Explanation Calculation: The differential equation is given by, y ‴ − y ″ + y = sin t ... ( 1 ) Let the particular solution is, y p ( t ) = A sin t Differentiate the above equation. y ′ p ( t ) = A cos t Again differentiate the above equation. y ″ p ( t ) = − A sin t Again differentiate the above equation. y ‴ p ( t ) = − A cos t Substitute − A cos t for y ‴ , − A sin t for y ″ and A sin t for y in the equation ( 1 ) . − A cos t − ( − A sin t ) + A sin t = sin t − A cos t + 2 A sin t = sin t ... ( 2 ) Compare the coefficients of the sin t in the above equation. 2 A = 1 A = 1 2 Compare the coefficients of the cos t in the equation ( 2 ) . − A = 0 A = 0 Hence, two values of the constant A is obtained. Therefore, the assumed particular solution is wrong. Now let the particular solution as, y p ( t ) = A sin t + B cos t ... ( 3 ) Differentiate the above equation. y ′ p ( t ) = A cos t − B sin t Again differentiate the above equation. y ″ p = − A sin t − B cos t Again differentiate the above equation. y ‴ p = − A cos t − B ( − sin t ) y ‴ p = − A cos t + B sin t Substitute − A cos t + B sin t for y ‴ , − A sin t − B cos t for y ″ and A sin t + B cos t for y in the equation ( 1 )

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 4.4, Problem 33E

To determine

To find:

The particular solution to the differential equation y‴−y″+y=sint.

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Chapter 4.4, Problem 32E

Chapter 4.4, Problem 34E

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Chapter 4 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

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