Explanation Given: The given differential equation is, x 2 y ' ' + 2 x ( 2 + x ) y ' + 2 ( 1 + x ) y = 0 Approach: Consider the differential equation, x 2 y ' ' + x p ( x ) y ' + q ( x ) y = 0 where p and q are analytic at x = 0 . Suppose that, p ( x ) = ∑ n = 0 ∞ p n x n , q ( x ) = ∑ n = 0 ∞ q n x n . for | x | < R . Let r 1 and r 2 denote the roots of the indicial equation and assume that r 1 ≥ r 2 if these roots are real. Then has two linearly independent solutions valid on the interval ( 0 , R ) . The form of solution is determined as follows: 1. r 1 − r 2 not an integer: y 1 ( x ) = x r 1 ∑ n = 0 ∞ a n x n , a 0 ≠ 0 y 2 ( x ) = x r 2 ∑ n = 0 ∞ b n x n , b 0 ≠ 0 2. r 1 = r 2 = r : y 1 ( x ) = x r ∑ n = 0 ∞ a n x n , a 0 ≠ 0 y 2 ( x ) = y 1 ( x ) ln x + x r ∑ n = 1 ∞ b n x n 3. r 1 − r 2 equals a positive integer: y 1 ( x ) = x r 1 ∑ n = 0 ∞ a n x n , a 0 ≠ 0 y 2 ( x ) = A y 1 ( x ) ln x + x r 2 ∑ n = 0 ∞ b n x n , b 0 ≠ 0 The coefficient in each of these solutions can be determined by direct substitution in the considered differential equation. Finally, if x r 1 and x r 2 are replaced by | x | r 1 and | x | r 2 , respectively, we obtain the linearly independent solutions that are valid for ( 0 , R ) . Calculation: The standard form of differential equation is given as, x 2 y ' ' + x p ( x ) y ' + q ( x ) y = 0 … ( 1 ) The indicial equation is given as, r ( r − 1 ) + p ( 0 ) r + q ( 0 ) = 0 … ( 2 ) where r is a root of the indicial equation. Consider the given differential equation, x 2 y ' ' + 2 x ( 2 + x ) y ' + 2 ( 1 + x ) y = 0 … ( 3 ) Compare equation ( 1 ) and ( 3 ) . p ( x ) = 2 ( 2 + x ) And, q ( x ) = 2 ( 1 + x ) Thus, p ( 0 ) = 2 ( 2 + 0 ) = 4 And, q ( 0 ) = 2 ( 1 + 0 ) = 2 Substitute 4 for p ( 0 ) and 2 for q ( 0 ) in equation ( 2 ) . r ( r − 1 ) + ( 4 ) r + ( 2 ) = 0 r 2 − r + 4 r + 2 = 0 r 2 + 3 r + 2 = 0 r 2 + 2 r + r + 2 = 0 Further solve the above equation. r ( r + 2 ) + 1 ( r + 2 ) = 0 ( r + 2 ) ( r + 1 ) = 0 r = − 1 , − 2 Thus, roots are r 1 = − 1 and r 2 = − 2 . As, r 1 − r 2 = − 1 − ( − 2 ) = − 1 + 2 = 1 Since r 1 − r 2 is a positive integer, the general form of two linearly independent solutions are, y 1 ( x ) = x r 1 ∑ n = 0 ∞ a n x n , a 0 ≠ 0 y 2 ( x ) = A y 1 ( x ) ln x + x r 2 ∑ n = 0 ∞ b n x n , b 0 ≠ 0 Substitute − 1 for r 1 and − 2 for r 2 in the above expressions to obtain the general form of two linearly independent solutions to the differential equation on an interval ( 0 , ∞ ) . y 1 ( x ) = x − 1 ∑ n = 0 ∞ a n x n And, y 2 ( x ) = A y 1 ( x ) ln x + x − 2 ∑ n = 0 ∞ b n x n Now consider, y = ∑ n = 0 ∞ a n x n + r Determine the coefficients. y ( x ) = x r ∑ n = 0 ∞ a n x n = ∑ n = 0 ∞ a n x n + r Differentiate above equation twice with respect to x . y ′ ( x ) = ∑ n = 0 ∞ ( r + n ) a n x r + n − 1 y ″ ( x ) = ∑ n = 0 ∞ ( r + n ) ( r + n − 1 ) a n x r + n − 2 Substitute the value of y ″ , y ′ and y in the given equation. [ x 2 ∑ n = 0 ∞ ( r + n ) ( r + n − 1 ) a n x r + n − 2 + 2 x ( 2 + x ) ∑ n = 0 ∞ ( r + n ) a n x r + n − 1 + 2 ( 1 + x ) ∑ n = 0 ∞ a n x n + r = 0 ] [ ∑ n = 0 ∞ ( r + n ) ( r + n − 1 ) a n x r + n + 2 ( 2 + x ) ∑ n = 0 ∞ ( r + n ) a n x r + n + 2 ( 1 + x ) ∑ n = 0 ∞ a n x r + n = 0 ] ∑ n = 0 ∞ [ ( r + n ) ( r + n − 1 ) + 4 ( r + n ) + 2 ] a n x n + r + ∑ n = 0 ∞ ( 2 r + 2 n + 2 ) a n x r + n + 1 = 0 Dividing by x r and replace n by n − 1 in the second term. ∑ n = 0 ∞ [ ( r + n ) ( r + n + 3 ) + 2 ] a n x n + ∑ n = 1 ∞ ( 2 r + 2 n ) a n − 1 x n = 0 The remaining coefficient must satisfy the recurrence relation. [ ( r + n ) ( r + n + 3 ) + 2 ] a n + 2 ( r + n ) a n − 1 = 0 , n = 1 , 2 , 3 , ..... . a n = − 2 ( r + n ) [ ( r + n ) ( r + n + 3 ) + 2 ] a n − 1 For r = − 1 in the above equation, a n = 2 n a n − 1 , n = 1 , 2 , 3 , .... … ( 4 ) Substitute n = 1 in equation ( 4 ) . a 1 = 2 a 0 Substitute n = 2 in equation ( 4 ) . a 2 = 2 2 a 1 = 2 a 0 Substitute n = 3 in equation ( 4 ) . a 2 = 2 3 a 2 = 4 3 a 0 For a 0 = 1 , a Frobenius series solution is, y 1 ( x ) = a 0 x − 1 [ 1 + 2 x + 2 x 2 + 4 3 x 3 +

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 11.5, Problem 24P

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The two linearly independent solutions to the given differential equation x2y''+2x(2+x)y'+2(1+x)y=0 on (0,∞).

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The two linearly independent solutions to the given differential equation x 2 y ' ' + 2 x ( 2 + x ) y ' + 2 ( 1 + x ) y = 0 on ( 0 , ∞ ) .

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The two linearly independent solutions to the given differential equation x2y''+2x(2+x)y'+2(1+x)y=0 on (0,∞).

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