Calculus 10th Edition

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

Chapter 6 - Differential Equations - 6.4 Exercises - Page 428: 15

Answer

$$y = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right)$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = {e^x} - y,{\text{ }}\left( {0,1} \right) \cr & \frac{{dy}}{{dx}} + y = {e^x} \cr & {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{With }}P\left( x \right) = 1{\text{ and }}Q\left( x \right) = {e^x} \cr & {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {dx} }} = {e^x} \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & {e^x}\left( {\frac{{dy}}{{dx}} + y} \right) = {e^x}\left( {{e^x}} \right) \cr & {e^x}\frac{{dy}}{{dx}} + {e^x}y = {e^{2x}} \cr & {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr & \frac{d}{{dx}}\left[ {{e^x}y} \right] = {e^{2x}} \cr & d\left[ {{e^x}y} \right] = \left( {{e^{2x}}} \right)dx \cr & {\text{Integrate both sides}} \cr & {e^x}y = \int {{e^{2x}}} dx \cr & {e^x}y = \frac{1}{2}{e^{2x}} + C \cr & {\text{Solve for }}y \cr & y = \frac{1}{{2{e^x}}}{e^{2x}} + \frac{C}{{{e^x}}} \cr & y = \frac{{{e^x}}}{2} + \frac{C}{{{e^x}}} \cr & y = \frac{{{e^x}}}{2} + C{e^{ - x}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}\left( {0,1} \right) \cr & 1 = \frac{{{e^0}}}{2} + C{e^{ - 0}} \cr & 1 = \frac{1}{2} + C \cr & C = \frac{1}{2} \cr & {\text{Substituting the initial condition into }}\left( {\bf{1}} \right) \cr & y = \frac{{{e^x}}}{2} + \frac{1}{2}{e^{ - x}} \cr & {\text{Factoring}} \cr & y = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right) \cr & \cr & {\text{Graph}} \cr} $$
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