Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.3 Exercises: 17

Answer

$$y=e^{-\frac{x^2}{2}-x}$$

Work Step by Step

$y(x+1)+y'=0$ $y'=-y(x+1)$ $\frac{y'}{y}=-(x+1)$ $\int\frac{1}{y}dy=-\int (x+1)dx$ $\ln{y}=-\frac{x^2}{2}-x+C$ $e^{\ln{y}}=y=e^{-\frac{x^2}{2}-x+C}=e^Ce^{-\frac{x^2}{2}-x}$ Let $k=e^C$ so we get $y=ke^{-\frac{x^2}{2}-x}$ Now applying our initial condition we have $1=ke^{-\frac{(-2)^2}{2}+2}=ke^0$ Therefore $k=1$ Our Particular solution then becomes $y=e^{-\frac{x^2}{2}-x}$
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