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Q. 27

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 the given differential equations. Your answers will involve unsolved constants.
$\frac{d y}{d x} = x e^{- y}$

Ans: The solution of the differential equation $\frac{d y}{d x} = x e^{- y}$ is $y = \ln |\frac{1}{2} x^{2} + C|$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

given,

$\frac{d y}{d x} = x e^{- y}$

Step 2. Consider the differential equation defined by equation (1) given below and solve it by using antidifferentiation and/or separation of the variable method.

$\frac{d y}{d x} = x e^{- y} . . . . (1)$

Step 3. Solution

Note that the differential equation (1) is of the form of $\frac{d y}{d x} = p (x) q (y)$ in which $p (x) = x$ and $q (y) = e^{- y}$ . So the differential equation can be solved by applying the variable separable method. Separate the variables and integrate both the sides

$\begin{aligned} \int \frac{1}{e^{- y}} d y = \int x d x \\ \int e^{y} d y = \frac{1}{2} x^{2} + C \\ e^{y} = \frac{1}{2} x^{2} + C \\ y = \ln |\frac{1}{2} x^{2} + C| \end{aligned}$

Hence a solution to the differential equation $\frac{d y}{d x} = x e^{- y}$ isrole="math" localid="1649178838912" $y = \ln |\frac{1}{2} x^{2} + C|$

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