JEE Main 2020 (Online) 3rd September Morning Slot | Differential Equations Question 96 | Mathematics

JEE Main 2020 (Online) 3rd September Morning Slot

MCQ (Single Correct Answer)

-1

The solution curve of the differential equation,

(1 + e^-x)(1 + y²)$${{dy} \over {dx}}$$ = y²,

which passes through the point (0, 1), is :

y² + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$

y² + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$

y² = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$

y² = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$

JEE Main 2020 (Online) 2nd September Evening Slot

MCQ (Single Correct Answer)

-1

If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x²dy= (2xy + y²)dx, then $$f\left( {{1 \over 2}} \right)$$ is equal to :

$${1 \over {1 - {{\log }_e}2}}$$

$${1 \over {1 + {{\log }_e}2}}$$

$${{ - 1} \over {1 + {{\log }_e}2}}$$

$${1 + {{\log }_e}2}$$

JEE Main 2020 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x satisfying y(x) = e is :

$$\sqrt 2 e$$

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

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

$$\sqrt 3 e$$

JEE Main 2020 (Online) 8th January Evening Slot

MCQ (Single Correct Answer)

-1

The differential equation of the family of curves, x² = 4b(y + b), b $$ \in $$ R, is

x(y')² = x – 2yy'

x(y')² = 2yy' – x

xy" = y'

x(y')² = x + 2yy'

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