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 4 - Higher Order Linear Equations - 4.4 The Method of Variation of Parameters - Problems - Page 242: 4

Answer

$y=C_{1}+C_{2}cos(t)+C_{3}sin(t)+ln|sec(t)+tan(t)|-tcos(t)+(sin(t))ln|cos(t)|$

Work Step by Step

Let $\;\;\;\;\;y=e^{rt}\\\\$ ${y}'''+{y}'=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}+re^{rt}=0\\\\$ $r^3+r=r(r^2+1)=0 $ $ \rightarrow\;\;\;\;\; r_{1}=0\;\;\;\;\;\;\;or\;\;\;\;,r_{2,3}=\pm i\;\;\;\;\;\;\;\;\;\\\\$ $\boxed{y_{c}(t)= C_{1}+C_{2}cos(t)+C_{3}sin(t)}$ $y_{1}=1\;\;\;\;\;,\;\;y_{2}=cos(t)\;\;\;\;\;\;\;,\;\;\;y_{3}=sin(t)$ $W(1,e^t,e^{-t})=\begin{vmatrix} 1 & cos(t) & sin(t) \\ 0 & -sin(t) & -cos(t) \\ 0 & -cos(t) & -sin(t) \end{vmatrix}\;=\;1$ $W_{1}=\begin{vmatrix} 0 & cos(t) & sin(t) \\ 0 & -sin(t) & -cos(t) \\ sec(t) & -cos(t) & -sin(t) \end{vmatrix}\;=\;sec(t)$ $W_{2}=\begin{vmatrix} 1 & 0 & sin(t) \\ 0 & 0 & -cos(t) \\ 0 & sec(t) & -sin(t) \end{vmatrix}\;=\;-1$ $W_{3}=\begin{vmatrix} 1 & cos(t) & 0 \\ 0 & -sin(t) & 0 \\ 0 & -cos(t) & sec(t) \end{vmatrix}\;=\;-tan(t)$ ${u}'_{1}=\frac{w_{1}}{w}=sec(t)$ $u_(1)=\int sec(t)= ln|sec(t)+tan(t)|$ ${u}'_{2}=\frac{w_{2}}{w}=-1$ $u_{2}=\int -1= -t$ ${u}'_{3}=\frac{w_{3}}{w}=-tan(t)$ $u_{3}=\int -tan(t)= ln|cos(t)|$ $y_{p}=y_{1}u_{1}+y_{2}u_{2}+y_{3}u_{3}$ $y_{p}=ln|sec(t)+tan(t)|-tcos(t)+(sin(t))ln|cos(t)|$ $y=y_{c}+y_{p}$ $y=C_{1}+C_{2}cos(t)+C_{3}sin(t)+ln|sec(t)+tan(t)|-tcos(t)+(sin(t))ln|cos(t)|$
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