Explanation Result used: If y 1 is a solution to the differential equation a y ″ + b y ′ + c y = f 1 ( t ) and y 2 is a solution to a y ″ + b y ′ + c y = f 2 ( t ) , then for any constants k 1 and k 2 , the function k 1 y 1 + k 2 y 2 is a solution to the differential equation a y ″ + b y ′ + c y = k 1 f 1 ( t ) + k 2 f 2 ( t ) . Description: Given that, y 1 is a solution to y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = g 1 ( t ) on the interval I . Then, y ″ 1 ( t ) + p ( t ) y ′ 1 ( t ) + q ( t ) y 1 ( t ) = g 1 ( t ) . Also, y 2 is a solution to y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = g 2 ( t ) on the interval I . Then, y ″ 2 ( t ) + p ( t ) y ′ 2 ( t ) + q ( t ) y 2 ( t ) = g 2 ( t ) . Let y = k 1 y 1 + k 2 y 2 . The derivatives will be, y ′ = k 1 y ′ 1 + k 2 y ′ 2 and y ″ = k 1 y ″ 1 + k 2 y ″ 2 . Substitute y = k 1 y 1 + k 2 y 2 , y ′ = k 1 y ′ 1 + k 2 y ′ 2 and y ″ = k 1 y ″ 1 + k 2 y ″ 2 in y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) , y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) <

Student's Solutions Manual for Fundamentals of Differential Equations and Fundamentals of Differential Equations and Boundary Value Problems

9th Edition

ISBN: 9780321977212

Author: Nagle, R. Kent; Saff, Edward B.; Snider, Arthur David

Publisher: PEARSON

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Chapter 4.7, Problem 30E

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To show: For any constants k1 and k2, the function k1y1+k2y2 is a solution on I to 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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Chapter 4 Solutions

Student's Solutions Manual for Fundamentals of Differential Equations and Fundamentals of Differential Equations and Boundary Value Problems

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

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For any constants k 1 and k 2 , the function k 1 y 1 + k 2 y 2 is a solution on I to y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = k 1 g 1 ( t ) + k 2 g 2 ( t ) .

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To show: For any constants k1 and k2, the function k1y1+k2y2 is a solution on I to y″(t)+p(t)y′(t)+q(t)y(t)=k1g1(t)+k2g2(t).

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Chapter 4 Solutions