Answer
See below
Work Step by Step
Given: $y'''-3y''-y'+3y=0$
Substitute: $y'''(x)-3y''-y'(x)+3y(x)\\
=r^3e^{rx}-3r^2e^{rx}-re^{rx}+3e^{rx}\\
=e^{rx}(r^3-3r^2-r+3)\\
=0$
Since $e^{rx} \ne0 \rightarrow r^3-3r^2-r+3=0\\
\rightarrow (r-1)(r+1)(r-3)=0\\
\rightarrow r=-3, r=-1,r=1$
The solutions to the given problem are: $y_1(x)=e^{-3x} \\
y_2(x)=e^{-x} \\
y_3(x)=e^x$
Obtain: $W_{[y_1,y_2]}(x)=\begin{vmatrix}
e^{-3x} & e^{-x} & e^x\\
-3e^{-3x} & -e^{-x} & e^x \\
9e^{-3x} & e^{-x} & e^x
\end{vmatrix}=e^{-3x}\begin{vmatrix}
-e^{-x} & e^{x}\\
e^{-x} & e^{x}
\end{vmatrix} - e^{-x}\begin{vmatrix}
-3e^{-3x} & e^{x}\\
9e^{-3x} & e^{x}
\end{vmatrix} + e^{x}\begin{vmatrix}
-3e^{-3x} & -e^{-x}\\
9e^{-3x} & e^{-x}
\end{vmatrix}=-2e^{-3x}+12e^{-3x}+6e^{-3x}=16e^{-3x}$
Since $16e^{-3x} \ne0$, $y_1,y_2,y_3$ are linearly independent solutions to the equation.
Hence, the solutions is $y(x)=C_1e^{-3x}+C_2e^{-x}+C_3e^x$
