Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 7


$$y= x^{\frac{1}{3}}\left( -\frac{1}{4} x^{-4/3}+C\right) .$$

Work Step by Step

Rewrite the equation $$y'-(3x)^{-1}y=\frac{1}{3}x^{-2}$$ This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{\int -(3x)^{-1} dx}=e^{- \frac{1}{3}\ln x}=x^{-\frac{1}{3}}$$ Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=x^{\frac{1}{3}}\left( \int \frac{1}{3}x^{-\frac{7}{3}} dx +C\right)\\ &=x^{\frac{1}{3}}\left( -\frac{1}{4} x^{-4/3}+C\right) \end{align} Thus, the general solution is $$y= x^{\frac{1}{3}}\left( -\frac{1}{4} x^{-4/3}+C\right) .$$
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.