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: 9


$$y= \ \frac{1}{5} x^{2}+\frac{1}{3}+Cx^{-3} .$$

Work Step by Step

This is a linear equation and it has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{\int 3x^{-1} dx}=e^{3\ln x}=x^{3}$$ Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=x^{-3}\left( \int x^{4} +x^2dx +C\right)\\ &=x^{-3}\left( \frac{1}{5} x^{5}+\frac{1}{3}x^3+C\right) \end{align} Thus, the general solution is $$y= \ \frac{1}{5} x^{2}+\frac{1}{3}+Cx^{-3} .$$
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.