Question

Let \(y=y(x)\) be the solution of the differential equation

\(\cos x\,(3\sin x+\cos x+3)\,dy=\left(1+y\sin x\,(3\sin x+\cos x+3)\right)dx,\ 0\le x\le\frac{\pi}{2},\ y(0)=0\)

. Then,

\(y\left(\frac{\pi}{3}\right)\)

is equal to :

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

Hint:

Let \(g=3\sin x+\cos x+3\). The equation is \(g\,(\cos x\,dy-y\sin x\,dx)=dx\), i.e. \(d(y\cos x)=\frac{dx}{g}\). Integrate with \(u=\tan\frac{x}{2}\).

Answer 1

The Correct Option is C

Let \(g=3\sin x+\cos x+3\). Then \(g\cos x\,dy-g\,y\sin x\,dx=dx\), i.e. \(g\,(\cos x\,dy-y\sin x\,dx)=dx\), i.e. \(d(y\cos x)=\frac{dx}{g}\).
So \(y\cos x=\int_0^x\frac{dt}{3\sin t+\cos t+3}\) (using \(y(0)=0\)).
Put \(u=\tan\frac{t}{2}\): \(\sin t=\frac{2u}{1+u^2}\), \(\cos t=\frac{1-u^2}{1+u^2}\), \(dt=\frac{2du}{1+u^2}\). Then \(g=\frac{6u+1-u^2+3+3u^2}{1+u^2}=\frac{2(u+1)(u+2)}{1+u^2}\).
\(\frac{dt}{g}=\frac{du}{(u+1)(u+2)}=\left(\frac{1}{u+1}-\frac{1}{u+2}\right)du\).
\(y\cos x=\left[\ln\frac{u+1}{u+2}\right]_0^{\tan(x/2)}=\ln\frac{2(\tan\frac{x}{2}+1)}{\tan\frac{x}{2}+2}\).
At \(x=\frac{\pi}{3}\): \(u=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\), \(\frac{2(u+1)}{u+2}=\frac{2(1+\sqrt{3})}{1+2\sqrt{3}}=\frac{2(1+\sqrt{3})(2\sqrt{3}-1)}{11}=\frac{2(5+\sqrt{3})}{11}=\frac{2\sqrt{3}+10}{11}\).
With \(\cos\frac{\pi}{3}=\frac{1}{2}\): \(\frac{y}{2}=\ln\frac{2\sqrt{3}+10}{11}\), so \(y\left(\frac{\pi}{3}\right)=2\log_e\left(\frac{2\sqrt{3}+10}{11}\right)\).
Answer: (3) \(2\log_e\left(\frac{2\sqrt{3}+10}{11}\right)\)