Explanation The given differential equation is y ″ + 9 y = 9 sec 2 3 t , 0 < t < π 6 . “Note that, the general solution of the differential equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) is in the form of Y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) , where y 1 ( t ) and y 2 ( t ) form a fundamental set of solution of the homogenous differential equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = 0 , here u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t and u 2 ( t ) = ∫ y 1 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t .” Obtain the complementary solution of the given differential equation. The characteristic equation is r 2 + 9 = 0 . r 2 + 9 = 0 r 2 = − 3 r = ± 3 i The characteristic roots are r 1 = 0 − 3 i or r 2 = 0 + 3 i . Since the roots are imaginary, the complementary solution is y c ( t ) = c 1 cos 3 t + c 2 sin 3 t . Note that, the functions y 1 = cos 3 t and y 2 = sin 3 t are the fundamental set of solutions of the homogeneous equation. Find the wronskian of the functions y 1 = cos 3 t and y 2 = sin 3 t . W [ y 1 , y 2 ] = | y 1 y 2 y ′ 1 y ′ 2 | = | cos 3 t sin 3 t − 3 sin 3 t 3 cos 3 t | = cos t ⋅ 3 cos 3 t − ( sin 3 t ⋅ − 3 sin 3 t ) = 3 cos 2 3 t + 3 sin 2 3 t On further simplification, W [ y 1 , y 2 ] = 3 ( 1 ) = 3 Obtain the value of u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t . u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t = − ∫ sin 3 t ( 9 sec 2 3 t ) 3 d t = − 9 3 ∫ sin 3 t sec 2 3 t d t = − 3 ∫ sin 3 t cos 2 3 t d t Let v = cos 3 t then d v d t = − 3 sin 3 t and sin 3 t d t = − d v 3 . Use the above substitution to find the value of u 1 ( t )

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3.6, Problem 6P

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The particular solution of the given differential equation by using the method of variation of parameters and check the result by using the method of undetermined coefficient.

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