Question:Find the general solution of the given differential equation.
y′′+9y=9sec2(3t) 0<t<π/6
Answer:
$r^{2}+9=0 \quad r^{2}=-9 \quad r=\pm 3 i$
$y_{C}(t)=C_{1} \cos (3 t)+C_{2} \sin (3 t)$
$y^{\prime \prime}+9 y=9 \sec ^{2}(3 t)$
$w\left(y_1, y_{2}\right)=\left|\begin{array}{ll}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_2^{\prime}(t)}\end{array}\right|=\left|\begin{array}{cc}{\cos (3 t)} & {\sin (3 t)} \\ {-3 \sin (3 t)} & {3 \cos (3 t)}\end{array}\right|=3 \cos ^{2}(3 t)+3 \sin ^{2}(3 t)=3$
$w_{1}=\left|\begin{array}{ll}{0} & {\sin (3t)} \\ {1} & {3 \cos (3t)}\end{array}\right|=-\sin (3t) \quad \quad \quad w_{2}=\left|\begin{array}{cc}{\cos (3t)} & {0} \\ {-3 \sin (3t)} & {1}\end{array}\right|=\cos (3t)$
$y_{p}(t)=y_{1}(t) \underbrace{\int \frac{q(s)w_1(s)}{w(s)} d s}_{u_{1}(s)}+y_{2}(t) \underbrace{\frac{q(s) \cdot w_{2}(s)}{w(s)} d s}_{u_{2}(s)}$
$\begin{aligned} u_{1}(s) &=-\int \frac{9 \sec ^{2}(3 s) \sin (3 s)}{3} d s \\ &=-3 \int \frac{\sin (3 s)}{\cos ^{2}(3 s)} d s \\ &=-3 \int \sec (3 s)\tan(3s) d s \\ &=-\sec 3 s \end{aligned}$
$\begin{aligned} u_{2}(s) &=\int \frac{9 \cos (3 s) \cdot \sec ^{2}(3s)}{3} d s \\ &=\int 3 \frac{\cos (3s)}{\cos ^{2}(3s)} d s \\ &=\int 3 \sec (3 s) d s \\ &=\ln|\sec (3s)+\tan (3s)|\end{aligned}$
$\begin{aligned} y_{p}(t) &=\cos (3 t)(-\sec (3t))+\sin (3t) \cdot \ln |\sec (3 t)+\tan (3 t)| \\ &=\sin (3 t)[\ln |\sec (3 t)+\tan (3 t)|]-1 \end{aligned}$
$ y(t)=y_{p}(t)+y_{C}(t)= C_{1} \cos (3 t)+C_{2} \sin (3 t)+ \sin (3 t)[\ln |\sec (3 t)+\tan (3 t)|]-1$
