Calculus (3rd Edition)

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

Chapter 10 - Introduction to Differential Equations - Chapter Review Exercises - Page 526: 26


$$ 4\sqrt y+2\ln y= e^{t^2}+3 .$$

Work Step by Step

By separation of variables, we have $$(y^{-1/2}+y^{-1})dy=te^{t^2} dt $$ then by integration, we get $$ \frac{1}{1/2}\sqrt y+\ln y=\frac{1}{2}e^{t^2}+c \Longrightarrow4\sqrt y+2\ln y= e^{t^2}+ A .$$ Now, since $y(0)=1$, then $A=3$. So we have $$ 4\sqrt y+2\ln y= e^{t^2}+3 .$$
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