Program Explanation Given information: The differential equation is x 2 y ″ + 5 x y ′ + 4 y = 0 and the function is y 1 = 1 x 2 and y 2 = ln x x 2 . Calculation: Label the given function as, y 1 = 1 x 2 ........ (1) Label the given function as, y 2 = ln x x 2 ........ (2) Label the given differential equation as, x 2 y ″ + 5 x y ′ + 4 y = 0 ........ (3) Differentiate equation (1) with respect to x as shown below. y ′ 1 = d d x ( 1 x 2 ) = − 2 x ( x 2 ) 2 = − 2 x 3 Again differentiate y ′ 1 = − 2 x 3 with respect to x as, y ″ 1 = d d x ( − 2 x 3 ) = − 2 ( − 3 x 2 ( x 3 ) 2 ) = 6 x 4 Substitute 6 x 4 for y ″ , − 2 x 3 for y ′ and 1 x 2 for y in the left hand side ( L .H .S ) of theequation (3) as, L .H .S = x 2 ( 6 x 4 ) + 5 x ( − 2 x 3 ) + 4 1 x 2 = 6 x 2 − 10 x 2 + 4 x 2 = 10 x 2 − 10 x 2 = 0 The right hand side ( R .H .S ) of theequation (3) is 0 . Hence, the left hand side and right hand side of the differential equation is same. Hence, by substitution the function y 1 = 1 x 2 satisfies the given differential equation x 2 y ″ + 5 x y ′ + 4 y = 0 . Differentiate equation (2) with respect to x as shown below

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Problems on Growing Functions

For a given set of functions f and g, the growth of functions states that the function g grows as fast as the function f. The growth rate of algorithms helps determine the algorithm's efficiency and compare its performance to other algorithms.

Question

Chapter 1.1, Problem 11P

Program Plan Intro

To show: That the given function is a solution of the given differential equation by substitution.

Expert Solution & Answer

Chapter 1 Solutions

EBK DIFFERENTIAL EQUATIONS

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Show that the substitution...Ch. 1.6 - Prob. 57P Ch. 1.6 - Prob. 58P Ch. 1.6 - Solve the differential equation dydx=xy1x+y+3 by...Ch. 1.6 - Prob. 60P Ch. 1.6 - Prob. 61P Ch. 1.6 - Prob. 62P Ch. 1.6 - Prob. 63P Ch. 1.6 - Prob. 64P Ch. 1.6 - Prob. 65P Ch. 1.6 - Prob. 66P Ch. 1.6 - Prob. 67P Ch. 1.6 - Prob. 68P Ch. 1.6 - Prob. 69P Ch. 1.6 - As in the text discussion, suppose that an...Ch. 1.6 - Prob. 71P Ch. 1.6 - Prob. 72P Ch. 1 - Prob. 1RP Ch. 1 - Prob. 2RP Ch. 1 - Prob. 3RP Ch. 1 - Prob. 4RP Ch. 1 - Prob. 5RP Ch. 1 - Prob. 6RP Ch. 1 - Prob. 7RP Ch. 1 - Prob. 8RP Ch. 1 - Prob. 9RP Ch. 1 - Prob. 10RP Ch. 1 - Prob. 11RP Ch. 1 - Prob. 12RP Ch. 1 - Prob. 13RP Ch. 1 - Prob. 14RP Ch. 1 - Prob. 15RP Ch. 1 - Prob. 16RP Ch. 1 - Prob. 17RP Ch. 1 - Prob. 18RP Ch. 1 - Prob. 19RP Ch. 1 - Prob. 20RP Ch. 1 - Prob. 21RP Ch. 1 - Prob. 22RP Ch. 1 - Prob. 23RP Ch. 1 - Prob. 24RP Ch. 1 - Prob. 25RP Ch. 1 - Prob. 26RP Ch. 1 - Prob. 27RP Ch. 1 - Prob. 28RP Ch. 1 - Prob. 29RP Ch. 1 - Prob. 30RP Ch. 1 - Prob. 31RP Ch. 1 - Prob. 32RP Ch. 1 - Prob. 33RP Ch. 1 - Prob. 34RP Ch. 1 - Prob. 35RP Ch. 1 - Prob. 36RP

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That the given function is a solution of the given differential equation by substitution.

Solution Summary: The author explains that the given function is a solution of the differential equation by substitution.

Concept explainers

To show: That the given function is a solution of the given differential equation by substitution.

Chapter 1 Solutions