Given differential equation,
$\sin x \frac{d y}{d x}+y \cos x=e^{2 x}, x \in(0, \pi)$
or $\quad \frac{d y}{d x}+y \cot x=\frac{e^{2 x}}{\sin x}$
This is linear differential equation and general form is given by
$\frac{d y}{d x}+P(x) y=Q(x)
Here $P(x)=\cot x, Q(x)=\frac{e^{2 x}}{\sin x}$
Integrating factor, IF $=e^{\int P(x) d x}=e^{\int \cot x d x}=e^{\ln \sin x}$
$\mathrm{IF}=\sin x$
Solution of linear differential equation
$y(\mathrm{IF})=\int Q(x) \operatorname{IF} d x$
$y(\sin x)=\int \frac{e^{2 x}}{\sin x} \cdot \sin x d x$
$y \sin x=\int e^{2 x} d x \Rightarrow y \sin x=\frac{e^{2 x}}{2}+C \quad \ldots (i)$ $\{$ where, $C=$ integrating constant $\}$
Given that, $y\left(\frac{\pi}{2}\right)=0$
$\therefore \quad 0 \times \sin \frac{\pi}{2}=\frac{e^{2 \times \pi / 2}}{2}+C \Rightarrow C=-\frac{e^\pi}{2}$
Putting the value of $c$ in Eq. (i), we get
$y \sin x=\frac{e^{2 x}}{2}-\frac{e^\pi}{2}$
At, $x=\frac{\pi}{6}, y\left(\frac{\pi}{6}\right) \sin \frac{\pi}{6}=\frac{e^{2 \times \frac{\pi}{6}}}{2}-\frac{e^\pi}{2}$
$y\left(\frac{\pi}{6}\right) \times \frac{1}{2}=\frac{1}{2}\left(e^{\pi / 3}-e^\pi\right) \Rightarrow y\left(\frac{\pi}{6}\right)=e^{\pi / 3}-e^\pi$