JEE Main 2020 (Online) 7th January Morning Slot | Differential Equations Question 151 | Mathematics

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:

2 + log_e2

log_e2

1 + log_e2

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MCQ (Single Correct Answer)

-1

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

y² + 2x³ + cx² = 0

y² + 2x² + cx³ = 0

y² – 2x + cx³ = 0

y² – 2x³ + cx² = 0

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MCQ (Single Correct Answer)

-1

Consider the differential equation, $${y^2}dx + \left( {x - {1 \over y}} \right)dy = 0$$, If value of y is 1 when x = 1, then the value of x for which y = 2, is :

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

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

$${5 \over 2} + {1 \over {\sqrt e }}$$

$${3 \over 2} - \sqrt e $$

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MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation,
$${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, such that y(0) = 1. Then :

$$y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2 $$

$$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $$

$$y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$$

$$y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2 $$

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