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: 21


$$y= \frac{1}{2} +e^{-x}-\frac{11}{2} e^{-2x} .$$

Work Step by Step

This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ \int 2 dx}=e^{2x}.$$ Now the general solution is \begin{align} y& =\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ & =e^{-2x}\left( \int e^{2x}(1+e^{-x})dx+C\right)\\ & =e^{-2x}\left( \frac{1}{2} e^{2x}+ e^{x}+C\right)\\ & =\frac{1}{2} +e^{-x}+C e^{-2x} \end{align} Since, $y(0)=-4$, then $C=-\frac{11}{2}$ Thus, the general solution is: $$y= \frac{1}{2} +e^{-x}-\frac{11}{2} e^{-2x} .$$
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