Question

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

• A. $y=sinx+\frac{1}{2}cosecx$
• B. $y=tan(x/2)+cot\left(x/2\right)$
• 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, $\frac{dy}{dx}=2cosx-ycosxcosecx$
$\Rightarrow$ $\frac{dy}{dx}+ycotx=2cosx$
$\therefore$ $IF=e^{\int cotxdx}=e^{ln\left(sinx\right)}=sinx$
$\therefore$ Solution is $ysinx=\int 2cosxsinxdx+c$
$\Rightarrow$ $ysinx=\int sin2xdx+c$
$\Rightarrow$ $ysinx=\frac{-cos2x}{2}+c$
At $x=\frac{\pi }{4},y=\sqrt{2}$
$\therefore$ $\sqrt{2}sin\frac{\pi }{4}=\frac{-cos2(\pi /4)}{2}+c$
$\Rightarrow$ $c=1$
$\therefore$ $ysinx=-\frac{1}{2}cos2x+1$
$\Rightarrow$ $y=-\frac{1}{2}.\frac{cos2x}{sinx}+cosecx$
$\Rightarrow$ $y=-\frac{1}{2sinx}\left(1-2{\left(sin\right)}^{2}x\right)+cosecx$
$\Rightarrow$ $y=\frac{1}{2}cosecx+sinx$