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Q. 39

Expert-verified

Calculus

Found in: Page 570

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52
$\frac{d y}{d x} = e^{x + y}, y (0) = 2$

The solution of the initial-value problem descibed by $\frac{d y}{d x} = e^{x + y}, y (0) = 2$ is $y (x) = - \ln | 1 + e^{- 2} - e^{x} |$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

The given initial value problem $\frac{d y}{d x} = e^{x + y}, y (0) = 2 . . . . . . . . . . . . . . (1)$

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value

Use the property of exponential function and rewrite the differential equation in (1) as

\frac{d y}{d x} = e^{x} . e^{y} (∵ e^{x} . e^{y} = e^{x + y})

Note that the differentlal equation is now in the form

\frac{d y}{d x} = p (x) q (y)

In which

p (x) = e^{x}

, and

q (y) = e^{y}

. So, the differential equation can be solved by applying variable separable method. Thus, the solution of the differential equation involved in the initial- value problem is given by

$\int \frac{1}{e^{y}} d y = \int e^{x} d x \int e^{- y} d y = e^{x} + C - e^{- y} = e^{x} + C e^{- y} = - e^{x} - C$

Now, use the given initial condition

y (0) = 2

, that is take

x = 0, y = 2

in the above result and evaluate the constant C.

$e^{- 2} = - 1 - C C = - e^{- 2} - 1$

Substitute this value of the constant C in the solution of the differential equation and obtain the solution of the initial - value problem

\frac{d y}{d x} = e^{x + y}, y (0) = 2

as

$e^{- y} = - e^{x} + 1 + e^{- 2} - y = \ln | 1 + e^{- 2} - e^{x} | y = - \ln | 1 + e^{- 2} - e^{x} |$

Therefore,the solution of the initial-value problem descibed by equation(1) is

$y (x) = - \ln | 1 + e^{- 2} - e^{x} |$

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