Explanation Procedure used: General solution of differential equation using variation of parameter. 1 If L is a differential equation a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = g ( x ) . Then convert it into standard form, which is given below. S F = a 2 ( x ) ′ ′ y + a 1 ( x ) y ′ + a 0 ( x ) y = g ( x ) a 2 2 Therefore, y ″ + P ( x ) y ′ + Q ( x ) y = f ( x ) is standard equation. Now determine the complementary solution to the homogeneous equation associated with y ″ + P ( x ) y ′ + Q ( x ) y = f ( x ) . 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. W is Wronskian. u ′ 1 = W 1 W = − y 2 f ( x ) W and u ′ 2 = W 2 W = y 1 f ( x ) W Here, 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 4 y ″ − y = e x / 2 + 3 . Rewrite the differential equation in standard form y ″ − 1 4 y = 1 4 ( e x / 2 + 3 ) . Compare the differential equation to y ″ + P ( x ) y + Q ( x ) y = f ( x ) . Therefore, f ( x ) = 1 4 ( e x / 2 + 3 ) . Homogeneous equation associated to 4 y ″ − y = e x / 2 + 3 is 4 y ″ − y = 0 . The auxiliary equation is 4 m 2 − 1 = 0 . Therefore, 4 m 2 − 1 = 0 m 2 = 1 2 m = ± 1 2 Roots of the auxiliary equation are m 1 = − 1 2 and m 2 = 1 2 . Hence, the complementary solution is y c = c 1 e − x 2 + c 2 e x 2 . Consider that y 1 ( x ) = e − x 2 and y 2 ( x ) = e x 2 . Now compute Wronskian of y 1 ( x ) and y 2 ( x ) . Therefore, W ( y 1 , y 2 ) = | y 1 y 2 y ′ 1 y ′ 2 | = | e − x 2 e x 2 − 1 2 e − x 2 1 2 e x 2 | = 1 2 e x 2 ( e − x 2 ) − ( e x 2 ) ( − 1 2 e − 1 2 x ) = 1 2 ( e 0 ) + 1 2 e 0 Therefore, W = 1 Since, W ≠ 0 the solutions y 1 ( x ) and y 2 ( x ) are linearly independent and forms a set of solutions for the homogeneous equation for y ″ − y 4 = 0 . Compute W 1 and W 2

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.5, Problem 10E

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To solve: The differential equation 4y″−y=ex/2+3 by variation of parameters.

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Chapter 3 Solutions

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Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

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The differential equation 4 y ″ − y = e x / 2 + 3 by variation of parameters.

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Chapter 3 Solutions