Explanation Procedure used: Solution of the differential equation with undetermined coefficients-Annihilator Approach: The Annihilator Approach represents the differential equation L ( y ) = g ( x ) , where, the function g ( x ) may consist any or the entire finite sum, product of multiple polynomials, sine and cosine functions and exponential functions as well. Steps: 1. Look for the complementary solution y c for the homogeneous equation L ( y ) = 0 . 2. Operate both sides of the equation L ( y ) = g ( x ) by the differential operator which annihilates the function g ( x ) . 3. Solve for a homogeneous differential equation by looking for the general solution of the differential equation obtained L 1 L ( y ) = 0 . 4. Solve for the particular solution of the non-homogeneous differential equation L ( y ) = g ( x ) by eliminating the duplicated term in the complementary solution of L 1 L ( y ) = 0 . 5. Determine the undetermined coefficients by substituting the particular solution into non-homogeneous differential equation L ( y ) = g ( x ) . 6. Write the general solution from complementary and particular solution in the form y = y c + y p . Calculation: Consider the differential equation y ‴ − 3 y ″ + 3 y ′ − y = e x − x + 16 . Find the roots of auxiliary equation of homogeneous DE y ‴ − 3 y ″ + 3 y ′ − y = 0 associate to y ‴ − 3 y ″ + 3 y ′ − y = e x − x + 16 . Therefore, y ‴ − 3 y ″ + 3 y ′ − y = 0 m 3 − 3 m 2 + 3 m − 1 = 0 ( m − 1 ) 3 = 0 [ Since ( m − 1 ) 2 = m 3 − 3 m 2 + 3 m − 1 ] Thus, the solutions of the auxiliary equation are m 1 = 1 , m 2 = 1 and m 3 = 1 . Thus, the complimentary function is y c ( x ) = c 1 e x + c 2 x e x + c 3 x 2 e x . The term − x + 16 can be annihilated by the differential operator D and the operator ( D − 1 ) annihilates e x . D 2 ( D − 1 ) ( e x − x + 16 ) = 0 Apply D 2 ( D − 1 ) to the given differential equation. D 2 ( D − 1 ) ( y ‴ − 3 y ″ + 3 y ′ − y ) = 0 D 2 ( D − 1 ) ( D 3 − 3 D 2 + 3 D − 1 ) ( y ) = 0 The auxiliary equation is m 2 ( m − 1 ) ( m 3 − 3 m 2 + 3 m − 1 ) = 0 . Thus, the roots of the auxiliary equation are m 1 = 1 = m 2 = m 3 = m 4 , m 5 = 0 and m 6 = 0 . Thus, the solution of the DE is y ( x ) = c 1 e x + c 2 x e x + c 3 x 2 e x + c 4 x 3 e x + c 5 + c 6 x . Since the terms c 1 e x , c 2 x e x and c 3 x 2 e x are repeated. Therefore, the particular solution is y p ( x ) = A x 3 e x + B + C x , Differentiate the above particular solution with respect to x

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

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 4.5, Problem 61E

To determine

To solve: The differential equation y‴−3y″+3y′−y=ex−x+16 by undetermined coefficients.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.5, Problem 60E

Chapter 4.5, Problem 62E

Students have asked these similar questions

11.9 Recall from Prob. 10.8, that the following system of equations is designed to determine concentrations (the e's in g/m³) in a series of coupled reactors as a function of amount of mass input to each reactor (the right-hand sides are in g/day): 15c3cc33300 -3c18c26c3 = 1200 -4c₁₂+12c3 = 2400 Solve this problem with the Gauss-Seidel method to & = 5%.

9.8 Given the equations 10x+2x2-x3 = 27 -3x-6x2+2x3 = -61.5 x1 + x2 + 5x3 = -21.5 (a) Solve by naive Gauss elimination. Show all steps of the compu- tation. (b) Substitute your results into the original equations to check your answers.

Tangent planes Find an equation of the plane tangent to the following surfaces at the given points (two planes and two equations).

Chapter 4 Solutions

Differential Equations with Boundary-Value Problems (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 - Prob. 7E 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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The differential equation y ‴ − 3 y ″ + 3 y ′ − y = e x − x + 16 by undetermined coefficients.

Solution Summary: The author explains the Annihilator Approach, which solves the differential equation with undetermined coefficients.

Concept explainers

Videos

To solve: The differential equation y‴−3y″+3y′−y=ex−x+16 by undetermined coefficients.

Want to see the full answer?

Chapter 4 Solutions