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: 22

Answer

$${\text{Separable}}$$

Work Step by Step

$$\eqalign{ & \frac{x}{y}\frac{{dy}}{{dx}} = 4 + {x^{3/2}} \cr & {\text{multiply both sides by }}\frac{y}{x} \cr & \frac{{dy}}{{dx}} = \frac{{y\left( {4 + {x^{3/2}}} \right)}}{x} \cr & or \cr & \frac{{dy}}{{dx}} = \frac{{{x^{ - 1}}\left( {4 + {x^{3/2}}} \right)}}{y} \cr & \cr & {\text{The equation can be written in the form }}\frac{{dy}}{{dx}} = \frac{{p\left( x \right)}}{{q\left( y \right)}},{\text{ then}} \cr & {\text{the given equation is separable}} \cr & \cr & {\text{The equation cannot be written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\,\,\,\left( {{\text{linear form}}} \right), \cr & {\text{ then}}{\text{, the given equation is not linear}} \cr & \cr & {\text{Separable}} \cr} $$
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