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