JEE Main 2021 (Online) 20th July Morning Shift | Differential Equations Question 124 | Mathematics

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 :

$${1 \over 8}(\pi - 1)$$

$${1 \over {12}}(\pi - 3)$$

$${1 \over 4}(\pi - 2)$$

$${1 \over 6}(\pi - 1)$$

JEE Main 2021 (Online) 20th July Morning Shift

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = $$-$$1. Then the value of (y(3))² is equal to :

1 $$-$$ 4e³

1 $$-$$ 4e⁶

1 + 4e³

1 + 4e⁶

JEE Main 2021 (Online) 18th March Evening Shift

MCQ (Single Correct Answer)

-1

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

$${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$$, 0 < x < 2.1, with y(2) = 0. Then the value of $${{dy} \over {dx}}$$ at x = 1 is equal to :

$${{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}$$

$${{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}$$

$$ - {{2{e^2}} \over {{{(1 + {e^2})}^2}}}$$

$${{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$$

JEE Main 2021 (Online) 18th March Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

The differential equation satisfied by the system of parabolas

y² = 4a(x + a) is :

$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) + y = 0$$

$$y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$