Question

The solution of $dy=cosx\left(2 - y \, c o s e c \, x\right)dx,$ where $y=\sqrt{2}$ when $x=\pi / 4$ , is

• A. $y=sin x+\frac{1}{2} \, cosec \, x$
• B. $y=tan \left(\right. x / 2 \left.\right)+cot ⁡ \left(x / 2\right)$
• C. $y=\left(1 / \sqrt{2}\right)sec \left(x / 2\right)+\sqrt{2}cos ⁡ \left(x / 2\right)$
• D. None of the above

Answer 1

Answer: (A)

Given, $\frac{d y}{d x}=2cos x-ycos ⁡ x \, cosec \, x$
$\Rightarrow $ $\frac{d y}{d x}+ycot x=2cos ⁡ x$
$\therefore $ $IF=e^{\displaystyle \int cot x \, d x}=e^{ln ⁡ \left(sin ⁡ x\right)}=sin ⁡ x$
$\therefore $ Solution is $ysin x=\displaystyle \int 2 cos ⁡ xsin ⁡ x \, dx+c$
$\Rightarrow $ $ysin x=\displaystyle \int sin ⁡ 2 x \, dx+c$
$\Rightarrow \, \, $ $ysin x=\frac{- cos ⁡ 2 x}{2}+c$
At $x=\frac{\pi }{4},y=\sqrt{2}$
$\therefore $ $\sqrt{2}sin \frac{\pi }{4}=\frac{- cos ⁡ 2 \left(\right. \pi / 4 \left.\right)}{2}+c$
$\Rightarrow $ $c=1$
$\therefore $ $ysin x=-\frac{1}{2}cos ⁡ 2 x+1$
$\Rightarrow $ $y=-\frac{1}{2}. \, \frac{cos 2 x}{sin ⁡ x}+cosec \, x$
$\Rightarrow $ $y=-\frac{1}{2 sin x}\left(1 - 2 \left(sin\right)^{2} ⁡ x\right)+cosec \, x$
$\Rightarrow $ $y=\frac{1}{2} \, cosec \, x+sin x$