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: 15


The general solution is $$ y=e^{-\frac{1}{3}x^3+c}.$$

Work Step by Step

We have $$ y'+x^2y=0\Longrightarrow \frac{dy}{y}=-x^2dx\\ \Longrightarrow\int \frac{dy}{y}=\int-x^2dx+c\\ \Longrightarrow \ln y=-\frac{1}{3}x^3+c.$$ Hence, the general solution is $$ y=e^{-\frac{1}{3}x^3+c}.$$
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