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: Given that, the differential equation is y ″ − 4 y ′ + 8 y = x 3 . First solve the homogeneous equation y ″ − 4 y ′ + 8 y = 0 . The auxiliary equation for the homogenous differential equation y ″ − 4 y ′ + 8 y = 0 is m 2 − 4 m + 8 = 0 . So, the values of m = − b ± b 2 − 4 a c 2 a . m 1 = − b + b 2 − 4 a c 2 a = − ( − 4 ) + − 16 2 ( 1 ) = 4 + 4 i 2 m 1 = 2 + 2 i Thus, the value of m 1 = 2 + 2 i . m 2 = − b − b 2 − 4 a c 2 a = − ( − 4 ) − − 16 2 ( 1 ) = 4 − 4 i 2 = 2 − 2 i Thus, the solutions of the auxiliary equation are m 1 = 2 + 2 i and m 2 = 2 − 2 i . So, the complementary solution of the homogeneous differential equation y ″ − 4 y ′ + 8 y = 0 is, y c = e 2 x ( C 1 cos 2 x + C 2 sin 2 x ) . Here, the differential operator D 4 annihilates of x 3 . Now, operate D 4 on both sides of the differential equation ( D 2 − 4 D + 8 ) y = x 3 as follows. D 4 ( D 2 − 4 D + 8 ) y = D 4 x 3 D 4 ( D 2 − 4 D + 8 ) y = 0 Auxiliary equation for the fourth order differential equation D 4 ( D 2 − 4 D + 8 ) y = 0 is m 4 ( m 2 − 4 m + 8 ) = 0 . The solutions of the auxiliary equation m 4 ( m 2 − 4 m + 8 ) = 0 are, m 1 = m 2 = m 3 = m 4 = 0 , m 5 = 2 + 2 i and m 6 = 2 − 2 i . Hence, the general solution to the nonhomogeneous differential equation y ″ − 4 y ′ + 8 y = x 3 will be in form, y = C 1 + C 2 x + C 3 x 2 + C 4 x 3 + e 2 x ( C 5 cos 2 x + C 6 sin 2 x ) . Because e 2 x ( C 5 cos 2 x + C 6 sin 2 x ) are repeated terms. So, eliminate these terms, it gives a particular solution that is y p = A + B x + C x 2 + D x 3 . Substitute y p = A + B x + C x 2 + D x 3 , y ′ p = B + 2 C x + 3 D x 2 and y ″ = 2 C + 6 D x into the fundamental differential equation y ″ − 4 y ′ + 8 y = x 3 as, 2 C + 6 D x − 4 ( B + 2 C x + 3 D x 2 ) + 8 ( A + B x + C x 2 + D x 3 ) = x 3

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4.5, Problem 71E

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To solve: The initial value problem y″−4y′+8y=x3,y(0)=2,y′(0)=4.

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Chapter 4.5, Problem 70E

Chapter 4.5, Problem 72E

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Chapter 4 Solutions

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - Given that y=c1+c2x2 is a two-parameter family of...Ch. 4.1 - Find two members of the family of solutions in...Ch. 4.1 - Given that x(t) = c1 cos t + c2 sin t is the...Ch. 4.1 - Use the general solution of x + 2x = 0 given in...Ch. 4.1 - In Problems 9 and 10 find an interval centered...Ch. 4.1 - In Problems 9 and 10 find an interval centered...

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The initial value problem y ″ − 4 y ′ + 8 y = x 3 , y ( 0 ) = 2 , y ′ ( 0 ) = 4 .

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

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To solve: The initial value problem y″−4y′+8y=x3,y(0)=2,y′(0)=4.

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Chapter 4 Solutions