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: 6

Answer

The solution of the given differential equation $$ \frac{d x}{d t}=1-t+x-t x $$ is $$ x=-1+K e^{t-t^{2} / 2} $$ where $K$ is any nonzero constant.

Work Step by Step

$$ \frac{d x}{d t}=1-t+x-t x $$ $$ \begin{split} \frac{d x}{d t} &=1-t+x-t x\\ &=1(1-t)+x(1-t) \\ & = (1+x)(1-t) \end{split} $$ $ \Rightarrow $ $$ \frac{d x}{1+x}=(1-t) d t \quad \Rightarrow \quad \int \frac{d x}{1+x}=\int(1-t) d t $$ integrate both sides to get: $$ \ln |1+x|=t-\frac{1}{2} t^{2}+C \Rightarrow |1+x|=e^{t-t^{2} / 2+C} $$ where $ C $ is an arbitrary constant. this implies that $$ 1+x=\pm e^{t-t^{2} / 2} \cdot e^{C} \Rightarrow x=-1+K e^{t-t^{2} / 2} $$ where $K=\pm e^{C}$ is an arbitrary constant. So the solution of the differential equation $$ \frac{d x}{d t}=1-t+x-t x $$ is $$ x=-1+K e^{t-t^{2} / 2} $$ where $K$ is any nonzero constant.
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