1
WB JEE 2023
+1
-0.25 Given $${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$. Changing the independent variable x to z by the substitution $$z = \log \tan {x \over 2}$$, the equation is changed to

A
$${{{d^2}y} \over {d{z^2}}} + {3 \over y} = 0$$
B
$$2{{{d^2}y} \over {d{z^2}}} + {e^y} = 0$$
C
$${{{d^2}y} \over {d{z^2}}} - 4y = 0$$
D
$${{{d^2}y} \over {d{z^2}}} + 4y = 0$$
2
WB JEE 2023
+2
-0.5 The family of curves $$y = {e^{a\sin x}}$$, where 'a' is arbitrary constant, is represented by the differential equation

A
$$y\log y = \tan x{{dy} \over {dx}}$$
B
$$y\log x = \cot x{{dy} \over {dx}}$$
C
$$\log y = \tan x{{dy} \over {dx}}$$
D
$$\log y = \cot x{{dy} \over {dx}}$$
3
WB JEE 2022
+1
-0.25 The solution of

$$\cos y{{dy} \over {dx}} = {e^{x + \sin y}} + {x^2}{e^{\sin y}}$$ is $$f(x) + {e^{ - \sin y}} = C$$ (C is arbitrary real constant) where f(x) is equal to

A
$${e^x} + {1 \over 2}{x^3}$$
B
$${e^{ - x}} + {1 \over 3}{x^3}$$
C
$${e^{ - x}} + {1 \over 2}{x^3}$$
D
$${e^x} + {1 \over 3}{x^3}$$
4
WB JEE 2022
+2
-0.5 If the transformation $$z = \log \tan {x \over 2}$$ reduces the differential equation

$${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$ into the form $${{{d^2}y} \over {d{z^2}}} + ky = 0$$ then k is equal to

A
$$-$$4
B
4
C
2
D
$$-$$2
WB JEE Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
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Calculus
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