Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.2 Basic Ideas and Terminology - Problems: 30

Answer

$y = \frac{\ln(x)}{x}$

Work Step by Step

$e^{xy} + x = c$ Implicit differentiation: differentiating the equation in terms of $x$: $e^{xy}(x\frac{dy}{dx} + y) - 1 = 0$ $e^{xy}(x\frac{dy}{dx} + y) = 1$ Rearranging the equation, we get: $\frac{1 - ye^{xy}}{xe^{xy}}$ = 1 Since $y(1) = 0$, therefore $c = 0$. As such, $e^{xy} - x = 0$ $e^{xy} = x$ $xy = \ln(x)$ $y = \frac{\ln(x)}{x}$
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