Explanation Calculation: Consider the equation x y ″ + ( 1 − 2 v ) y ′ + x y = 0... ( 1 ) Write the indicial equation of ( 1 ) and solve for r . r ( r − 1 ) + ( 1 − 2 v ) r = 0 r 2 − r + r − 2 v r = 0 r ( r − 2 v ) = 0 This implies that, r 1 = 0 and r 2 = 2 v are the roots of the indicial equation. If 2 v is not an integer then by the method of frobenius, a linearly independent solution is given by, y ( x ) = a 0 ∑ n = 0 ∞ ( − 1 ) n 2 2 n ( n ! ) ( 1 + v ) n x 2 n + 2 v ... ( 2 ) Take a 0 = 1 2 v Γ ( 1 + v ) in equation ( 2 ) , then by the relation ( t ) n = Γ ( t + n ) Γ ( t ) , the equation ( 2 ) becomes,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 8.8, Problem 21E

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That xvJv(x) satisfies the equation xy″+(1−2v)y′+xy=0 and use this result to find a solution for the equation xy″−2y′+xy=0.

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Chapter 8.8, Problem 20E

Chapter 8.8, Problem 22E

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That x v J v ( x ) satisfies the equation x y ″ + ( 1 − 2 v ) y ′ + x y = 0 and use this result to find a solution for the equation x y ″ − 2 y ′ + x y = 0 .

Solution Summary: The author explains how to solve the indicial equation y′′+(1-2v) — if 2v is not an integer, a linearly independent solution is given

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To show:

That xvJv(x) satisfies the equation xy″+(1−2v)y′+xy=0 and use this result to find a solution for the equation xy″−2y′+xy=0.

Want to see the full answer?

Chapter 8 Solutions

That x v J v ( x ) satisfies the equation x y ″ + ( 1 − 2 v ) y ′ + x y = 0 and use this result to find a solution for the equation x y ″ − 2 y ′ + x y = 0 .

Solution Summary: The author explains how to solve the indicial equation y′′+(1-2v) — if 2v is not an integer, a linearly independent solution is given

Concept explainers

Videos

To show: That xvJv(x) satisfies the equation xy″+(1−2v)y′+xy=0 and use this result to find a solution for the equation xy″−2y′+xy=0.

Want to see the full answer?

Chapter 8 Solutions

To show:

That xvJv(x) satisfies the equation xy″+(1−2v)y′+xy=0 and use this result to find a solution for the equation xy″−2y′+xy=0.