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. $y\left(x\right)$ increases in interval $(\frac{\pi }{6},\frac{\pi }{3})$
• D. The value of definite integral $\underset{-\pi /2}{\overset{\pi /2}{\int }}y\left(x\right)dx$ is equal to $\pi$

Answer 1

Answer: (A), (C)

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

Given,$\frac{dy}{dx}=\sin 2x+3y\cot x$ and $y\left(\frac{\pi }{2}\right)=2$
$\Rightarrow \frac{dy}{dx}-3y\cot x=\sin 2x$
This is a linear differential equation.
$\therefore I.F.=e^{-\int 3\cot xdx}=e^{-3\ln |\sin x|}=\frac{1}{{\sin }^{3}x}$

$\therefore$ General solution is
$y\left(\frac{1}{{\sin }^{3}x}\right)=\int \frac{2\sin x\cdot \cos x}{{\sin }^{3}x}dx+C\Rightarrow \frac{y}{{\sin }^{3}x}=2\int \text{cosec}x\cot xdx+C$
$\Rightarrow \frac{y}{{\sin }^{3}x}=-2\text{cosec}x+C$

Now, $y\left(\frac{\pi }{2}\right)=2$
$\Rightarrow \frac{2}{1}=-2+C\Rightarrow C=4$
$\therefore y=4{\sin }^{3}x-2{\sin }^{2}x$

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

$y^{\prime }\left(x\right)=12{\sin }^{2}x\cos x-4\sin x\cos x$
$\Rightarrow y^{\prime }\left(\frac{\pi }{3}\right)=(\frac{12\times 3}{4}\times \frac{1}{2}-4\times \frac{\sqrt{3}}{2}\times \frac{1}{2})\Rightarrow y^{\prime }\left(\frac{\pi }{3}\right)=\left(\frac{9-2\sqrt{3}}{2}\right)$

$\Rightarrow y^{\prime }\left(x\right)=2\sin 2x(3\sin x-1)$
Now, $y^{\prime }\left(x\right)=0$ gives
$\sin 2x=0\text{or}3\sin x=1$
$\Rightarrow x=\frac{\pi }{2}\text{or}x={\sin }^{-1}\left(\frac{1}{3}\right)<\frac{\pi }{6}$

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

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}y\left(x\right)dx$
$=\underset{-\pi /2}{\overset{\pi /2}{\int }}(4{\sin }^{3}x-2{\sin }^{2}x)dx$
$=-2\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{2}xdx(\because {\sin }^{3}x\text{is an odd function})$
$=-4\underset{0}{\overset{\pi /2}{\int }}\left(\frac{1-\cos 2x}{2}\right)dx$
$=-\pi$