Question

If $y\left(x\right)$ satisfies the differential equation $\frac{dy}{dx}=\sin 2x+3y\cot x$ and $y\left(\frac{\pi }{2}\right)=2,$ then which of the following statements is(are) CORRECT ?

• A. $y\left(\frac{\pi }{6}\right)=0$
• B. $y^{\prime }\left(\frac{\pi }{3}\right)=\frac{9-3\sqrt{2}}{2}$
• C. $y\left(x\right)$ strictly increases in interval $(\frac{\pi }{6},\frac{\pi }{3})$
• D. The value of definite integral $\underset{-\pi /2}{\overset{\pi /2}{\int }}y\left(x\right)dx$ equals $\pi$

Answer 1

Answer: (A), (C)

$y\left(x\right)$ strictly increases in interval $(\frac{\pi }{6},\frac{\pi }{3})$

$\frac{dy}{dx}=\sin 2x+3y\cot x$
$\Rightarrow \frac{dy}{dx}-3y\cot x=\sin 2x$
which is linear differential equation.
$I.F.=e^{-\int 3\left(\cot x\right)dx}=e^{-3\ln \left(\sin x\right)}=\frac{1}{{\sin }^{3}x}$
Now, the general solution is
$y\left(\frac{1}{{\sin }^{3}x}\right)=\int \frac{2\sin x\cdot \cos x}{{\sin }^{3}x}dx+C$
$\Rightarrow y\left(\frac{1}{{\sin }^{3}x}\right)=2\int \text{cosec}x\cot xdx+C$
$\Rightarrow y\left(\frac{1}{{\sin }^{3}x}\right)=-2\text{cosec}x+C$
Since $y\left(\frac{\pi }{2}\right)=2,$ therefore $C=4$
$\therefore y=4{\sin }^{3}x-2{\sin }^{2}x$

$y\left(\frac{\pi }{6}\right)=4{\left(\frac{1}{2}\right)}^3-2{\left(\frac{1}{2}\right)}^2$
$\Rightarrow y\left(\frac{\pi }{6}\right)=\frac{1}{2}-\frac{1}{2}=0$


$y^{\prime }\left(x\right)=12{\sin }^{2}x\cos x-4\sin x\cos x$
$\Rightarrow y^{\prime }\left(\frac{\pi }{3}\right)=\frac{9}{2}-\frac{2\sqrt{3}}{2}=\left(\frac{9-2\sqrt{3}}{2}\right)$


We have, $y^{\prime }\left(x\right)=4\sin x\cos x(3\sin x-1)$
$\Rightarrow y^{\prime }\left(x\right)=2\sin 2x(3\sin x-1)$
$\therefore y^{\prime }\left(x\right)>0$ for interval $(\frac{\pi }{6},\frac{\pi }{3})$
Hence, $y\left(x\right)$ strictly increases in interval $(\frac{\pi }{6},\frac{\pi }{3})$


$\underset{-\pi /2}{\overset{\pi /2}{\int }}(4{\sin }^{3}x-2{\sin }^{2}x)dx$
$=0-4\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}xdx$
$=-2\underset{0}{\overset{\pi /2}{\int }}\left(1-\cos 2x\right)dx$
$=-2{[x-\frac{\sin 2x}{2}]}_0^{\pi /2}=-\pi$