JEE Main 2018 (Online) 15th April Morning Slot | Differential Equations Question 128 | Mathematics | JEE Main - ExamSIDE.com

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1

JEE Main 2018 (Online) 15th April Morning Slot

MCQ (Single Correct Answer)

+4

-1

Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + 2y = f\left( x \right),$$

where $$f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$$

If y(0) = 0, then $$y\left( {{3 \over 2}} \right)$$ is :

A

$${{{e^2} + 1} \over {2{e^4}}}$$

B

$${1 \over {2e}}$$

C

$${{{e^2} - 1} \over {{e^3}}}$$

D

$${{{e^2} - 1} \over {2{e^3}}}$$

2

JEE Main 2017 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

+4

-1

If 2x = y$${^{{1 \over 5}}}$$ + y$${^{ - {1 \over 5}}}$$ and

(x² $$-$$ 1) $${{{d^2}y} \over {d{x^2}}}$$ + $$\lambda $$x $${{dy} \over {dx}}$$ + ky = 0,

then $$\lambda $$ + k is equal to :

A

$$-$$ 23

B

$$-$$ 24

C

26

D

$$-$$ 26

3

JEE Main 2017 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

+4

-1

If y = $${\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},$$

then (x² $$-$$ 1) $${{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is equal to :

A

125 y

B

124 y²

C

225 y²

D

225 y

4

JEE Main 2017 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

+4

-1

The curve satisfying the differential equation, ydx $$-$$(x + 3y²)dy = 0 and passing through the point (1, 1), also passes through the point :

A

$$\left( {{1 \over 4}, - {1 \over 2}} \right)$$

B

$$\left( { - {1 \over 3},{1 \over 3}} \right)$$

C

$$\left( {{1 \over 3}, - {1 \over 3}} \right)$$

D

$$\left( {{1 \over 4}, {1 \over 2}} \right)$$

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