Explanation Calculation: Consider the differential equation y x ′ − x = 2 y 2 . The differential equation can be written as, x ′ − x y = 2 y The differential equation is in its standard form with f ( y ) = 2 y and P ( y ) = 1 y . Therefore the integrating factor is given by, e ∫ P ( x ) = e ∫ − 1 / y = e − ln y = 1 y . Multiply both sides of the differential equation by the integrating factor. 1 y ( x ′ − x y ) = 1 y ( 2 y ) ( x y ) ′ = 2 Integrate the above obtained result

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 2.3, Problem 26E

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The solution of the given differential equation ydxdy−x=2y2 with initial value y(1)=5.

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Chapter 2.3, Problem 25E

Chapter 2.3, Problem 27E

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Temperature measurements are based on the transfer of heat between the sensor of a measuring device (such as an ordinary thermometer or the gasket of a thermocouple) and the medium whose temperature is to be measured. Once the sensor or thermometer is brought into contact with the medium, the sensor quickly receives (or loses, if warmer) heat and reaches thermal equilibrium with the medium. At that point the medium and the sensor are at the same temperature. The time required for thermal equilibrium to be established can vary from a fraction of a second to several minutes. Due to its small size and high conductivity it can be assumed that the sensor is at a uniform temperature at all times, and Newton's cooling law is applicable. Thermocouples are commonly used to measure the temperature of gas streams. The characteristics of the thermocouple junction and the gas stream are such that λ = hA/mc 0.02s-1. Initially, the thermocouple junction is at a temperature Ti and the gas stream at…

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The solution of the given differential equation y d x d y − x = 2 y 2 with initial value y ( 1 ) = 5 .

Solution Summary: The author explains the solution of the differential equation ydx,dy, and the integrating factor.

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The solution of the given differential equation ydxdy−x=2y2 with initial value y(1)=5.

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Chapter 2 Solutions