JEE Main 2021 (Online) 22th July Evening Shift | Differential Equations Question 68 | Mathematics

JEE Main 2021 (Online) 22th July Evening Shift

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation $$\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$$, with $$y\left( {{\pi \over 4}} \right) = 0$$. Then, the value of $${(y(0) + 1)^2}$$ is equal to :

e^1/2

e^$$-$$1/2

e^$$-$$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 $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $$ - 1 \le x \le 1$$, $$y\left( {{1 \over 2}} \right) = {\pi \over 6}$$. 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 :

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

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

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

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

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