Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Question: In Problems 1 - 30, solve the equation.
$\frac{dy}{dx} = \frac{e^{x + y}}{y - 1}$

$e^{x} + {ye}^{- y} = C$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information and simplification

Given that, $\frac{dy}{dx} = \frac{e^{x + y}}{y - 1} \cdot \cdot \cdot \cdot \cdot \cdot (1)$

Evaluate the equation (1).

$\begin{matrix} \frac{dy}{dx} = \frac{e^{x + y}}{y - 1} \\ \frac{dy}{dx} = \frac{e^{x} e^{y}}{y - 1} \\ = \frac{e^{x}}{(y - 1) e^{- y}} \\ (y - 1) e^{- y} dy = e^{x} dx \cdot \cdot \cdot \cdot \cdot \cdot (2) \end{matrix}$

Now integrate the equation (2) on both sides.

$\begin{matrix} \int (y - 1) e^{- y} dy = \int e^{x} dx \\ \int (y - 1) e^{- y} dy = e^{x} + C \end{matrix}$

Step 2: Evaluation method

Find the value of $\int (y - 1) e^{- y} dy$ separately.

Let us take $u = y - 1, dv = e^{- y} dy$ .

$du = 1 dy, v = - e^{- y}$ .

Use the integration by parts formula.

$\begin{matrix} \int (y - 1) e^{- y} dy = - e^{- y} (y - 1) + \int e^{- y} dy \\ = - {ye}^{- y} + e^{- y} - e^{- y} \\ = - {ye}^{- y} \end{matrix}$

Then,

$\begin{matrix} \int (y - 1) e^{- y} dy = e^{x} + C \\ - {ye}^{- y} = e^{x} + C \\ e^{x} + {ye}^{- y} = C \end{matrix}$

So, the solution is $e^{x} + {ye}^{- y} = C$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems 1 - 30, solve the equation.
$\frac{dy}{dx} = \frac{e^{x + y}}{y - 1}$

Step by Step Solution

Step 1: Given information and simplification

Step 2: Evaluation method

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Question: In Problems 1 - 30, solve the equation.
$\frac{dy}{dx} = \frac{e^{x + y}}{y - 1}$