Question

If $y\left(x\right)$ satisfies the differential equation $\frac{dy}{dx}+2y\tan x=\sin x$ and $y\left(\frac{\pi }{3}\right)=0,$ then which of the following statements is(are) CORRECT ?

• A. $y^{\prime }\left(\frac{\pi }{6}\right)=\frac{2\sqrt{3}-1}{2}$
• B. $y\left(\pi \right)=3$
• C. Number of values of $x$ for which $y\left(x\right)=-1$ in $(0,2\pi )$ is two.
• D. $\underset{0}{\overset{\pi /2}{\int }}y\left(x\right)dx=1-\frac{\pi }{4}$

Answer 1

Answer: (A), (C)

Number of values of $x$ for which $y\left(x\right)=-1$ in $(0,2\pi )$ is two.

$\frac{dy}{dx}+\left(2\tan x\right)y=\sin x$ (Linear differential equation)
$\therefore I.F.=e^{2\int \tan xdx}=e^{2\ln \left(\sec x\right)}={\sec }^{2}x$
So, general solution is
$y\cdot {\sec }^{2}x=\int \frac{\sin x}{{\cos }^{2}x}dx+C\Rightarrow y\cdot {\sec }^{2}x=\sec x+C$
Now, $y\left(\frac{\pi }{3}\right)=0\Rightarrow C=-2$
$\Rightarrow y\left(x\right)=\cos x-2{\cos }^{2}x$
Differentiate w.r.t. to $x$
$y^{\prime }\left(x\right)=-\sin x+4\cos x\cdot \sin x$
$y^{\prime }\left(\frac{\pi }{6}\right)=\frac{-1}{2}+4\times \frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{2\sqrt{3}-1}{2}$

$y\left(\pi \right)=\cos \pi -2{\cos }^2\pi =-3$

$y\left(x\right)=-1\Rightarrow \cos x-2{\cos }^{2}x=-1\Rightarrow 2{\cos }^{2}x-\cos x-1=0$
$\Rightarrow (2\cos x+1)(\cos x-1)=0$
$\Rightarrow \cos x=\frac{-1}{2}$ or $\cos x=1$
But $x\in (0,2\pi ),$ so $x=\frac{2\pi }{3},\frac{4\pi }{3}$

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