1
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
$${e^x} + {1 \over 3}{x^3}$$
2
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
3
WB JEE 2021
+1
-0.25
The differential equation of all the ellipses centred at the origin and have axes as the co-ordinate axes is where $$y^{\prime}\equiv{{{dx}\over {dy}}},y^{\prime\prime}\equiv{{{d^2}y\over {dx^2}}}$$
A
y2 + xy'2 $$-$$ yy' = 0
B
xyy'' + xy'2 $$-$$ yy' = 0
C
yy' + xy'2 $$-$$ xy' = 0
D
x2y' + xy'' $$-$$ 3y = 0
4
WB JEE 2021
+1
-0.25
If $$x{{dy} \over {dx}} + y = {{xf(xy)} \over {f'(xy)'}}$$, then | f(xy) | is equal to (where k is an arbitrary positive constant).
A
$$k{e^{{x^2}/2}}$$
B
$$k{e^{{y^2}/2}}$$
C
$$k{e^{{x^2}}}$$
D
$$k{e^{{y^2}}}$$
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