Answer
See below
Work Step by Step
Given: $y''-36y=0$
Substitute: $y''(x)-36y(x)\\
=r^2e^{rx}-36e^{rx}\\
=e^{rx}(r^2-36)\\
=0$
Since $e^{rx} \ne0 \rightarrow r^2-36=0\\
\rightarrow (r+6)(r-6)=0\\
\rightarrow r=-6, r=6$
The solutions to the given problem are: $y_1(x)=e^{-6x}\\
y_2(x)=e^{6x}$
Obtain: $W_{[y_1,y_2]}(x)=\begin{vmatrix}
e^{-6x} & e^{6x}\\
-6e^{-6x}& 6e^{6x}
\end{vmatrix}=6-(-6)=12$
Since $12 \ne0$, $y_1,y_2$ are linearly independent solutions to the equation.
Hence, the solutions is $y(x)=C_1e^{-6x}+C_2e^{6x}$
