Let y = y(x) be the solution of the differential equation x tan (y/x)dy = (y tan (y/x) - x)dx, - 1 ≤ x ≤ 1, y(1/2) = (π)/6. Then the area of the region bounded by the curves x = 0, x = 1/(2) and y = y(x) in the upper half plane is:

Differential Equations: JEE Main 2021 Mathematics question

JEE Main 2021 · 20 Jul, Morning shift · Q3609

JEE Main 202120 Jul, Morning shiftMathematicsMediumDifferential Equations163× since 2002

Let y = y(x) be the solution of the differential equation

x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx

- 1 \le x \le 1

y\left( {{1 \over 2}} \right) = {\pi \over 6}

. Then the area of the region bounded by the curves x = 0,

x = {1 \over {\sqrt 2 }}

and y = y(x) in the upper half plane is :

JEE Main 2021 · 20 Jul, Morning shift · Q3609

JEE Main 202120 Jul, Morning shiftMathematicsMediumDifferential Equations163× since 2002

Let y = y(x) be the solution of the differential equation

x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx

- 1 \le x \le 1

y\left( {{1 \over 2}} \right) = {\pi \over 6}

. Then the area of the region bounded by the curves x = 0,

x = {1 \over {\sqrt 2 }}

and y = y(x) in the upper half plane is :