Explanation Given: The initial value problem is y ″ − 4 x y ′ + 5 y = 0 ; y ( 0 ) = − 1 , y ′ ( 0 ) = 1 Approach: Rewrite the given equation a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = 0 in the form y ″ + p ( x ) y ′ + q ( x ) y = 0 Here, p ( x ) = a 1 ( x ) a 2 ( x ) and q ( x ) = a 0 ( x ) a 2 ( x ) If x = 0 is an ordinary point for this equation then, the general solution of the given equation is y ( x ) = ∑ n = 0 ∞ a n x n . Calculation: The initial value problem is y ″ − 4 x y ′ + 5 y = 0 … ( 1 ) Because p ( x ) = − 4 x and q ( x ) = 5 are analytic at x = 0 therefore, write the general solution of the given problem as y ( x ) = ∑ n = 0 ∞ a n x n This gives y ′ ( x ) = ∑ n = 1 ∞ n a n x n − 1 y ″ ( x ) = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 Substitute these values in equation ( 1 ) ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − 4 x ∑ n = 1 ∞ n a n x n − 1 + 5 ∑ n = 0 ∞ a n x n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − 4 ∑ n = 1 ∞ n a n x n + 5 ∑ n = 0 ∞ a n x n = 0 … ( 2 ) In order to make each term in the above equation like term, put k = n − 2 in the first summation, k = n in the second summation, k = n in the third summation Then, equation ( 2 ) becomes ∑ k = 0 ∞ ( k + 1 ) ( k + 2 ) a k + 2 x k − 4 ∑ k = 1 ∞ k a k x k + 5 ∑ k = 0 ∞ a k x k = 0 2 a 2 + ∑ k = 1 ∞ ( k + 1 ) ( k + 2 ) a k + 2 x k − 4 ∑ k = 1 ∞ k a k x k + 5 a 0 + 5 ∑ k = 1 ∞ a k x k = 0 ( 2 a 2 + 5 a 0 ) + ∑ k = 1 ∞ [ ( k + 1 ) ( k + 2 ) a k + 2 − 4 k a k + 5 a k ] x k = 0 This will make all the coefficients equal to zero

Fundamentals of Differential Equations [With CDROM] - 7th Edition

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ISBN: 9780321410481

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Publisher: Addison Wesley

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Chapter 8.3, Problem 30E

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Chapter 8.3, Problem 29E

Chapter 8.3, Problem 31E

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To find: The cubic polynomial approximation for the solution to the given problem and plot the graph of the linear, quadratic and cubic polynomial.

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Chapter 8 Solutions

To find:

The cubic polynomial approximation for the solution to the given problem and plot the graph of the linear, quadratic and cubic polynomial.