University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 16 - Section 16.2 - First-Order Linear Equations - Exercises - Page 16-15: 31

Answer

$y=(1+\dfrac{c}{x^3})^{1/3}$

Work Step by Step

Here, we have $xy'+y=y^{-2}$ Compare the above equation with the Bernoulli equation with $n=-2$ Plug $u=y^{1-(-2)}=y^{3}$ or, $y=u^{1/3}$ Then $\dfrac{dy}{dx}=(\dfrac{1}{3})u^{-2/3}\dfrac{du}{dx}$ or, $\dfrac{du}{dx}+\dfrac{3}{x}u=\dfrac{3}{x}$ The integrating factor is: $e^{\int (\frac{3}{x})dx}=x^{3}$ In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides. $\int [x^3u]'=\int 3x^2 dx$ or, $u=1+cx^{-3} \implies y^3=1+cx^{-3}$ Thus, $y=(1+\dfrac{c}{x^3})^{1/3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.