Explanation Given: y 1 is a solution of the equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = g 1 ( t ) ... ( 1 ) And y 2 is solution of the equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = g 2 ( t ) ... ( 2 ) and the other equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = k 1 g 1 ( t ) + k 2 g 2 ( t ) ... ( 3 ) Calculation: As y 1 is a solution of equation ( 1 ) on the interval I then y 1 satisfies the equation ( 1 ) . That is, y ″ 1 ( t ) + p ( t ) y ′ 1 ( t ) + q ( t ) y 1 ( t ) = g 1 ( t ) ... ( 4 ) And y 2 is a solution of equation ( 2 ) on the interval I then y 2 satisfies the equation ( 2 ) . That is, y ″ 2 ( t ) + p ( t ) y ′ 2 ( t ) + q ( t ) y 2 ( t ) = g 2 ( t ) ... ( 5 ) Multiply equation ( 4 ) by k 1 and multiply equation ( 5 ) by k 2 and then add both equations

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 4.7, Problem 30E

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k1y1+k2y2 is a solution on I to the equation

y″(t)+p(t)y′(t)+q(t)y(t)=k1g1(t)+k2g2(t).

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Chapter 4.7, Problem 29E

Chapter 4.7, Problem 31E

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k 1 y 1 + k 2 y 2 is a solution on I to the equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = k 1 g 1 ( t ) + k 2 g 2 ( t ) .

Solution Summary: The author explains that y_1 is a solution of the equation (1) on the interval I.

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To show:

k1y1+k2y2 is a solution on I to the equation

y″(t)+p(t)y′(t)+q(t)y(t)=k1g1(t)+k2g2(t).

Want to see the full answer?

Chapter 4 Solutions

k 1 y 1 + k 2 y 2 is a solution on I to the equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = k 1 g 1 ( t ) + k 2 g 2 ( t ) .

Solution Summary: The author explains that y_1 is a solution of the equation (1) on the interval I.

Concept explainers

Videos

To show: k1y1+k2y2 is a solution on I to the equation y″(t)+p(t)y′(t)+q(t)y(t)=k1g1(t)+k2g2(t).

Want to see the full answer?

Chapter 4 Solutions

To show:

k1y1+k2y2 is a solution on I to the equation

y″(t)+p(t)y′(t)+q(t)y(t)=k1g1(t)+k2g2(t).