Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 2 - First Order Differential Equations - 2.1 Linear Equations; Method of Integrating Factors - Problems - Page 39: 5

Answer

$y=c e^{2 t}-3 e^{t}$

Work Step by Step

$y^{\prime}-2 y=3 e^{t}$ We solve this by integrating factor method. First, find the integrating factor. $p(t)=-2$ $\mu(t)=e^{\int p(t)} d t$ $\mu(t)=e^{\int -2dt}$ $\mu(t)=e^{-2 t}$ Multiplying both sides with $\mu(t)$ $$ \begin{array}{c} e^{-2 t}\left(y^{\prime}-2 y\right)=3 \cdot e^{-2 t} \cdot e^{t} \\ y^{\prime} e^{-2 t}-2 e^{-2 t} y=3 e^{-t} \end{array} $$ simplifying the left hand side $$\left(y e^{-2 t}\right)^{\prime}=3 e^{-t}$$ integrating on both sides. $\int\left(y e^{-2 t}\right)^{\prime} d t=\int 3 e^{-t} d t$ $y e^{-2 t}=-3 e^{-t}+c$ Further dividing both sides by $e^{-2 t}$ $ \quad y=-3 e^{-t+2 t}+c e^{2 t}$ $y=-3 e^{t}+c e^{2 t}$ $y=c e^{2 t}-3 e^{t}$
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