$\text{ A function }\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\text{ satisfies the differential equation }$$\mathrm{f}\left(\mathrm{x}\right)\sin 2\mathrm{x}-\cos \mathrm{x}+\left(1+{\sin }^2\mathrm{x}\right){\mathrm{f}}^1\left(\mathrm{x}\right)=0\text{ with initial condition }\mathrm{y}\left(0\right)=0\text{ then the value of }$

$5\mathrm{f}(\pi /6)\text{ is }$

$y\sin 2x-\cos x+(1+{\sin }^{2}x)\frac{dy}{dx}=0$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\sin 2\mathrm{x}}{1+{\sin }^2\mathrm{x}}\right)\mathrm{y}=\frac{\cos \mathrm{x}}{1+{\sin }^2\mathrm{x}}\\ \mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \frac{\sin 2\mathrm{x}}{1+{\sin }^{2}x}\mathrm{d}x}={\mathrm{e}}^{\log \left|1+{\sin }^{2}x\right|}\\ =1+{\sin }^{2}x\\ \text{ Solution of the differential equation }y\left(1+{\sin }^{2}x\right)=\int \cos xdx=\sin x+c\\ \text{ Given y}\left(0\right)=0\Rightarrow c=0\\ \Rightarrow \mathrm{y}=\frac{\sin \mathrm{x}}{1+{\sin }^2\mathrm{x}}\\ \mathrm{y}\left(\frac{\pi }{6}\right)=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{5}{4}}}=\frac{2}{5}\\ \Rightarrow 5\mathrm{y}\left(\frac{\pi }{6}\right)=2\end{array}$