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: 40


$$ y= \ln(\frac{1}{2}t^2+\frac{1}{2}).$$

Work Step by Step

By separation of variables, we have $$e^{y}dy=tdt $$ then by integration, we get $$e^y=\frac{1}{2}t^2+c\Longrightarrow y= \ln(\frac{1}{2}t^2+c).$$ Now, since $y(1)=0$, then $c=\frac{1}{2}$. So the general solution is given by $$ y= \ln(\frac{1}{2}t^2+\frac{1}{2}).$$
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