Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 13


(a) $e^{y} d y=-x^{-2} d x$ (b) $e^{y} =x^{-1}+C $ (c) $y =\ln \left(x^{-1}+C\right)$ (d) $y =\ln \left(x^{-1}+ e^4-0.5\right)$

Work Step by Step

Given $$ x^{2} y'+e^{-y}=0$$ (a) Then \begin{aligned} x^{2} \cdot \frac{d y}{d x} &=-e^{-y} \\ \frac{d y}{-e^{-y}} &=\frac{d x}{x^{2}} \\ e^{y} d y &=-x^{-2} d x \end{aligned} (b) Integrate both sides \begin{aligned} \int e^{y} d y &=\int-x^{-2} d x \\ e^{y} &=-\frac{x^{-1}}{(-1)}+C \\ e^{y} &=x^{-1}+C \end{aligned} (c) Then \begin{align*} \ln \left(e^{y}\right)&=\ln \left(x^{-1}+C\right)\\ y&=\ln \left(x^{-1}+C\right) \end{align*} (d) At $x= 2 $, $y=4$, then $c= e^4-0.5$ and $$y =\ln \left(x^{-1}+ e^4-0.5\right)$$
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