JEE Main 2022 (Online) 25th June Morning Shift | Differential Equations Question 100 | Mathematics

If the solution curve $$y = y(x)$$ of the differential equation $${y^2}dx + ({x^2} - xy + {y^2})dy = 0$$, which passes through the point (1, 1) and intersects the line $$y = \sqrt 3 x$$ at the point $$(\alpha ,\sqrt 3 \alpha )$$, then value of $${\log _e}(\sqrt 3 \alpha )$$ is equal to :

$${\pi \over 3}$$

$${\pi \over 2}$$

$${\pi \over 12}$$

$${\pi \over 6}$$

JEE Main 2022 (Online) 24th June Morning Shift

MCQ (Single Correct Answer)

-1

If x = x(y) is the solution of the differential equation

$$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$$; then x(e) is equal to :

$${e^3}({e^e} - 1)$$

$${e^e}({e^3} - 1)$$

$${e^2}({e^e} + 1)$$

$${e^e}({e^2} - 1)$$

JEE Main 2021 (Online) 1st September Evening Shift

MCQ (Single Correct Answer)

-1

If y = y(x) is the solution curve of the differential equation $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0 and y(1) = 1, then $$y\left( {{1 \over 2}} \right)$$ is equal to :

$${3 \over 2} - {1 \over {\sqrt e }}$$

$$3 + {1 \over {\sqrt e }}$$

3 + e

3 $$-$$ e

JEE Main 2021 (Online) 31st August Evening Shift

MCQ (Single Correct Answer)

-1

If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :

(1, 2)

$$\left( {{1 \over 2},1} \right]$$

(2, 3)

$$\left( {0,{1 \over 2}} \right]$$

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