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: 23


$$ y =\pm\sqrt{x^2+2c}.$$

Work Step by Step

We have $$ yy'= x\Longrightarrow ydy=xdx\\ \Longrightarrow\int ydy=\int xdx+c\\ \Longrightarrow \frac{1}{2}y^2=\frac{1}{2}x^2+c \Longrightarrow y =\pm\sqrt{x^2+2c}.$$ Hence, the general solution is $$ y =\pm\sqrt{x^2+2c}.$$
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