Explanation Procedure used: Particular solution to a linear DE by greens 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 4 y ″ − 4 y ′ + y = arctan x . The associate homogeneous differential equation to 4 y ″ − 4 y ′ + y = arctan x is 4 y ″ − 4 y ′ + y = 0 . The auxiliary equation to 4 y ″ − 4 y ′ + y = 0 is 4 m 2 − 4 m + 1 = 0 . Thus, calculate the solution to the auxiliary equation, 4 m 2 − 4 m + 1 = 0 ( 2 m − 1 ) 2 = 0 Hence, the roots of the auxiliary equation are m 1 = 1 2 and m 2 = 1 2 . Thus, the two solutions to the DE are y 1 = e 1 2 x and y 2 = x e 1 2 x . Hence, the solution to the homogeneous differential equation 4 y ″ − 4 y ′ + y = 0 is y h = c 1 e 1 2 x + c 2 x e 1 2 x . Since, there are no boundary values or initial values, leave the 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, [ y 1 ( x ) = e 1 2 x defines y 1 ( t ) = e 1 2 t , y 2 ( x ) = x e 1 2 x defines y 2 ( t ) = t e 1 2 t and f ( x ) = arctan x defines f ( t ) = arctan t ] W ( t ) = | y 1 ( t ) y 2 ( t ) y ′ 1

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.10, Problem 10E

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The particular solution of the differential equation 4y″−4y′+y=arctanx in the integral form.

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The particular solution of the differential equation 4 y ″ − 4 y ′ + y = arctan x in the integral form.

Solution Summary: The author explains the particular solution of the differential equation 4y′′-4xprime +y=mathrmarctan

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The particular solution of the differential equation 4y″−4y′+y=arctanx in the integral form.

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