1
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x $$\ne$$ 0) is : (where c is a constant of integration)
A
y2 + 2x3 + cx2 = 0
B
y2 + 2x2 + cx3 = 0
C
y2 – 2x + cx3 = 0
D
y2 – 2x3 + cx2 = 0
2
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
Consider the differential equation, $${y^2}dx + \left( {x - {1 \over y}} \right)dy = 0$$, If value of y is 1 when x = 1, then the value of x for which y = 2, is :
A
$${3 \over 2} - {1 \over {\sqrt e }}$$
B
$${1 \over 2} + {1 \over {\sqrt e }}$$
C
$${5 \over 2} + {1 \over {\sqrt e }}$$
D
$${3 \over 2} - \sqrt e$$
3
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
Let y = y(x) be the solution of the differential equation,
$${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, such that y(0) = 1. Then :
A
$$y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2$$
B
$$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2$$
C
$$y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$$
D
$$y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2$$
4
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
If y = y(x) is the solution of the differential equation
$${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
such that y (0) = 0, then $$y\left( { - {\pi \over 4}} \right)$$ is equal to :
A
$${1 \over 2} - e$$
B
$$e - 2$$
C
$$2 + {1 \over e}$$
D
$${1 \over e} - 2$$
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination