Explanation Given: The given differential equation is, x 2 y ' ' + 2 x 2 y ' + ( x − 3 4 ) y = 0 Approach: Consider the differential equation, x 2 y ″ + x p ( x ) y ′ + q ( x ) y = 0 Here, 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 if 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 ) Here, r is a root of the indicial equation. Consider the given differential equation, x 2 y ″ + 2 x 2 y ′ + ( x − 3 4 ) y = 0... ( 3 ) Compare equation ( 1 ) and ( 3 ) . p ( x ) = 2 x And, q ( x ) = ( x − 3 4 ) Thus, p ( 0 ) = 2 ( 0 ) = 0 And, q ( 0 ) = ( 0 − 3 4 ) = − 3 4 Substitute 0 for p ( 0 ) and − 3 4 for q ( 0 ) in equation ( 2 ) . r ( r − 1 ) + ( 0 ) r + ( − 3 4 ) = 0 r ( r − 1 ) − 3 4 = 0 r 2 − r − 3 4 = 0 4 r 2 − 4 r − 3 = 0 Further solve the above equation. 4 r 2 + 2 r − 6 r − 3 = 0 2 r ( 2 r + 1 ) − 3 ( 2 r + 1 ) = 0 ( 2 r − 3 ) ( 2 r + 1 ) = 0 r = 3 2 , − 1 2 Thus, roots are r 1 = 3 2 and r 2 = − 1 2 . As, r 1 − r 2 = 3 2 − ( − 1 2 ) = 3 2 + 1 2 = 4 2 = 2 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 3 2 for r 1 and − 1 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 , R ) . y 1 ( x ) = x 3 2 ∑ n = 0 ∞ a n x n And, y 2 ( x ) = A y 1 ( x ) ln x + x − 1 2 ∑ n = 0 ∞ b n x n . As, r 1 − r 2 = 3 2 − ( − 1 2 ) = 3 2 + 1 2 = 2 = positive integer Thus, as r 1 − r 2 is equals a positive integer, obtain the recurrence relation. The general solution to the given differential equation can be expressed in the form. y ( x ) = x r ∑ n = 0 ∞ a n x n ... ( 4 ) Differentiate the equation ( 4 ) twice with respect to x . y ′ ( x ) = ∑ n = 0 ∞ ( r + n ) a n x r + n − 1 And, y ″ ( x ) = ∑ n = 0 ∞ ( r + n ) ( r + n − 1 ) a n x r + n − 2 Substitute x r ∑ n = 0 ∞ a n x n for y ( x ) , ∑ n = 0 ∞ ( r + n ) a n x r + n − 1 for y ′ ( x ) and ∑ n = 0 ∞ ( r + n ) ( r + n − 1 ) a n x r + n − 2 for y ″ ( x ) in equation ( 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 5P

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The roots of the indicial equation and obtain the general form of two linearly independent solutions to the differential equation on an interval (0,R) and if r1−r2 equals a positive integer, obtain the recurrence relation and check whether the constant A in y2(x)=Ay1(x)lnx+xr2∑n=0∞bnxn is zero or nonzero.

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Differential Equations and Linear Algebra (4th Edition)

Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 2TFR Ch. 11.1 - Prob. 3TFR Ch. 11.1 - Prob. 4TFR Ch. 11.1 - Prob. 5TFR Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 7TFR Ch. 11.1 - Prob. 8TFR Ch. 11.1 - Prob. 9TFR Ch. 11.1 - Prob. 10TFR

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Solution Summary: The author explains the roots of the indicial equation and the general form of two linearly independent solutions to the differential equation on an interval.

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