Question

Let $y=y\left(x\right)$ be the solution of the differential equation $x\tan \left(\frac{y}{x}\right)dy=(y\tan \left(\frac{y}{x}\right)-x)dx,$$-1\le x\le 1,$ $y\left(\frac{1}{2}\right)=\frac{\pi }{6}.$ Then the area of region bounded by the curves $x=0,$ $x=\frac{1}{\sqrt{2}}$ and $y=y\left(x\right)$ in the upper half plane is

• A. $\frac{1}{6}(\pi -1)$
• B. $\frac{1}{12}(\pi -3)$
• C. $\frac{1}{8}(\pi -1)$
• D. $\frac{1}{4}(\pi -2)$

Answer 1

The correct option is C $\frac{1}{8}(\pi -1)$

$\tan \left(\frac{y}{x}\right)\frac{dy}{dx}=\left(\frac{y}{x}\right)\tan \left(\frac{y}{x}\right)-1$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\cot \left(\frac{y}{x}\right)\dots \left(1\right)$
Put $\frac{y}{x}=u\Rightarrow y=ux$
$\Rightarrow \frac{dy}{dx}=u+x\frac{du}{dx}$
From equation $\left(1\right)$
$\Rightarrow x\frac{du}{dx}=-\cot u$
$\Rightarrow -\tan udu=\frac{dx}{x}$
$\Rightarrow -\ln \left|\sec \frac{y}{x}\right|=\ln x+\ln c$
$\Rightarrow \cos \frac{y}{x}=xc$
Given $y\left(\frac{1}{2}\right)=\frac{\pi }{6}$
$\cos \frac{\pi }{3}=\frac{c}{2}\Rightarrow \frac{1}{2}=\frac{c}{2}\Rightarrow c=1$
Now we have
$\cos \frac{y}{x}=x$
$\Rightarrow y=x{\text{cos}}^{-1}x$

Area$\left(A\right)={\int }_0^{1/\sqrt{2}}x{\text{cos}}^{-1}xdx$
put $x=\text{cos}\theta$
$\therefore A={\int }_{\pi /2}^{\pi /4}\text{cos}\theta .\theta .(-\text{sin}\theta )d\theta$
$={\int }_{\pi /4}^{\pi /2}\left(\frac{\theta }{2}\right)\cdot \sin 2\theta d\theta$
$A={[\frac{-\theta }{2}\cdot \frac{\cos 2\theta }{2}]}_{\pi /4}^{\pi /2}+{\int }_{\pi /4}^{\pi /2}\frac{1}{2}\cdot \frac{\cos 2\theta }{2}d\theta$
$A=\left(\frac{-\pi }{8}\left(\cos \pi \right)\right)-\left(\frac{-\pi }{16}\left(\cos \frac{\pi }{2}\right)\right)+{\left[\frac{\sin 2\theta }{8}\right]}_{\pi /4}^{\pi /2}$
$A=\frac{-\pi }{8}\times (-1)-0+[\frac{\sin \pi }{8}-\frac{\sin \frac{\pi }{2}}{8}]$
$A=\frac{\pi }{8}+[0-\frac{1}{8}]$
$A=\frac{\pi -1}{8}$