Explanation From Problem 22, the two solutions are y 1 ( x ) = [ 1 − α ( α + 1 ) x 2 2 ! + α ( α − 2 ) ( α + 1 ) ( α + 3 ) x 4 4 ! + … + ∑ m = 3 ∞ ( − 1 ) m ( α ) ⋯ ( α − 2 m + 2 ) ( α + 1 ) ⋯ ( α + 2 m − 1 ) x 2 m ( 2 m ) ! ] and y 2 ( x ) = [ x − ( α − 1 ) ( α + 2 ) x 3 3 ! + ( α − 1 ) ( α − 3 ) ( α + 2 ) ( α + 4 ) x 5 5 ! + … + ∑ m = 3 ∞ ( − 1 ) m ( α − 1 ) ⋯ ( α − 2 m + 1 ) ( α + 2 ) ⋯ ( α + 2 m ) x 2 m + 1 ( 2 m + 1 ) ! ] . Consider α = 2 n as even integer. Then n can be written as 2 m for m ∈ ℕ 0 which results in elimination of all the terms after m th term of the solution y 1 ( x ) and it can have the factor α − 2 m . Obtain the polynomials for α = 0 , 2 , 4 Obtain polynomial solution for α = n = 0 . y = y 1 ( x ) = 1 − 0 ( 0 + 1 ) x 2 2 ! + 0 ( 0 − 2 ) ( 0 + 1 ) ( 0 + 3 ) x 4 4 ! + … + = 1 Obtain polynomial solution for α = n = 2 . y = y 1 ( x ) = 1 − 2 ( 2 + 1 ) x 2 2 ! + 2 ( 2 − 2 ) ( 2 + 1 ) ( 2 + 3 ) x 4 4 ! + … + = 1 − 2 ⋅ 3 x 2 2 ! = 1 − 3 x 2 Obtain polynomial solution for α = n = 4 . y = y 1 ( x ) = 1 − 4 ( 4 + 1 ) x 2 2 ! + 4 ( 4 − 2 ) ( 4 + 1 ) ( 4 + 3 ) x 4 4 ! + … + = 1 − 4 ⋅ 5 x 2 2 ! + 4 ⋅ 2 ⋅ 5 ⋅ 7 4 ! + 0 = 1 − 10 x 2 + 35 4 x 4 Thus, the polynomial solution for α = n = 0 , 2 , 4 are 1 , 1 − 3 x 2 and 1 − 10 x 2 + 35 4 x 4 , respectively

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 5.3, Problem 23P

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To show: If a constant α is zero or positive even integer 2n, then series solution y1 reduces to polynomial of degree 2n which only has even powers of x and similarly show for positive odd integer 2n + 1.

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Chapter 5.3, Problem 22P

Chapter 5.3, Problem 24P

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If a constant α is zero or positive even integer 2 n , then series solution y 1 reduces to polynomial of degree 2 n which only has even powers of x and similarly show for positive odd integer 2 n + 1.

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To show: If a constant α is zero or positive even integer 2n, then series solution y1 reduces to polynomial of degree 2n which only has even powers of x and similarly show for positive odd integer 2n + 1.

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