Question

If y(x) satisfies the differential equation $\frac{dy}{dx}=sin2x+3ycotx$ and $y\left(\frac{\pi }{2}\right)=2,$ then which of the following statement(s) 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(x) increases in interval $(\frac{\pi }{6},\frac{\pi }{3})$
• D. The value of definite integral ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}y\left(x\right)$ dx equals $\pi .$

Answer 1

Answer: (A), (C)

The correct options are
A

$y\left(\frac{\pi }{6}\right)=0$

C

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

$\frac{dy}{dx}-3ycotx=sin2x$

$\Rightarrow I.F.=e^{-\int 3cotxdx}=e^{-3ln\left(sinx\right)}=\frac{1}{si{n}^{3}x}$

$\therefore$ General solution

$\therefore \frac{y}{si{n}^{3}x}=\int \frac{2sinxcosx}{si{n}^{3}x}dx+C$

$=\frac{y}{si{n}^{3}x}=-2cosecx+C$ ......(i)

$\Rightarrow y\left(\frac{\pi }{2}\right)=2$ (given)

$\Rightarrow$ Putting the above value in (i), we get $\frac{2}{(1{)}^3}=-2+C\Rightarrow C=4$

$\therefore y=4si{n}^{3}x-2si{n}^{2}x$

$\Rightarrow y\left(\frac{\pi }{6}\right)=0$

$\Rightarrow y^{\prime }\left(x\right)=12si{n}^{2}xcosx-4sinxcosx$

$=y^{\prime }\left(\frac{\pi }{3}\right)=\frac{9-2\sqrt{3}}{2}$

$\Rightarrow y^{\prime }\left(x\right)=2sin2x(3sinx-1)$

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

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

$\Rightarrow {\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(4si{n}^{3}x-2si{n}^{2}x)dx=0-4{\int }_0^{\frac{\pi }{2}}si{n}^{2}xdx=-\pi$