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: 25


$$ y^2-2y= t^2+14.$$

Work Step by Step

By separation of variables, we have $$(y-1)dy=t dt $$ then by integration, we get $$ \frac{1}{2}y^2-y=\frac{1}{2}t^2+c \Longrightarrow y^2-2y= t^2+A .$$ Now, since $y(1)=-3$, then $A=14$. So we have $$ y^2-2y= t^2+14.$$
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