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: 34


$$y = \frac{{\ln \left| x \right|}}{{{x^3}}} + \frac{C}{{{x^3}}}$$

Work Step by Step

$$\eqalign{ & {x^4}\frac{{dy}}{{dx}} + 3{x^3}y = 1 \cr & {\text{this equation is not written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right),{\text{ then}} \cr & {\text{divide both sides of the equation by }}{x^4} \cr & \frac{{dy}}{{dx}} + \frac{3}{x}y = \frac{1}{{{x^4}}} \cr & {\text{the equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{ we can note that }}P\left( x \right){\text{ is }}\frac{3}{x} \cr & {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {\frac{3}{x}} dx}} = {e^{3\ln x}} = {x^3} \cr & {\text{multiplying both sides of the differential equation }}\frac{{dy}}{{dx}} + \frac{3}{x}y = \frac{1}{{{x^4}}}{\text{ by }}{x^3} \cr & {x^3}\frac{{dy}}{{dx}} + 3{x^2}y = \frac{1}{x} \cr & {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr & {D_x}\left[ {{x^3}y} \right] = \frac{1}{x} \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {x^3}y = \int {\frac{1}{x}} dx \cr & {x^3}y = \ln \left| x \right| + C \cr & y = \frac{{\ln \left| x \right|}}{{{x^3}}} + \frac{C}{{{x^3}}} \cr} $$
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