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: 20


$$ y =\left(\frac{8t+c}{2}\right)^2.$$

Work Step by Step

We have $$\frac{dy}{dt}=8\sqrt{y}\Longrightarrow y^{-1/2}dy=8dt\\ \Longrightarrow\int y^{-1/2}dy=8\int dt+c\\ \Longrightarrow 2y^{1/2}=8t+c \Longrightarrow y =\left(\frac{8t+c}{2}\right)^2.$$ Hence, the general solution is $$ y =\left(\frac{8t+c}{2}\right)^2.$$
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