Question: $$y'' + 9y' = 9\sec^2(3t), 0<t<\frac{\pi}{6}$$
We first solve the homogeneous part: $$r^2+9=0$$,
where $$r=\pm3i$$,
so $$y_c(t) = c_1\cos{3t}+c_2\sin{3t}$$
Then we use wronskian to solve non-homoogeneous part:
\begin{align*}
w_1 &=
\begin{vmatrix}
0 & \sin{3t} \\
1 & 3\cos{3t} \\
\end{vmatrix} = -\sin{3t} \end{align*}
\begin{align*}
w_2 &=
\begin{vmatrix}
\cos{3t} & 0 \\
-3\sin{3t} & 1 \\
\end{vmatrix} = \cos{3t}
\end{align*}
\begin{align*}
w &=
\begin{vmatrix}
\cos{3t} & \sin{3t} \\
-3\sin{3t} & 3\cos{3t} \\
\end{vmatrix} = 3\cos^2{3t} + 3\sin^2{3t} = 3
\end{align*}
Subsititues above into formula we get:
$$\begin{align*}
Y_p(t) &= \cos{3t} \int{\frac{-\sin{3s}\times 9\sec^2{3s}}{3}ds} + \sin{3t} \int{\frac{\cos{3s}\times9\sec^2{3s}}{3}ds} \\
&= \cos{3t} \cdot -3\int{\sin{3s}\frac{1}{\cos^2{3s}}ds} + \sin{3t} \cdot 3\int{\cos{3s}\frac{1}{\cos^2{3s}}ds}\\
&= \cos{3t} \cdot -3\int{\tan{3s}\cdot\sec{3s}ds} + \sin{3t} \cdot 3\int{\sec{3s}ds}\\
&= \cos{3t} \cdot -\sec{3t} + \sin{3t} \cdot \ln{|\sec{3t}+\tan{3t}|} \\
&= -1 + \sin{3t} \ln{|\sec{3t}+\tan{3t}|}
\end{align*}$$
Finally, $$Y(t) = c_1\cos{3t}+c_2\sin{3t} + \sin{3t} \ln{|\sec{3t}+\tan{3t}|} -1 $$
