# Let y = y(x) be the solution of the

Question:

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

1. $\frac{1}{8}(\pi-1)$

2. $\frac{1}{12}(\pi-3)$

3. $\frac{1}{4}(\pi-2)$

4. $\frac{1}{6}(\pi-1)$

Correct Option: 1

Solution:

We have

$\frac{d y}{d x}=\frac{x\left(\frac{y}{x} \cdot \tan \frac{y}{x}-1\right)}{x \tan \frac{y}{x}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}-\cot \left(\frac{\mathrm{y}}{\mathrm{x}}\right)$

Put $\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{V}$

$\Rightarrow \mathrm{y}=\mathrm{Vn}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\frac{\mathrm{ndv}}{\mathrm{dx}}$

Now, we get

$v+n \frac{d v}{d x}=v-\cot (v)$

$\Rightarrow \int(\tan ) \mathrm{dv}=-\int \frac{\mathrm{dx}}{\mathrm{x}}$

$\therefore \ell \mathrm{n}\left|\sec \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right|=-\ell \mathrm{n}|\mathrm{x}|+\mathrm{c}$

$\operatorname{As}\left(\frac{1}{2}\right)=\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \Rightarrow \mathrm{C}=0$

$\therefore \sec \left(\frac{y}{x}\right)=\frac{1}{x}$

$\Rightarrow \cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\mathrm{x}$

$\therefore y=x \cos ^{-1}(x)$

So, required bounded area

$=\int_{0}^{1 / \sqrt{2}} \underset{\text { (II) }}{x}\left(\underset{\text { (I) }}{\cos ^{-1} x}\right) \mathrm{dx}=\left(\frac{\pi-1}{8}\right)$

(I.B.P.)

$\therefore$ option $(1)$ is correct.