Answer
See below
Work Step by Step
Given: $y'''-y''-2y'=0$
Substitute: $y'''(x)-y''(x)-2y'(x)\\
=r^3e^{rx}-r^2e^{rx}-2re^{rx}\\
=e^{rx}(r^3-r^2-2r)\\
=0$
Since $e^{rx} \ne0 \rightarrow r^3-r^2-2r=0\\
\rightarrow r(r-2)(r+1)=0\\
\rightarrow r=0, r=2,r=-1$
The solutions to the given problem are: $y_1(x)=1\\
y_2(x)=e^{2x} \\
y_3(x)=e^{-x}$
Obtain: $W_{[y_1,y_2]}(x)=\begin{vmatrix}
1 & e^{2x} & e^{-x}\\
0 & 2e^{2x} & -e^{-x} \\
0 & 4e^{2x} & e^{-x}
\end{vmatrix}=\begin{vmatrix}
2e^{2x} & -e^{-x}\\
4e^{2x} & e^{-x}
\end{vmatrix} +0\begin{vmatrix}
2e^{2x} & e^{-x}\\
4e^{2x} & e^{-x}
\end{vmatrix} + 0\begin{vmatrix}
e^{2x} & e^{-x}\\
2e^{2x} & -e^{-x}
\end{vmatrix}=2e^{x}+4e^{x}=6e^{x}$
Since $6e^{x} \ne0$, $y_1,y_2,y_3$ are linearly independent solutions to the equation.
Hence, the solutions is $y(x)=C_1+C_2e^{2x}+C_3e^{-x}$
