Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 10 - Differential Equations - Chapter Review - Review Exercises - Page 561: 8



Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} + 3\frac{y}{x} = \frac{1}{{{x^2}}} \cr & or \cr & \frac{{dy}}{{dx}} + \left( {\frac{3}{x}} \right)y = \frac{1}{{{x^2}}} \cr & {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{comparing }}\frac{{dy}}{{dx}} + \left( {\frac{3}{x}} \right)y = \frac{1}{{{x^2}}}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is }}\frac{3}{x}{\text{ }} \cr & {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int {\frac{3}{x}dx} }} \cr & I\left( x \right) = {e^{3\ln x}} = {e^{\ln {x^3}}} \cr & I\left( x \right) = {x^3} \cr & {\text{then, the statement is true}} \cr} $$
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