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
Let xk + yk = ak, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:
A
$${1 \over 3}$$
B
$${2 \over 3}$$
C
$${4 \over 3}$$
D
$${3 \over 2}$$
3
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
4
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
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