1
NDA 2019 Paper 2
MCQ (Single Correct Answer)
+2.5
-0.83
The differential equation which represents the family of curves given by $$\tan y = C(1 - {e^x})$$ is
A
$${e^x}\tan ydx + (1 - {e^x})dy = 0$$
B
$${e^x}\tan ydx + (1 - {e^x}){\sec ^2}ydy = 0$$
C
$${e^x}(1 - {e^x})dx + \tan ydy = 0$$
D
$${e^x}\tan ydy + (1 - {e^x})dx = 0$$
2
NDA 2019 Paper 1
MCQ (Single Correct Answer)
+2.5
-0.83
The solution of the differential equation $${{dy} \over {dx}} = \cos (y - x) + 1$$ is
A
$${e^x}[\sec (y - x) - \tan (y - x)] = c$$
B
$${e^x}[\sec (y - x) + \tan (y - x)] = c$$
C
$${e^x}\sec (y - x)\tan (y - x) = c$$
D
$${e^x} = c\sec (y - x)\tan (y - x)$$
3
NDA 2019 Paper 1
MCQ (Single Correct Answer)
+2.5
-0.83
If $$y = a\cos 2x + b\sin 2x$$, then
A
$${{{d^2}y} \over {d{x^2}}} + y = 0$$
B
$${{{d^2}y} \over {d{x^2}}} + 2y = 0$$
C
$${{{d^2}y} \over {d{x^2}}} - 4y = 0$$
D
$${{{d^2}y} \over {d{x^2}}} + 4y = 0$$
4
NDA 2019 Paper 1
MCQ (Single Correct Answer)
+2.5
-0.83
The differential equation of the system of circles touching the Y-axis at the origin is
A
$${x^2} + {y^2} - 2xy{{dy} \over {dx}} = 0$$
B
$${x^2} + {y^2} + 2xy{{dy} \over {dx}} = 0$$
C
$${x^2} - {y^2} + 2xy{{dy} \over {dx}} = 0$$
D
$${x^2} - {y^2} - 2xy{{dy} \over {dx}} = 0$$
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