Explanation Procedure used: General solution of differential equation using variation of parameter: 1. If L is a differential equation for a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = g ( x ) , then convert it into standard form as, S F = a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = g ( x ) a 2 2. The standard equation is y ″ + P ( x ) y ′ + Q ( x ) y = f ( x ) . Now, determine the complementary solution to the homogeneous equation associated with y ″ + P ( x ) y ′ + Q ( x ) y = f ( x ) as, C S : y c = c 1 y 1 ( x ) + c 2 y 2 ( x ) 3. Replace the coefficients c 1 and c 2 with functions u 1 ( x ) and u 2 ( x ) . 4. Determine the differential of functions u 1 ( x ) and u 2 ( x ) by Cramer’s Rule, where W is Wronskian, u ′ 1 = W 1 W = − y 2 f ( x ) W and u ′ 2 = W 2 W = y 1 f ( x ) W and W = | y 1 y 2 y ′ 1 y ′ 2 | , W 1 = | 0 y 2 f ( x ) y ′ 2 | and W 2 = | y 1 0 y ′ 1 f ( x ) | 5. Integrate the derivatives u ′ 1 and u ′ 2 and obtain the particular solution from y p ( x ) = u 1 ( x ) y 1 ( x ) + u 2 ( x ) y 2 ( x ) . 6. Obtain the general solution from y = y c + y p . Calculation: Consider the differential equation y ″ − y = 2 e x e x + e − x . Compare the differential equation to y ' ' + P ( x ) y ' + Q ( x ) y = f ( x ) , f ( x ) = 2 e x e x + e − x The homogeneous equation associated to y ″ − y = 2 e x e x + e − x is y ″ − y = 0 and the auxiliary equation is m 2 − 1 = 0 . Calculate the roots of auxiliary equation as below, m 2 − 1 = 0 m 2 = 1 m = ± 1 Hence, the roots of the auxiliary equation are m 1 = 1 and m 2 = − 1 . The complementary solution is y c = c 1 e x + c 2 e − x . Consider that y 1 ( x ) = e x and y 2 ( x ) = e x . Compute Wronskian of y 1 and y 2 as, W ( y 1 , y 2 ) = | y 1 y 2 y ′ 1 y ′ 2 | = | e x e − x e x − e − x | = − e x e − x − e x e − x = − 2 Since W ≠ 0 , the solutions y 1 ( x ) and y 2 ( x ) are linearly independent and form a set of solutions for the homogeneous equation for y ″ − y = 0

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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Chapter 4, Problem 28RE

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To find: The general solution of the differential equation y″−y=2exex+e−x.

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Chapter 4, Problem 27RE

Chapter 4, Problem 29RE

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Differential Equations with Boundary-Value Problems

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The general solution of the differential equation y ″ − y = 2 e x e x + e − x .

Solution Summary: The author explains how to find the general solution of the differential equation using variation of parameter.

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To find: The general solution of the differential equation y″−y=2exex+e−x.

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Chapter 4 Solutions