JEE Main 2020 (Online) 7th January Evening Slot | Differential Equations Question 102 | Mathematics | JEE Main - ExamSIDE.com

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1

JEE Main 2020 (Online) 7th January Evening Slot

MCQ (Single Correct Answer)

+4

-1

Let y = y(x) be the solution curve of the differential equation,
$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is:

A

2 + e

B

-e

C

2

D

2 - e

2

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

+4

-1

Let x^k + y^k = a^k, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:

A

$${1 \over 3}$$

B

$${2 \over 3}$$

C

$${4 \over 3}$$

D

$${3 \over 2}$$

3

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

+4

-1

If y = y(x) is the solution of the differential equation, $${e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x}$$ such that y(0) = 0, then y(1) is equal to:

A

2 + log_e2

B

log_e2

C

1 + log_e2

D

2e

4

JEE Main 2019 (Online) 12th April Evening Slot

MCQ (Single Correct Answer)

+4

-1

The general solution of the differential equation (y² – x³)dx – xydy = 0 (x $$ \ne $$ 0) is : (where c is a constant of integration)

A

y² + 2x³ + cx² = 0

B

y² + 2x² + cx³ = 0

C

y² – 2x + cx³ = 0

D

y² – 2x³ + cx² = 0

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