1
WB JEE 2019
+1
-0.25 General solution of $${(x + y)^2}{{dy} \over {dx}} = {a^2},a \ne 0$$ is (C is an arbitrary constant)
A
$${x \over a} = \tan {y \over a} + C$$
B
$$\tan xy = C$$
C
$$\tan (x + y) = C$$
D
$$\tan {{y + C} \over a} = {{x + y} \over a}$$
2
WB JEE 2018
+1
-0.25 The differential equation representing the family of curves $${y^2} = 2d(x + \sqrt d )$$, where d is a parameter, is of
A
order 2
B
degree 2
C
degree 3
D
degree 4
3
WB JEE 2018
+1
-0.25 Let y(x) be a solution of

$$(1 + {x^2}){{dy} \over {dx}} + 2xy - 4{x^2} = 0$$. Then y(1) is equal to
A
$${1 \over 2}$$
B
$${1 \over 3}$$
C
$${1 \over 6}$$
D
$$-$$1
4
WB JEE 2017
+1
-0.25 If $$y = {e^{m{{\sin }^{ - 1}}x}}$$ then $$(1 - {x^2}){{{d^2}y} \over {d{x^2}}} - x{{dy} \over {dx}} -$$ky = 0, where k is equal to
A
m2
B
2
C
$$-$$ 1
D
$$-$$ m2
WB JEE 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