1
JEE Main 2020 (Online) 7th January Evening Slot
+4
-1
Let y = y(x) be the solution curve of the differential equation,

$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
A
2 + e
B
-e
C
2
D
2 - e
2
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
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:
A
2 + loge2
B
loge2
C
1 + loge2
D
2e
3
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x $$\ne$$ 0) is : (where c is a constant of integration)
A
y2 + 2x3 + cx2 = 0
B
y2 + 2x2 + cx3 = 0
C
y2 – 2x + cx3 = 0
D
y2 – 2x3 + cx2 = 0
4
JEE Main 2019 (Online) 12th April Morning Slot
+4
-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 :
A
$${3 \over 2} - {1 \over {\sqrt e }}$$
B
$${1 \over 2} + {1 \over {\sqrt e }}$$
C
$${5 \over 2} + {1 \over {\sqrt e }}$$
D
$${3 \over 2} - \sqrt e$$
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