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: 20


$$y= 3 t+3+2e^{t+1} .$$

Work Step by Step

This is a linear equation and has the integrating factor as follows $$\alpha(t)= e^{\int P(t)dt}=e^{- \int dt}=e^{-t}.$$ Now the general solution is \begin{align} y& =\alpha^{-1}(t)\left( \int\alpha(t) Q(t)dt +C\right)\\ & =e^t\left( \int e^{-t}(-3t)dt+C\right)\\ & = e^t\left( 3e^{-t}t+3 e^{-t} +C\right)\\ & =3 t+3+Ce^{t} \end{align} where we used integration by parts. Since, $y(-1)=2$, then $C=2e$ Thus, the general solution is: $$y= 3 t+3+2e^{t+1} .$$
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