Explanation Procedure used: Solution to the differential equation with undetermined Coefficients-Annihilator Approach. The differential equation L ( y ) = g ( x ) , where the function g ( x ) may consists the finite sum, product of multiple polynomials, sine and cosine functions and exponential functions as well. 1 Find the complementary solution y c for the homogeneous equation L ( y ) = 0 . 2 Operate on both sides of the equation L ( y ) = g ( x ) by the differential operator that will annihilate the function g ( x ) . 3 Find the general solution to the differential equation obtained L 1 L ( y ) = 0 . This is a homogeneous differential equation. 4 Remove the duplicated term in the complementary solution of L 1 L ( y ) = 0 . This gives the particular solution to the nonhomogeneous differential equation L ( y ) = g ( x ) . 5 Substitute the particular solution into nonhomogeneous differential equation L ( y ) = g ( x ) and determine the undetermined coefficients. 6 Write the general solution from complementary and particular solution in form y = y c + y p . 7 Substitute the given initial values into the general solution to determine the undetermined coefficients of complementary solution. 8 Substitute the solution from step 7 into general solution. Calculation: Consider the differential equation y ″ + y = 8 cos 2 x − 4 sin x . First solve the homogeneous equation y ″ + y = 0 . The auxiliary equation for the homogenous differential equation y ″ + y = 0 is m 2 + 1 = 0 . m 2 + 1 = 0 m = ± i Thus, the solutions to the auxiliary equation are m 1 = i and m 2 = − i . Since both the roots are conjugate complex, the complementary solution to the homogeneous differential equation y ″ + y = 0 is y c = C 1 cos x + C 2 sin x . Differential operator ( D 2 + 4 ) annihilates 8 cos 2 x and ( D 2 + 1 ) annihilates 4 sin x . Thus, operate on both sides with ( D 2 + 4 ) ( D 2 + 1 ) . ( D 2 + 4 ) ( D 2 + 1 ) ( D 2 + 1 ) y = ( D 2 + 4 ) ( D 2 + 1 ) 8 cos 2 x − 4 sin x ( D 2 + 4 ) ( D 2 + 1 ) ( D 2 + 1 ) y = 0 Auxiliary equation for the fourth order differential equation ( D 2 + 4 ) ( D 2 + 1 ) ( D 2 + 1 ) y = 0 is ( m 2 + 4 ) ( m 2 + 1 ) ( m 2 + 1 ) = 0 . The solutions to the auxiliary equation ( m 2 + 4 ) ( m 2 + 1 ) ( m 2 + 1 ) = 0 are m 1 = i , m 2 = − i , m 3 = i , m 4 = − i , m 5 = 2 i and m 6 = − 2 i . Hence, the general solution to the nonhomogeneous differential equation y ″ + y = 8 cos 2 x − 4 sin x will be in form, y = C 1 cos x + C 2 sin x + C 3 x cos x + C 4 x sin x + C 5 cos 2 x + C 6 sin 2 x . Since C 1 cos x and C 2 sin x are repeated terms, eliminate these. Thus, it gives a particular solution that is y p = A x cos x + B x sin x + C cos 2 x + D sin 2 x

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4.5, Problem 69E

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To solve: The initial value problem y″+y=8cos2x−4sinx, y(π2)=−1,y′(π2)=0.

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Chapter 4.5, Problem 68E

Chapter 4.5, Problem 70E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

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The initial value problem y ″ + y = 8 cos 2 x − 4 sin x , y ( π 2 ) = − 1 , y ′ ( π 2 ) = 0 .

Solution Summary: The author describes the solution to the differential equation with undetermined Coefficients-Annihilator Approach.

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To solve: The initial value problem y″+y=8cos2x−4sinx, y(π2)=−1,y′(π2)=0.

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Chapter 4 Solutions