Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Review - Exercises - Page 635: 9


The solution of the given initial-value problem $$ \frac{d r}{d t}+2 t r=r , \quad \quad r(0)=5 $$ is $$ r(t)=5 e^{t-t^{2}} $$

Work Step by Step

The initial-value problem is: $$ \frac{d r}{d t}+2 t r=r $$ $$ \frac{d r}{d t}+2 t r=r \Rightarrow \frac{d r}{d t}=r-2 t r=r(1-2 t) $$ separating and integrating both sides, we get: $$ \int \frac{d r}{r}=\int(1-2 t) d t \Rightarrow \ln |r|=t-t^{2}+C $$ $\Rightarrow$ $$ |r|=e^{t-t^{2}+C}=k e^{t-t^{2}} $$ where where $ k $ is an arbitrary constant. Since $$ r(0)=5,5=k e^{0}=k . $$ Thus the solution of the given initial-value problem is $$ r(t)=5 e^{t-t^{2}} $$
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