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 as, 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 where, 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 form y p ( x ) = u 1 ( x ) y 1 ( x ) + u 2 ( x ) y 2 ( x ) . 6. Obtain the general solution form y = y c + y p . Calculation: Consider the differential equation x y ″ + y ′ + x = 0 and compare the differential equation to y ″ + P ( x ) y ′ + Q ( x ) y = f ( x ) . f ( x ) = − 1 x The homogeneous equation associated to x y ″ + y ′ + x = 0 is x y ″ + y ′ = 0 . The auxiliary equation associated to x y ″ + y ′ = 0 is m 2 = 0 . Calculate the roots of the auxiliary equation m 2 = 0 as, m 2 = 0 Roots of the auxiliary equation are m 1 = 0 and m 2 = 0 . Hence, the complementary function is y c = c 1 + c 2 ln x . Use variation of parameters to determine the particular solution. Consider that y 1 = 1 and y 2 = ln x . Now compute Wronskian of y 1 and y 2 as, W ( y 1 , y 2 ) = | 1 ln x 0 1 x | = 1 x 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 x y ″ + y ′ = 0 . Compute W 1 . W 1 = | 0 ln x − 1 x 1 x | = 1 x ln x Compute W 2 . W 2 = | 1 0 0 − 1 x | = − 1 x Divide W 1 by W to find u ′ 1 . u ′ 1 = 1 x ln x 1 x = x ln x Integrate both sides

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 46RE

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To find: A member of the family of solution of xy″+y′+x=0 whose graph is tangent on the x-axis at x=1. Also, project the graph with the help of a graphing utility.

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Chapter 4, Problem 44RE

Chapter 4, Problem 47RE

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Chapter 4 Solutions

Differential Equations with Boundary-Value Problems

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A member of the family of solution of x y ″ + y ′ + x = 0 whose graph is tangent on the x -axis at x = 1 . Also, project the graph with the help of a graphing utility.

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

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To find: A member of the family of solution of xy″+y′+x=0 whose graph is tangent on the x-axis at x=1. Also, project the graph with the help of a graphing utility.

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Chapter 4 Solutions