Explanation Procedure used: Particular solution to a linear DE by Green’s function: 1. Consider the differential equation and determine the complementary function to the homogeneous differential equation. 2. If y 1 ( x ) and y 2 ( x ) are two fundamental solution of the differential equation, then particular solution is given by y p = u 1 y 1 ( x ) + u 2 y 2 ( x ) . 3. Use, y p = ∫ x 0 x G ( x , t ) f ( t ) d t to find the particular solution. 4. The general solution to a nonhomogeneous differential equation is given by y ( x ) = y h + y p , where, y h is the solution to the homogeneous equation associate to the differential equation L ( x ) with nonhomogeneous boundary conditions and y p is the solution to the nonhomogeneous differential equation with homogeneous boundary values. Calculation: Consider the differential equation y ″ + 2 y ′ + y = e − x . The associate homogeneous differential equation to y ″ + 2 y ′ + y = e − x is y ″ + 2 y ′ + y = 0 . The auxiliary equation to y ″ + 2 y ′ + y = 0 is m 2 + 2 m + 1 = 0 . Thus, calculate the solution to the auxiliary equation as, y ″ + 2 y ′ + y = 0 m 2 + 2 m + 1 = 0 ( m + 1 ) 2 = 0 Hence, the roots of the auxiliary equation are m 1 = − 1 and m 2 = − 1 . Thus, the two solutions to the DE are y 1 = e − x and y 2 = x e − x . Thus, the solution to the homogeneous equation is y h = c 1 e − x + c 2 x e − x . Since, there are no boundary values or initial values, leave the values of coefficients unknown. The particular solution to the nonhomogeneous differential equation is given by y p = ∫ x 0 x G ( x , t ) f ( t ) d t . Now, to determine the Wronskian, use variable transformation. Thus, y 1 ( x ) = e − x defines y 1 ( t ) = e − t , y 2 ( x ) = x e − x defines y 2 ( t ) = t e − t and and f ( x ) = e − x defines f ( t ) = e − t

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 4.8, Problem 9E

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The particular solution of the differential equation y″+2y′+y=e−x in the integral form.

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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 particular solution of the differential equation y ″ + 2 y ′ + y = e − x in the integral form.

Solution Summary: The author explains the particular solution of the differential equation y′′+2e-x in the integral form and the general solution to a nonhomogeneous equation.

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The particular solution of the differential equation y″+2y′+y=e−x in the integral form.

Chapter 4 Solutions