Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 8 - Mathematical Modeling With Differential Equations - 8.1 Modeling With Differential Equations - Exercises Set 8.1 - Page 566: 4

Answer

$$\eqalign{ & \left( {\text{a}} \right){\text{First order}} \cr & \left( {\text{b}} \right){\text{Second order}} \cr} $$

Work Step by Step

$$\eqalign{ & \left( {\text{a}} \right)2\frac{{dy}}{{dx}} + y = x - 1;{\text{ }}y = c{e^{ - x/2}} + x - 3 \cr & {\text{The differential equation }}2\frac{{dy}}{{dx}} + y = x - 1{\text{ is the first order}} \cr & {\text{Checking}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {c{e^{ - x/2}} + x - 3} \right] \cr & \frac{{dy}}{{dx}} = - \frac{c}{2}{e^{ - x/2}} + 1,{\text{ then}} \cr & 2\left( { - \frac{c}{2}{e^{ - x/2}} + 1} \right) + c{e^{ - x/2}} + x - 3 = x - 1 \cr & - c{e^{ - x/2}} + 2 + c{e^{ - x/2}} + x - 3 = x - 1 \cr & x - 1 = x - 1 \cr & y = c{e^{ - x/2}} + x - 3{\text{ is a solution}}{\text{.}} \cr & \cr & \left( {\text{b}} \right)y'' - y = 0;{\text{ }}y = {c_1}{e^t} + {c_2}{e^{ - t}} \cr & {\text{The differential equation }}y'' - y = 0{\text{ is the second order}} \cr & {\text{Checking}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{c_1}{e^t} + {c_2}{e^{ - t}}} \right] \cr & \frac{{dy}}{{dx}} = {c_1}{e^t} - {c_2}{e^{ - t}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = {c_1}{e^t} + {c_2}{e^{ - t}},{\text{ then}} \cr & {c_1}{e^t} + {c_2}{e^{ - t}} - {c_1}{e^t} - {c_2}{e^{ - t}} = 0 \cr & 0 = 0 \cr & y = {c_1}{e^t} + {c_2}{e^{ - t}}{\text{ is a solution}}{\text{.}} \cr} $$
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