Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 625: 24


$y = (x~ln~x+Cx)^{-1}$

Work Step by Step

$xy'+y = -xy^2$ $y'+\frac{1}{x}y = -y^2$ Note that $n=2$ Let $u = y^{1-n} = y^{-1}$ A first-order linear differential equation is one that can be put into the form: $\frac{dy}{dx} + P(x)~y = Q(x)$ We can consider the given equation with $u$: $y'+\frac{1}{x}y = -y^2$ $\frac{du}{dx} +(-1)\frac{1}{x}u = (-1)(-1)$ $\frac{du}{dx} -\frac{1}{x}u = 1$ Let $P(x) = \frac{-1}{x}$ Let $Q(x) = 1$ We can find the integrating factor: $I(x) = e^{\int \frac{-1}{x}~dx} = e^{-1~ln~x} = e^{ln~x^{-1}} = \frac{1}{x}$ We can multiply by the integrating factor: $\frac{1}{x}~\frac{du}{dx}+(\frac{-1}{x^2})~u = \frac{1}{x}$ $\frac{d}{dx}(\frac{1}{x}~u) = \frac{1}{x}$ We can integrate both sides of the equation: $\frac{1}{x}~u = \int \frac{1}{x}~dx$ $\frac{1}{x}~u = ln~x+C$ $u = x~ln~x+Cx$ We can use the substitution $u = y^{-1}$: $y^{-1} = x~ln~x+Cx$ $y = (x~ln~x+Cx)^{-1}$
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