Question

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

• A. $2{\log }_e\left(\frac{2\sqrt{3}+10}{11}\right)$
• B. $2{\log }_e\left(\frac{\sqrt{3}+7}{2}\right)$
• C. $2{\log }_e\left(\frac{3\sqrt{3}-8}{4}\right)$
• D. $2{\log }_e\left(\frac{2\sqrt{3}+9}{6}\right)$

Answer 1

The correct option is A $2{\log }_e\left(\frac{2\sqrt{3}+10}{11}\right)$

$\cos x(3\sin x+\cos x+3)dy=(1+y\sin x(3\sin x+\cos x+3))dx\cdots \left(1\right)$
$\Rightarrow (3\sin x+\cos x+3)(\cos xdy-y\sin xdx)=dx$
$\Rightarrow \int d(y\cdot \cos x)=\int \frac{dx}{3\sin x+\cos x+3}$
$\Rightarrow y\cos x=\int \frac{dx}{3\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+3}$
$\Rightarrow y\cos x=\int \frac{{\sec }^2\frac{x}{2}dx}{6\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}+3+3{\tan }^2\frac{x}{2}}$
$\Rightarrow y\cos x=\int \frac{{\sec }^2\frac{x}{2}dx}{2{\tan }^2\frac{x}{2}+6\tan \frac{x}{2}+4}$
$=\int \frac{\frac{1}{2}{\sec }^2\frac{x}{2}dx}{{\tan }^2\frac{x}{2}+3\tan \frac{x}{2}+2}$
$\Rightarrow y\cos x=\ln \left|\frac{\tan \frac{x}{2}+1}{\tan \frac{x}{2}+2}\right|+C$
$\because y\left(0\right)=0$
$\Rightarrow C=-\ln \left(\frac{1}{2}\right)=\ln \left(2\right)$
Now, $y\left(\frac{\pi }{3}\right)=2\ln \left|\frac{1+\sqrt{3}}{1+2\sqrt{3}}\right|+\ln 2$
$=2\ln \left|\frac{5+\sqrt{3}}{11}\right|+\ln 2$
$=2\ln \left|\frac{2\sqrt{3}+10}{11}\right|$