Explanation Given: The differential equation is, y ″ − 2 x y ′ + 3 y = x 2 Approach: Assume the differential equation given as, y ″ ( x ) + p ( x ) y ′ ( x ) + q ( x ) y ( x ) = 0 Suppose x 0 is an ordinary point for given differential equation. Then the equation has two linearly independent analytic solutions of the form, y ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n Calculation: The differential equation is, y ″ − 2 x y ′ + 3 y = x 2 ... ( 1 ) Let y ( x ) = ∑ n = 0 ∞ a n x n be the solution to the equation ( 1 ) Differentiate y ( x ) with respect to x as, y ′ ( x ) = ∑ n = 0 ∞ n a n x n − 1 Again differentiate y ′ ( x ) with respect to x as, y ″ ( x ) = ∑ n = 0 ∞ n ( n − 1 ) a n x n − 2 Substitute ∑ n = 0 ∞ n a n x n − 1 for y ′ ( x ) and ∑ n = 0 ∞ n ( n − 1 ) a n x n − 2 for y ″ ( x ) in equation ( 1 ) as, y ″ − 2 x y ′ + 3 y = x 2 ∑ n = 0 ∞ n ( n − 1 ) a n x n − 2 − 2 x ∑ n = 0 ∞ n a n x n − 1 + 3 ∑ n = 0 ∞ a n x n = x 2 ( [ 2 a 2 + 6 a 3 x + 12 a 4 x 2 + 20 a 5 x 3 + 30 a 6 x 4 + … ] − 2 [ a 1 x + 2 a 2 x 2 + 3 a 3 x 3 + 4 a 4 x 4 + … ] + 3 [ a 0 + a 1 x + a 2 x 2 + a 3 x 3 + … ] ) = x 2 ( ( 2 a 2 + 3 a 0 ) + ( 6 a 3 + 2 a 1 + 3 a 1 ) x + ( 12 a 4 − 4 a + 2 3 a 2 ) x 2 + ( 20 a 5 − 6 a 3 + 3 a 3 ) x 3 + … ) = x 2 Equate the coefficients of like terms on both sides of the above equation as,

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 8.4, Problem 24E

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The first four nonzero terms in the power series expansion about x=0

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