Question

Suppose $\left(cos\right)^{2} y\cdot \frac{d y}{d x}=sin⁡\left(x + y\right)+sin⁡\left(x - y\right),\left|x\right|\leq \frac{\pi }{2}$ and $\left|y\right|\leq \frac{\pi }{2}$ . If $y\left(\frac{\pi }{3}\right)=-\frac{\pi }{2}$ , then $y\left(\frac{\pi }{2}\right)$ is

• A. $0$
• B. $\frac{\pi }{4}$
• C. $-\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$cos^{2} y\frac{d y}{d x}=2sin⁡xcos⁡y$
$\Rightarrow cos ydy=2sin ⁡ xdx$
$\Rightarrow sin y+2cos ⁡ x=c$
Now, $y\left(\frac{\pi }{3}\right)=-\frac{\pi }{2}\Rightarrow c=0$
So $sin y=-2cos ⁡ x$
At $x=\frac{\pi }{2},sin y=0\Rightarrow y\left(\frac{\pi }{2}\right)=0$