Explanation Given: The equation y ″ − 6 y 2 = 0 has three solution 1 / ( 1 − t ) 2 , 1 / ( 2 − t ) 2 and 1 / ( 3 − t ) 2 . Interval I = ( − 1 , 1 ) . Approach: Three functions y 1 ( t ) , y 2 ( t ) and y 3 ( t ) are linearly dependent on and interval I if there exist constants C 1 , C 2 , and C 3 , not all zero, such that C 1 y 1 ( t ) + C 2 y 2 ( t ) + C 3 y 3 ( t ) = 0 … ( 1 ) for all t ∈ I Calculation: Consider the given equation, y ″ − 6 y 2 = 0 Substitute 1 / ( 1 − t ) 2 for y 1 ( t ) , 1 / ( 2 − t ) 2 for y 2 ( t ) and 1 / ( 3 − t ) 2 for y 3 ( t ) in equation ( 1 ) . C 1 ( 1 − t ) 2 + C 2 ( 2 − t ) 2 + C 3 ( 3 − t ) 2 = 0 Simplify above equation.

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 4.8, Problem 4E

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To prove:

The three solutions 1/(1−t)2, 1/(2−t)2 and 1/(3−t)2 to y″−6y2=0 are linearly independent on (−1,1).

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Chapter 4.8, Problem 3E

Chapter 4.8, Problem 5E

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To prove: The three solutions 1 / ( 1 − t ) 2 , 1 / ( 2 − t ) 2 and 1 / ( 3 − t ) 2 to y ″ − 6 y 2 = 0 are linearly independent on ( − 1 , 1 ) .

Solution Summary: The author explains how the three solutions are linearly independent on (-1,1).

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To prove:

The three solutions 1/(1−t)2, 1/(2−t)2 and 1/(3−t)2 to y″−6y2=0 are linearly independent on (−1,1).

Want to see the full answer?

Chapter 4 Solutions

To prove: The three solutions 1 / ( 1 − t ) 2 , 1 / ( 2 − t ) 2 and 1 / ( 3 − t ) 2 to y ″ − 6 y 2 = 0 are linearly independent on ( − 1 , 1 ) .

Solution Summary: The author explains how the three solutions are linearly independent on (-1,1).

Concept explainers

Videos

To prove: The three solutions 1/(1−t)2, 1/(2−t)2 and 1/(3−t)2 to y″−6y2=0 are linearly independent on (−1,1).

Want to see the full answer?

Chapter 4 Solutions

To prove:

The three solutions 1/(1−t)2, 1/(2−t)2 and 1/(3−t)2 to y″−6y2=0 are linearly independent on (−1,1).