NDA 2019 Paper 2 | Differential Equations Question 29 | Mathematics

Which one of the following is the differential equation that represents the family of curves $$y = {1 \over {2{x^2} - C}}$$,where C is an arbitrary constant?

$${{dy} \over {dx}} = 4x{y^2}$$

$${{dy} \over {dx}} = {1 \over y}$$

$${{dy} \over {dx}} = {x^2}y$$

$${{dy} \over {dx}} = - 4x{y^2}$$

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

$${e^x}\tan ydx + (1 - {e^x})dy = 0$$

$${e^x}\tan ydx + (1 - {e^x}){\sec ^2}ydy = 0$$

$${e^x}(1 - {e^x})dx + \tan ydy = 0$$

$${e^x}\tan ydy + (1 - {e^x})dx = 0$$

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

$${e^x}[\sec (y - x) - \tan (y - x)] = c$$

$${e^x}[\sec (y - x) + \tan (y - x)] = c$$

$${e^x}\sec (y - x)\tan (y - x) = c$$

$${e^x} = c\sec (y - x)\tan (y - x)$$

NDA 2019 Paper 1

MCQ (Single Correct Answer)

+2.5

-0.83

If $$y = a\cos 2x + b\sin 2x$$, then

$${{{d^2}y} \over {d{x^2}}} + y = 0$$

$${{{d^2}y} \over {d{x^2}}} + 2y = 0$$

$${{{d^2}y} \over {d{x^2}}} - 4y = 0$$

$${{{d^2}y} \over {d{x^2}}} + 4y = 0$$

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