Let $y=y\left(x\right)$ be the solution of the differential equation $\cos x(3\sin x+\cos x+3)dy=(1+y\sin x(3\sin x+\cos x+3\left)\right)dx, 0\le x\le \frac{\mathrm{\pi }}{2}, y\left(0\right)=0.$ Then, $y\left(\frac{\pi }{3}\right)$ is equal to:

Given $\cos x(3\sin x+\cos x+3)dy=(1+y\sin x(3\sin x+\cos x+3\left)\right)dx$

$\Rightarrow \frac{dy}{dx}=\frac{(1+y\sin x(3\sin x+\cos x+3\left)\right)}{\cos x(3\sin x+\cos x+3)}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{\cos x(3\sin x+\cos x+3)}+\frac{y\sin x(3\sin x+\cos x+3))}{\cos x(3\sin x+\cos x+3)}$

$\Rightarrow \frac{dy}{dx}-\left(\tan x\right)y=\frac{1}{(3\sin x+\cos x+3)\cos x}$

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$ where $P=-\tan x$ and $Q=\frac{1}{\cos x\left(3\sin x+\cos x+3\right)}.$

Now, the integrating factor $\mathrm{I}.\mathrm{F}.=e^{\int Pdx}=e^{\int -\mathrm{tanxdx}}$

$={\mathrm{e}}^{\ln \left|\mathrm{cosx}\right|}=\left|\cos x\right|$

$=\cos x \forall 0\le x\le \frac{\pi }{2}.$

The solution of the given differential equation is $y\left(I.F.\right)=\int Q\left(I.F.\right)dx+C$

$\Rightarrow y\left(\cos x\right)=\int \left(\cos x\right)·\frac{1}{\cos x(3\sin x+\cos x+3)}dx+C$

$\Rightarrow y\left(\cos x\right)=\int \frac{dx}{3\sin x+\cos x+3}dx+C$

Using $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta } \& \cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$\Rightarrow y\left(\cos x\right)=\int \frac{1}{3\times {\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}+{\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}+3}dx+C$

$\Rightarrow y\left(\cos x\right)=\int \frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{{\displaystyle 6\tan \frac{{\displaystyle x}}{{\displaystyle 2}}}+{\displaystyle 1-{\tan }^2\frac{{\displaystyle x}}{{\displaystyle 2}}}+3+3{\tan }^2\frac{x}{2}}dx+C$

$\Rightarrow y\left(\cos x\right)=\int \frac{\left({\sec }^2{\displaystyle \frac{x}{2}}\right)}{2{\tan }^2{\displaystyle \frac{x}{2}}+6\tan {\displaystyle \frac{x}{2}}+4}dx+C$

Let $I_1=\int \frac{\left({\sec }^2{\displaystyle \frac{x}{2}}\right)}{2\left({\tan }^2{\displaystyle \frac{x}{2}}+3\tan {\displaystyle \frac{x}{2}}+2\right)}dx+C$

Put $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$

$\Rightarrow I_1=\int \frac{dt}{t^3+3t+2}=\int \frac{dt}{\left(t+2\right)(t+1)}$

$\Rightarrow I_1=\int \left(\frac{1}{t+1}-\frac{1}{t+2}\right)dt$

$\Rightarrow I_1={\log }_e\left(t+1\right)-{\log }_e\left(t+2\right)$

$\Rightarrow I_1={\log }_e\left(\frac{t+1}{t+2}\right)$

$\Rightarrow I_1={\log }_e\left|\left(\frac{\tan {\displaystyle \frac{x}{2}}+1}{\tan {\displaystyle \frac{x}{2}}+2}\right)\right|$

So, the solution is $y\left(\cos x\right)={\log }_e\left|\frac{1+\tan {\displaystyle \frac{x}{2}}}{2+\tan {\displaystyle \frac{x}{2}}}\right|+C$

$\Rightarrow y\left(\cos x\right)={\log }_e\left(\frac{1+\tan {\displaystyle \frac{x}{2}}}{2+\tan {\displaystyle \frac{x}{2}}}\right)+C$ for $0\le x\le \frac{\pi }{2}.$

Now, it is given $y\left(0\right)=0$

$\Rightarrow 0={\log }_e\left(\frac{1}{2}\right)+\mathrm{C} \Rightarrow \mathrm{C}={\log }_{e}2$

$\Rightarrow y\left(\cos x\right)={\log }_e\left(\frac{1+\tan {\displaystyle \frac{x}{2}}}{2+\tan {\displaystyle \frac{x}{2}}}\right)+{\log }_{e}2$

Now, for $x=\frac{\pi }{3},$ $y\left(\frac{1}{2}\right)={\log }_e\left(\frac{1+{\displaystyle \frac{1}{\sqrt{3}}}}{2+{\displaystyle \frac{1}{\sqrt{3}}}}\right)+{\log }_{e}2$

$\Rightarrow y·\left(\frac{1}{2}\right)={\log }_e\left(\frac{{\displaystyle \sqrt{3}+1}}{{\displaystyle 2\sqrt{3}+1}}\right)+{\log }_{e}2$

$\Rightarrow y·\left(\frac{1}{2}\right)={\log }_e\left(\left(\frac{{\displaystyle \sqrt{3}+1}}{{\displaystyle 2\sqrt{3}+1}}\right)\times \left(\frac{2\sqrt{3}-1}{2\sqrt{3}-1}\right)\right)+{\log }_{e}2$

$\Rightarrow y·\left(\frac{1}{2}\right)={\log }_e\left(\frac{{\displaystyle 5+\sqrt{3}}}{{\displaystyle 11}}\right)+{\log }_{e}2$

$\Rightarrow y=2{\log }_e\left(\frac{2\sqrt{3}+10}{11}\right).$