JEE Main 2021 (Online) 25th July Evening Shift | Differential Equations Question 65 | Mathematics

JEE Main 2021 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential
equation xdy = (y + x³ cosx)dx with y($$\pi$$) = 0, then $$y\left( {{\pi \over 2}} \right)$$ is equal to :

$${{{\pi ^2}} \over 4} + {\pi \over 2}$$

$${{{\pi ^2}} \over 2} + {\pi \over 4}$$

$${{{\pi ^2}} \over 2} - {\pi \over 4}$$

$${{{\pi ^4}} \over 4} - {\pi \over 2}$$

JEE Main 2021 (Online) 25th July Morning Shift

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0$$

then, the minimum value of $$y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$$ is equal to :

$$\left( {2 - \sqrt 3 } \right) - {\log _e}2$$

$$\left( {2 + \sqrt 3 } \right) + {\log _e}2$$

$$\left( {1 + \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$

$$\left( {1 - \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$

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 :

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

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

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

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

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