1
JEE Main 2022 (Online) 28th June Morning Shift
+4
-1

Let y = y(x) be the solution of the differential equation $$x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0$$, $$x > 1$$, with $$y(2) = - 2$$. Then y(3) is equal to :

A
$$-$$18
B
$$-$$12
C
$$-$$6
D
$$-$$3
2
JEE Main 2022 (Online) 27th June Evening Shift
+4
-1

If the solution curve of the differential equation

$$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

A
2e
B
$${2 \over e}$$
C
2
D
$${1 \over e}$$
3
JEE Main 2022 (Online) 27th June Morning Shift
+4
-1
Out of Syllabus

Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :

A
10
B
8
C
7
D
5
4
JEE Main 2022 (Online) 27th June Morning Shift
+4
-1

If $${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$$, x, y > 0, y(1) = 1, then y(2) is equal to :

A
$$2 + {\log _2}3$$
B
$$2 + {\log _3}2$$
C
$$2 - {\log _3}2$$
D
$$2 - {\log _2}3$$
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