Explanation Given: Differential equation is ( 1 − x ) y ″ + x y ′ − 2 y = 0 with initial value y ( 0 ) = 0 and y ′ ( 0 ) = 1 . Formula used: The power series solution for a differentiable equation y ″ + p ( x ) y ′ + q ( x ) y = 0 is y = ∑ n = 0 ∞ a n ( x − x 0 ) n where x 0 initial point. Calculation: First obtain the power series solution for the differential equation ( 1 − x ) y ″ + x y ′ − 2 y = 0 about the point x 0 = 0 and obtain the recurrence relation. Given differential equation is ( 1 − x ) y ″ + x y ′ − 2 y = 0 with initial point x 0 = 0 . Consider a power series solution about x 0 = 0 as follows. y = ∑ n = 0 ∞ a n ( x − x 0 ) n = ∑ n = 0 ∞ a n ( x − 0 ) n [ ∵ x 0 = 0 ] = ∑ n = 0 ∞ a n x n Compute the first derivative of y as follows. d y d x = d d x ( ∑ n = 0 ∞ a n x n ) y ′ = ∑ n = 1 ∞ n a n x n − 1 Compute the second derivative of y as follows. d 2 y d x 2 = d d x ( d d x ( ∑ n = 0 ∞ a n x n ) ) y ″ = d d x ( ∑ n = 1 ∞ n a n x n − 1 ) = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 Thus, y = ∑ n = 0 ∞ a n x n , y ′ = ∑ n = 1 ∞ n a n x n − 1 and y ″ = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 . Now substitute y , y ′ and y ″ in the differentiable equation ( 1 − x ) y ″ + x y ′ − 2 y = 0 as follows. ( 1 − x ) y ″ + x y ′ − 2 y = 0 ( 1 − x ) ∑ n = 2 ∞ n ( n − 1 ) a n x ( n − 2 ) + x ∑ n = 1 ∞ n a n x ( n − 1 ) − 2 ∑ n = 0 ∞ a n x n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x ( n − 2 ) − x ∑ n = 2 ∞ n ( n − 1 ) a n x ( n − 2 ) + ∑ n = 1 ∞ n a n x n − ∑ n = 0 ∞ 2 a n x n = 0 ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n − ∑ n = 1 ∞ ( n + 1 ) n a n + 1 x n + ∑ n = 1 ∞ n a n x n − ∑ n = 0 ∞ 2 a n x n = 0 On further simplifications, [ ( 0 + 2 ) ( 0 + 1 ) a ( 0 + 2 ) x 0 + ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) a ( n + 2 ) x n − ∑ n = 1 ∞ ( n + 1 ) n a ( n + 1 ) x n + ∑ n = 1 ∞ n a n x n − 2 a 0 x 0 − ∑ n = 1 ∞ 2 a n x n ] = 0 2 a 2 + ∑ n = 1 ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n − ∑ n = 1 ∞ ( n + 1 ) n a n + 1 x n + ∑ n = 1 ∞ n a n x n − 2 a 0 − ∑ n = 1 ∞ 2 a n x n = 0 2 a 2 − 2 a 0 + ∑ n = 1 ∞ x n [ ( n + 2 ) ( n + 1 ) a n + 2 − ( n + 1 ) n a n + 1 + n a n − 2 a n ] = 0 2 ( a 2 − a 0 ) + ∑ n = 1 ∞ x n [ ( n + 2 ) ( n + 1 ) a n + 2 − ( n + 1 ) n a n + 1 + a n ( n − 2 ) ] = 0 The coefficient of each power of x must be zero. Then ( n + 2 ) ( n + 1 ) a n + 2 − ( n + 1 ) n a n + 1 + a n ( n − 2 ) = 0 and a 2 − a 0 = 0 . Hence, the recurrence relation is ( n + 2 ) ( n + 1 ) a n + 2 − ( n + 1 ) n a n + 1 + a n ( n − 2 ) = 0 for n = 0 , 1 , 2 , 3 , … . On further simplification, ( n + 2 ) ( n + 1 ) a n + 2 = ( n + 1 ) n a n + 1 − a n ( n − 2 ) a n + 2 = ( n + 1 ) n a n + 1 − a n ( n − 2 ) ( n + 2 ) ( n + 1 ) a n + 2 = n a n + 1 n + 2 − a n ( n − 2 ) ( n + 2 ) ( n + 1 ) Simplify a 2 − a 0 = 0 as follows. a 2 − a 0 = 0 a 2 = a 0 Substitute 1 for n in a n + 2 = n a n + 1 n + 2 − a n ( n − 2 ) ( n + 2 ) ( n + 1 ) as follows. a 1 + 2 = a 2 3 − a 1 ( 1 − 2 ) ( 3 ) ( 2 ) a 3 = a 2 3 + a 1 6 = 1 3 ( a 0 ) + a 1 6 [ ∵ a 2 = a 0 ] = a 0 3 + a 1 6 Substitute 2 for n in a n + 2 = n a n + 1 n + 2 − a n ( n − 2 ) ( n + 2 ) ( n + 1 ) as follows. a 2 + 2 = 2 a 2 + 1 2 + 2 − a 2 ( 2 − 2 ) ( 2 + 2 ) ( 2 + 1 ) a 4 = 2 a 3 4 − a 2 ( 0 ) 12 = 1 2 ( a 0 3 + a 1 6 ) [ ∵ a 3 = a 0 3 + a 1 6 ] = a 0 6 + a 1 12 Substitute 3 for n in a n + 2 = n a n + 1 n + 2 − a n ( n − 2 ) ( n + 2 ) ( n + 1 ) as follows. a 3 + 2 = 3 a 3 + 1 3 + 2 − a 3 ( 3 − 2 ) ( 3 + 2 ) ( 3 + 1 ) a 5 = 3 a 4 5 − a 3 20 = 3 5 ( a 0 6 + a 1 12 ) − 1 20 ( a 0 3 + a 1 6 ) = a 0 10 + a 1 20 − a 0 60 − a 1 120 On further simplifications, a 5 = 12 a 0 120 + 6 a 1 120 − 2 a 0 120 − a 1 120 = 10 a 0 120 + 5 a 1 120 = a 0 12 + a 1 24 Substitute 4 for n in a n + 2 = n a n + 1 n + 2 − a n ( n − 2 ) ( n + 2 ) ( n + 1 ) as follows

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 5.2, Problem 28P

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To sketch: The partial sums in a series of the given initial value problem about x=0.

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Chapter 5.2, Problem 27P

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The partial sums in a series of the given initial value problem about x = 0 .

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To sketch: The partial sums in a series of the given initial value problem about x=0.

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