Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 400: 1

Answer

a) $e^{-x}$ b) $e^{-x}$ c) $e^{-x}$

Work Step by Step

a) Here $y'=-e^{-x}$ Thus, $2y'+3y=-2 (e^{-x}) +3 (e^{-x})=e^{-x}$ b) Here, $y'=-e^{-x}-\dfrac{3}{2}e^{-\frac{3}{2}x}$ Now, $2y'+3y=-2 (e^{-x}) -2[\dfrac{3}{2}e^{-\frac{3}{2}x}]+3 e^{-x}+3e^{-\frac{3}{2}x}=e^{-x}$ c) Here, $y'=-e^{-x}-\dfrac{3}{2} Ce^{-\frac{3}{2}x}$ Then, $2y'+3y=-2 (e^{-x}) -2[\dfrac{3}{2}Ce^{-\frac{3}{2}x}]+3 (e^{-x})+3C(e^{-\frac{3}{2}x})=e^{-x}$
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