# Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 16

$$y = \ln \left( {{e^x} + C} \right)$$

#### Work Step by Step

\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^y}}} \cr & {\text{Separating variables leads to}} \cr & {e^y}dy = {e^x}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {{e^y}dy} = \int {{e^x}dx} \cr & {\text{integrating by using }}\int {{e^u}du} = {e^u} + C{\text{ and the power rule }} \cr & {e^y} = {e^x} + C \cr & {\text{multiply both sides by }} - 1 \cr & {e^y} = {e^x} + C \cr & {\text{solve the equation for }}y \cr & \ln {e^y} = \ln \left( {{e^x} + C} \right) \cr & y = \ln \left( {{e^x} + C} \right) \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {{e^x} + C} \right)} \right] \cr & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^x} + C}} \cr & {\text{where }}{e^y} = {e^x} + C.{\text{ then}} \cr & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^y}}} \cr & {\text{The general solution is verified}} \cr}

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