## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 27

#### Answer

$$y = - \frac{3}{{3\ln \left| x \right| - 4}}$$

#### Work Step by Step

\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{y^2}}}{x};\,\,\,\,\,\,\,\,\,\,\,y\left( e \right) = 3 \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{{{y^2}}} = \frac{1}{x}dx \cr & {\text{use }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr & {y^{ - 2}}dy = \frac{{dx}}{x} \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {{y^{ - 2}}dydy} = \int {\frac{1}{x}} dx \cr & {\text{integrate by using the logarithmic rule and the power rule for integration}} \cr & \frac{{{y^{ - 1}}}}{{ - 1}} = \ln \left| x \right| + C \cr & - \frac{1}{y} = \ln \left| x \right| + C\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( e \right) = 3 \cr & \,y\left( e \right) = 3{\text{ implies that }}y = 3{\text{ when }}x = e \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & - \frac{1}{3} = \ln \left| e \right| + C \cr & - \frac{1}{3} = 1 + C \cr & C = - \frac{4}{3} \cr & {\text{substitute }}C = - \frac{4}{3}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & - \frac{1}{y} = \ln \left| x \right| - \frac{4}{3} \cr & {\text{Solve the equation for }}y \cr & - \frac{1}{y} = \frac{{3\ln \left| x \right| - 4}}{3} \cr & y = - \frac{3}{{3\ln \left| x \right| - 4}} \cr}

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