## Elementary Differential Equations and Boundary Value Problems 9th Edition

$W(e^t,e^{-t},e^{-2t})(t)=-6e^{-2t}$
we verify the given functions are the solutions of the differential equation by plugging them into it: $y=e^t\;\;\;\rightarrow \;\;\;({e^t})'''+2({e^t})''-(e^t)'-2(e^t)=\;e^t+2e^t-e^t-2e^t=0\\\\$ $y=e^{-t}\;\;\;\rightarrow \;\;\;({e^{-t}})'''+2({e^{-t}})''-(e^t{-t})'-2(e^t{-t})=\;-e^t+2e^t+e^t-2e^t=0\\\\$ $y=e^{-2t}\;\;\;\rightarrow \;\;\;({e^{-2t}})'''+2({e^{-2t}})''-(e^t{-2t})'-2(e^t{-2t})=\;-8e^t+8e^t+2e^t-2e^t=0\\\\$ $W(e^t,e^{-t},e^{-2t})(t)=\begin{vmatrix} e^t & e^{-t} & e^{-2t}\\ ({e^t})' & ({e^{-t}})' & ({e^{-2t}})'\\ ({e^t})''&({e^{-t}})'' & ({e^{-2t}})'' \end{vmatrix}\;\;=\;\;\begin{vmatrix} e^t & e^{-t} & e^{-2t}\\ e^t & -e^{-t} & -2e^{-2t}\\ e^t & e^{-t} & 4e^{-2t} \end{vmatrix}\;\;=$ $e^t\begin{vmatrix} -e^{-t} & -2e^{-2t} \\ e^{-t} & 4e^{-2t} \end{vmatrix}\;\;-\;\;e^t\begin{vmatrix} e^{-t} & e^{-2t} \\ e^{-t} & 4e^{-2t} \end{vmatrix}\;\;+\;\;e^t\begin{vmatrix} e^{-t} & e^{-2t} \\ -e^{-t} & -2e^{-2t} \end{vmatrix}\;\;=$ $e^t(-4e^{-3t}+2e^{-3t})\;\;-\;e^t(4e^{-3t}-e^{-3t})\;\;+\;e^t(-2e^{-3t}+e^{-3t})\;=\\$ $-2e^{-2t}\;-\;6e^{-2t}\;+\;2e^{-2t}\;=\;-6e^{-2t}$ $W(e^t,e^{-t},e^{-2t})(t)=-6e^{-2t}$