Explanation Given: The differential equation and function, x 2 y ″ − 3 x y ′ + 5 y = 0 y 1 ( x ) = x 2 cos ( ln x ) Formula used: If the general form the equation y ″ + P y ′ + Q y and one of the solutions is y 1 ( x ) then, y 2 ( x ) = y 1 ( x ) ∫ e − ∫ P d x y 1 2 d x Calculation: Consider differential equation and function, x 2 y ″ − 3 x y ′ + 5 y = 0 y 1 ( x ) = x 2 cos ( ln x ) Rewrite the differential equation in the standard form and then divide the differential equation x 2 y ″ − 3 x y ′ + 5 y = 0 by x 2 , x 2 y ″ − 3 x y ′ + 5 y = 0 y ″ − 3 x x 2 y ′ + 5 x 2 y = 0 y ″ − 3 x y ′ + 5 x 2 y = 0 Compare the differential equation y ″ − 3 x y ′ + 5 x 2 y = 0 with the standard form y ″ + P y ′ + Q y , y ″ − 3 x y ′ + 5 x 2 y = 0 y 1 = x 2 cos ( ln x ) After the comparison we get, y 1 = x sin ( ln x ) , P ( x ) = − 1 x Find the second solution, y 2 ( x ) = y 1 ( x ) ∫ e − ∫ P d x y 1 2 d x = x 2 cos ( ln x ) ∫ e − ∫ ( − 3 x ) d x ( x 2 cos ( ln x ) ) 2 d x = x

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 3.2, Problem 14E

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The second solution y2(x) of the given differential equation.

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Chapter 3.2, Problem 13E

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The second solution y 2 ( x ) of the given differential equation.

Solution Summary: The author calculates the second solution of the differential equation y_2(x).

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