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: 18


$$y= \left( - t^2-2( t+1 )+6e^t\right).$$

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}t^2dt+C\right)\\ & =e^t\left( -e^{-t}t^2-2( e^{-t}t+ e^{-t})+C\right)\\ & = \left( - t^2-2( t+1 )+Ce^t\right) \end{align} where we used integration by parts. Since $y(0)=4$, then $4=-2+C$ and hence $C=6$. Thus, the general solution is: $$y= \left( - t^2-2( t+1 )+6e^t\right).$$
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