Explanation Procedure used: General solution of a higher order differential equation: In order, to solve an n th order differential equation a n y n + a n − 1 y n − 1 + ... a 2 y ″ + a 1 ′ + a 0 y = 0 , initially solve the n th degree polynomial equation a n m n + a n − 1 m n − 1 + ...... + a 0 = 0 If all the roots of a n m n + a n − 1 m n − 1 + ...... + a 0 = 0 are real and distinct, then the general solution will be y = c 1 e m 1 x + c 2 e m 2 x + ⋯ + c n e m n x . If an auxiliary equation has k roots and all the roots of the auxiliary equation are equal, then the general solution will be c 1 e m 1 x + c 2 x e m 1 x + c 3 x 2 e m 1 x + ....... + c k x k − 1 e m 1 x . In case if higher order differential equation has repeated conjugate complex roots, then the general solution will be, y = c 1 e m 1 x + c 2 e − m 1 x + c 3 x e m 1 x + c 4 x e − m 1 x where m 1 represents an imaginary root. Calculation: Consider the given differential equation y ( 4 ) − y = 0 . Substitute, y = e m x and y ( 4 ) = m 4 e m x into the differential equation y ( 4 ) − y = 0 to obtain an auxiliary equation. m 4 e m x − e m x = 0 e m x ( m 4 − 1 ) = 0 For a real x , e m x ≠ 0

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.3, Problem 62E

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To verify: The solution y=sinhx−2cos(x+π6) is a particular solution of y(4)−y=0. Also, reconcile the particular solution with the general solution of the differential equation.

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Chapter 3.3, Problem 61E

Chapter 3.3, Problem 63E

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To verify: The solution y = sinh x − 2 cos ( x + π 6 ) is a particular solution of y ( 4 ) − y = 0 . Also, reconcile the particular solution with the general solution of the differential equation.

Solution Summary: The author explains how to reconcile the particular solution with the general solution of the differential equation.

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To verify: The solution y=sinhx−2cos(x+π6) is a particular solution of y(4)−y=0. Also, reconcile the particular solution with the general solution of the differential equation.

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