Explanation The given differential equation is 4 y ″ + y = 2 sec 2 t , − π < t < π . “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 4 r 2 + 1 = 0 . 4 r 2 + 1 = 0 4 r 2 = − 1 r 2 = − 1 4 r = ± 1 2 i The characteristic roots are r 1 = 0 + 1 2 i and r 2 = 0 − 1 2 i . Since the roots are imaginary, the complementary solution is y c ( t ) = c 1 cos t 2 + c 2 sin t 2 . Note that, the functions y 1 = cos t 2 and y 2 = sin t 2 are the fundamental set of solutions of the homogeneous equation. Find the wronskian of the functions y 1 = cos t 2 and y 2 = sin t 2 . W [ y 1 , y 2 ] = | y 1 y 2 y ′ 1 y ′ 2 | = | cos t 2 sin t 2 − 1 2 sin t 2 1 2 cos t 2 | = cos t 2 ( 1 2 cos t 2 ) − ( − 1 2 sin t 2 ) ( sin t 2 ) = 1 2 cos 2 t 2 + 1 2 sin 2 t 2 On further simplification, W [ y 1 , y 2 ] = 1 2 ( cos 2 t 2 + sin 2 t 2 ) = 1 2 ( 1 ) = 1 2 Obtain the value of u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d 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 9P

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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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Chapter 3.6, Problem 8P

Chapter 3.6, Problem 10P

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Elementary Differential Equations

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