Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 21


$$ y =e^{-\sqrt{1-x^2}+c}.$$

Work Step by Step

We have $$\sqrt{1-x^2}y'=xy\Longrightarrow \frac{dy}{y}=\frac{xdx}{\sqrt{1-x^2}}\\ \Longrightarrow\int \frac{dy}{y}=\int\frac{xdx}{\sqrt{1-x^2}}+c\\ \Longrightarrow \ln y=-\frac{1}{2} 2\sqrt{1-x^2}+c \Longrightarrow y =e^{-\sqrt{1-x^2}+c}.$$ Hence, the general solution is $$ y =e^{-\sqrt{1-x^2}+c}.$$
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