Answer
$\text{The general solution satisfies the differential equation.}$
$\text{The particular solution is $y = 2\sin {3x} - \frac{1}{3}\cos {3x}$ .}$
Work Step by Step
$\text{Let us find the second derivative of the general solution:}$
\begin{align}
& y = C_1 \sin {3x} + C_2 \cos {3x} \\
& y' = 3C_1 \cos {3x} - 3 C_2 \sin {3x} \\
& y'' = -9C_1 \sin {3x} - 9 C_2 \cos {3x}
\end{align}
$\text{Then, substitute $y''$ and y into the differential equation}$
\begin{align}
& y''+9y=0 \\
-9C_1 \sin {3x} - 9 C_2 &\cos {3x} + 9C_1 \sin {3x} + 9C_2 \cos {3x} = 0 \\
&0 = 0
\end{align}
$\text{Thus, the general solution satisfies the differential equation.}$
$\text{Now, we have to find the particular solution:}$
\begin{align}
& y = 2 \ \ when \ \ x =\frac{\pi}{6} \\
& 2 = C_1 \sin {\frac{\pi}{2}} + C_2 \cos {\frac{\pi}{2}} \Rrightarrow C_1 = 2
\end{align}
\begin{align}
& y' = 1 \ \ when x =\frac{\pi}{6} \\
1 = 3C_1 \cos {\frac{\pi}{2}}& - 3 C_2 \sin {\frac{\pi}{2}} = - 3C_2 \Rrightarrow C_2 = -\frac{1}{3}
\end{align}
$\text{Thus, the particular solution is}$
\begin{align}
y = 2\sin {3x} - \frac{1}{3}\cos {3x}
\end{align}
