1
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
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 :
A
1 $$-$$ 4e3
B
1 $$-$$ 4e6
C
1 + 4e3
D
1 + 4e6
2
JEE Main 2021 (Online) 18th March Evening Shift
MCQ (Single Correct Answer)
+4
-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 :
A
$${{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}$$
B
$${{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}$$
C
$$ - {{2{e^2}} \over {{{(1 + {e^2})}^2}}}$$
D
$${{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$$
3
JEE Main 2021 (Online) 18th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
The differential equation satisfied by the system of parabolas

y2 = 4a(x + a) is :
A
$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$
B
$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) + y = 0$$
C
$$y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$
D
$$y\left( {{{dy} \over {dx}}} \right) + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$
4
JEE Main 2021 (Online) 17th March Evening Shift
MCQ (Single Correct Answer)
+4
-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 :
A
$$4\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$$
B
$$\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$$
C
$$\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$$
D
$$4\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$$
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