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 - 1.4 Separable Differential Equations - Problems: 5

Answer

$y=C_1(x-2)$

Work Step by Step

Multiply the entire equation by $\frac{1}{y(x-2)}$ to separate variables. $$\frac{1}{y(x-2)}(y {dx}-(x-2)dy=0)$$ $$\frac{dx}{x-2}-\frac{dy}{y}=0$$ Since the dx and dy are in terms of one variable, you can integrate. $$\int \frac{dx}{x-2}-\int \frac{dy}{y} = \int 0 dx$$ $$ln|x-2|-ln|y|=C$$ Solve for $y$. $$ln|y|=ln|x-2|-C$$ $$|y|=e^{ln|x-2|-C}$$ $$|y|=e^{-C}e^{ln|x-2|}$$ $$|y|=C_1|x-2|$$ $$y=C_1(x-2)$$
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