Explanation Given: It is given that y 1 is a solution of y ( n ) + a n − 1 y ( n − 1 ) + a n − 2 y ( n − 2 ) + ⋯ + a 1 y ′ + a o y = f 1 ( t ) and y 2 is a solution of y ( n ) + a n − 1 ( t ) y ( n − 1 ) + a n − 2 ( t ) y ( n − 2 ) + ⋯ + a 1 ( t ) y ′ + a o ( t ) y = f 2 ( t ) . Calculation: Consider the n t h order differential equation. d n y d t n + a n − 1 d n − 1 y d t n − 1 + a n − 2 d n − 2 y d t n − 2 + ⋯ + a 1 d y d t + a o y = f ( t ) (1) Substitute y 1 for y in Equation (1). d n y 1 d t n + a n − 1 d n − 1 y 1 d t n − 1 + a n − 2 d n − 2 y 1 d t n − 2 + ⋯ + a 1 d y 1 d t + a o y 1 = f 1 ( t ) (2) Substitute y 2 for y in Equation (1). d n y 2 d t n + a n − 1 d n − 1 y 2 d t n − 1 + a n − 2 d n − 2 y 2 d t n − 2 + ⋯ + a 1 d y 2 d t + a o y 2 = f 2 ( t ) (3) Add Equation (2) and Equation (3)

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 4.6, Problem 35E

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To show: The expression y1+y2 is a solution of y(n)+an−1(t)y(n−1)+an−2(t)y(n−2)+......+a1(t)y′+ao(t)y=f1(t)+f2(t).

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Chapter 4.6, Problem 31E

Chapter 4.7, Problem 1E

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Chapter 4 Solutions

Introductory Differential Equations

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The expression y 1 + y 2 is a solution of y ( n ) + a n − 1 ( t ) y ( n − 1 ) + a n − 2 ( t ) y ( n − 2 ) + ...... + a 1 ( t ) y ′ + a o ( t ) y = f 1 ( t ) + f 2 ( t ) .

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To show: The expression y1+y2 is a solution of y(n)+an−1(t)y(n−1)+an−2(t)y(n−2)+......+a1(t)y′+ao(t)y=f1(t)+f2(t).

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Chapter 4 Solutions