Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley

Chapter 4 - Higher Order Linear Equations - 4.4 The Method of Variation of Parameters - Problems - Page 242: 2

Answer

$y=C_{1}+C_{2}e^t+C_{3}e^{-t}-\frac{1}{2}t^2$

Work Step by Step

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

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