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 :

Differential Equations — JEE Main · 2021 · 20 Jul 2021 · Shift 1 Question with Solution

Let

y = y (x)

be the solution of the differential equation

x tan (\frac{y}{x}) d y = (y tan (\frac{y}{x}) - x) d x, - 1 \leq x \leq 1, y (\frac{1}{2}) = \frac{π}{6}

Then the area of the region bounded by the curves

x = 0, x = \frac{1}{2}

and

y = y (x)

in the upper half plane is :

Answer: (B)

The equation is homogeneous, so putting

y = v x

with

d y = v d x + x d v

x tan v (v d x + x d v) = (v x tan v - x) d x \Rightarrow x^{2} tan v d v = - x d x

Separating and integrating gives

lo g_{e} ∣ sec v ∣ = - lo g_{e} ∣ x ∣ + lo g_{e} A

, that is

x sec \frac{y}{x} = A

. At

x = \frac{1}{2}

with

y = \frac{π}{6}

the ratio is

\frac{y}{x} = \frac{π}{3}

and

sec \frac{π}{3} = 2

, so

A = 1

and the solution curve is

cos \frac{y}{x} = x

, that is

y = x cos^{- 1} x

. This is positive on

(0, \frac{1}{2})

, so the required area is

\int_{0}^{1/ 2} x cos^{- 1} x d x = [\frac{x ^{2}}{2} cos^{- 1} x]_{0}^{1/ 2} + \frac{1}{2} \int_{0}^{1/ 2} \frac{x ^{2}}{1 - x ^{2}} d x

The bracket equals

\frac{1}{4} \cdot \frac{π}{4} = \frac{π}{16}

, and

x = sin θ

turns the remaining integral into

\int_{0}^{π /4} sin^{2} θ d θ = \frac{π}{8} - \frac{1}{4}

, which contributes

\frac{π}{16} - \frac{1}{8}

. Adding the two parts, the area is

\frac{π}{8} - \frac{1}{8} = \frac{1}{8} (π - 1)

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option