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: 22


$$ y =\frac{3}{3x-x^3+3c}.$$

Work Step by Step

We have $$ y'= y^2(1-x^2)\Longrightarrow \frac{dy}{y^2}=(1-x^2)dx\\ \Longrightarrow\int \frac{dy}{y^2}=\int (1-x^2)dx+c\\ \Longrightarrow -y^{-1}=x-\frac{1}{3}x^3+c \Longrightarrow y =\frac{3}{3x-x^3+3c}.$$ Hence, the general solution is $$ y =\frac{3}{3x-x^3+3c}.$$
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