞ c n t n = 0 ∑ n = 2 ∞ n ( n − 1 ) c n t n − 2 ︸ k = n − 2 + ∑ n = 1 ∞ n c n t n ︸ k = n + ∑ n = 1 ∞ n c n t n − 1 ︸ k = n − 1 + ∑ n = 0 ∞ c n t n ︸ k = n = 0 ∑ k = 0 ∞ ( k + 2 ) ( k + 1 ) c k + 2 t k + ∑ k = 1 ∞ k c k t k + ∑ k = 0 ∞ ( k + 1 ) c k + 1 t k + ∑ k = 0 ∞ c k t k = 0 Further solve the above equation. 2.1. c 2 t 0 + ∑ k = 1 ∞ ( k + 2 ) ( k + 1 ) c k + 2 t k + ∑ k = 1 ∞ k c k t k + 1 c 1 t 0 + ∑ k = 1 ∞ ( k + 1 ) c k + 1 t k + c 0 t 0 + ∑ k = 1 ∞ c k t k = 0 2 c 2 + c 1 + c 0 + ∑ k = 1 ∞ [ ( k + 2 ) ( k + 1 ) c k + 2 + ( k + 1 ) c k + 1 + ( k + 1 ) c k ] t k = 0 Compare the power of x of both sides. 2 c 2 + c 1 + c 0 = 0 ( k + 2 ) ( k + 1 ) c k + 2 + ( k + 1 ) c k + 1 + ( k + 1 ) c k = 0 Solve the above expression

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

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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The solution of the initial value problem y″+xy′+y=0, y(1)=−6,y′(1)=3.

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The solution of the initial value problem y ″ + x y ′ + y = 0 , y ( 1 ) = − 6 , y ′ ( 1 ) = 3 .

Solution Summary: The author explains that if y is a function of initial value problem, the solution of the given function is underset_y=-6left.

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The solution of the initial value problem y″+xy′+y=0, y(1)=−6,y′(1)=3.

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