Answer
See below
Work Step by Step
Given: $y'''+3y''-4y'-12y=0$
Substitute: $y'''(x)+3y''(x)-4y'(x)-12y(x)\\
=r^3e^{rx}+3r^2e^{rx}-4re^{rx}-12e^{rx}\\
=e^{rx}(r^3+3r^2-4r-12)\\
=0$
Since $e^{rx} \ne0 \rightarrow r^3+3r^2-4r-12=0\\
\rightarrow (r-1)(r+1)(r-3)=0\\
\rightarrow r=-3, r=-2,r=2$
The solutions to the given problem are: $y_1(x)=e^{-3x} \\
y_2(x)=e^{-2x} \\
y_3(x)=e^{2x}$
Obtain: $W_{[y_1,y_2]}(x)=\begin{vmatrix}
e^{-3x} & e^{-2x} & e^{2x}\\
-3e^{-3x} & -2e^{-2x} & 2e^{2x} \\
9e^{-3x} & 4e^{-2x} & 4e^{2x}
\end{vmatrix}=e^{-3x}\begin{vmatrix}
-2e^{-2x} & 2e^{2x}\\
2e^{-2x} & 2e^{2x}
\end{vmatrix} - e^{-2x}\begin{vmatrix}
-3e^{-3x} & 2e^{2x}\\
9e^{-3x} & 4e^{2x}
\end{vmatrix} + e^{2x}\begin{vmatrix}
-3e^{-3x} & -2e^{-2x}\\
9e^{-3x} & 4e^{-2x}
\end{vmatrix}=-8e^{-3x}+30e^{-3x}+6e^{-3x}=28e^{-3x}$
Since $28e^{-3x} \ne0$, $y_1,y_2,y_3$ are linearly independent solutions to the equation.
Hence, the solutions is $y(x)=C_1e^{-3x}+C_2e^{-2x}+C_3e^{2x}$
