JEE Main 2021 (Online) 20th July Morning Shift | Differential Equations Question 118 | Mathematics

Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = $$-$$1. Then the value of (y(3))² is equal to :

1 $$-$$ 4e³

1 $$-$$ 4e⁶

1 + 4e³

1 + 4e⁶

JEE Main 2021 (Online) 18th March Evening Shift

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation

$${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$$, 0 < x < 2.1, with y(2) = 0. Then the value of $${{dy} \over {dx}}$$ at x = 1 is equal to :

$${{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}$$

$${{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}$$

$$ - {{2{e^2}} \over {{{(1 + {e^2})}^2}}}$$

$${{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$$

JEE Main 2021 (Online) 18th March Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

The differential equation satisfied by the system of parabolas

y² = 4a(x + a) is :

$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) + y = 0$$

$$y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

$$y\left( {{{dy} \over {dx}}} \right) + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

JEE Main 2021 (Online) 17th March Evening Shift

MCQ (Single Correct Answer)

-1

If the curve y = y(x) is the solution of the differential equation

$$2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$$, x > 0 which

passes through the point $$\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$$, then the value of y(16) is equal to :

$$4\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$$

$$\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$$

$$\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$$

$$4\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$$

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