1
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
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
2
WB JEE 2017
MCQ (Single Correct Answer)
+1
-0.25
Change Language
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
3
WB JEE 2017
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Solution of $${(x + y)^2}{{dy} \over {dx}} = {a^2}$$ ('a' belong a constant) is
A
$${{(x + y)} \over a} = \tan {{y + C} \over a},C$$ is an arbitrary constant
B
$$xy = a\tan Cx,C$$ is an arbitrary constant
C
$${x \over a} = \tan {y \over C},C$$ is an arbitrary constant
D
$$xy = \tan (x + C),C$$ is an arbitrary constant
4
WB JEE 2017
MCQ (Single Correct Answer)
+1
-0.25
Change Language
The integrating factor of the first order differential equation $${x^2}({x^2} - 1){{dy} \over {dx}} + x({x^2} + 1)y = {x^2} - 1$$ is
A
ex
B
$$x - {1 \over x}$$
C
$$x + {1 \over x}$$
D
$${1 \over {{x^2}}}$$
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