Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 8 - Mathematical Modeling With Differential Equations - 8.2 Separation Of Variables - Exercises Set 8.2 - Page 575: 9

Answer

$$y = \frac{1}{{1 - k\left( {\csc x - \cot x} \right)}}$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} - \frac{{{y^2} - y}}{{\sin x}} = 0 \cr & {\text{separating the variables}} \cr & \frac{{dy}}{{dx}} = \frac{{{y^2} - y}}{{\sin x}} \cr & \frac{{dy}}{{{y^2} - y}} = \frac{{dx}}{{\sin x}} \cr & \frac{{dy}}{{y\left( {y - 1} \right)}} = \csc xdx \cr & \int {\frac{{dy}}{{y\left( {y - 1} \right)}}} = \int {\csc x} dx \cr & \int {\frac{{dy}}{{y\left( { - 1 + y} \right)}}} = \int {\csc x} dx \cr & \cr & {\text{Integrate by using the formulas }}\int {\frac{{du}}{{u\left( {a + bu} \right)}} = \frac{1}{a}\ln \left| {\frac{u}{{a + bu}}} \right| + C} \cr & \,\,\,\,\,\,\,\,\,\int {\csc x} dx = \ln \left| {\csc x - \cot x} \right| + C \cr & {\text{then}}{\text{,}} \cr & \frac{1}{{ - 1}}\ln \left| {\frac{y}{{ - 1 + y}}} \right| = \ln \left| {\csc x - \cot x} \right| + C \cr & - \ln \left| {\frac{y}{{y - 1}}} \right| = \ln \left| {\csc x - \cot x} \right| + C \cr & \ln \left| {\frac{{y - 1}}{y}} \right| = \ln \left| {\csc x - \cot x} \right| + C \cr & \cr & {\text{Solving the equation for }}y \cr & {e^{\ln \left| {\frac{{y - 1}}{y}} \right|}} = {e^{\ln \left| {\csc x - \cot x} \right|}}{e^C} \cr & \frac{{y - 1}}{y} = k\left( {\csc x - \cot x} \right) \cr & y - 1 = ky\left( {\csc x - \cot x} \right) \cr & y - ky\left( {\csc x - \cot x} \right) = 1 \cr & y = \frac{1}{{1 - k\left( {\csc x - \cot x} \right)}},\,\,\,{\text{where }}k{\text{ is a constant}} \cr} $$
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