1
COMEDK 2022
MCQ (Single Correct Answer)
+1
-0

The solution of the differential equation $$x \frac{d y}{d x}=\cot y$$ is

A
$$y \cos x=c$$
B
$$x \cos y=c$$
C
$$\log (x \cos y)=c$$
D
$$\log (y \cos x)=c$$
2
COMEDK 2021
MCQ (Single Correct Answer)
+1
-0

The solution of the differential equation $${\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0$$ is

A
$$\tan y\,.\,\tan x = C$$
B
$${{\tan y} \over {\tan x}} = C$$
C
$${{{{\tan }^2}x} \over {\tan y}} = C$$
D
None of these
3
COMEDK 2021
MCQ (Single Correct Answer)
+1
-0

The solution of the differential equation $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$ is (where, C is a constant)

A
$$\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = Cx$$
B
$$x\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = C$$
C
$$\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = C{x^2}$$
D
$${x^2}\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = C$$
4
COMEDK 2021
MCQ (Single Correct Answer)
+1
-0

The solution of the differential equation $$(1 + {y^2}) + (x - {e^{{{\tan }^{ - 1}}y}}){{dy} \over {dx}} = 0$$ is

A
$$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + C$$
B
$$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + C$$
C
$$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + C$$
D
$$(x - 2) = C{e^{ - {{\tan }^{ - 1}}y}}$$
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