1
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
Out of Syllabus
The differential equation representing the family of ellipse having foci eith on the x-axis or on the $$y$$-axis, center at the origin and passing through the point (0, 3) is :
A
xy y'' + x (y')2 $$-$$ y y' = 0
B
x + y y'' = 0
C
xy y'+ y2 $$-$$ 9 = 0
D
xy y' $$-$$ y2 + 9 = 0
2
JEE Main 2018 (Offline)
+4
-1
Let y = y(x) be the solution of the differential equation

$$\sin x{{dy} \over {dx}} + y\cos x = 4x$$, $$x \in \left( {0,\pi } \right)$$.

If $$y\left( {{\pi \over 2}} \right) = 0$$, then $$y\left( {{\pi \over 6}} \right)$$ is equal to :
A
$$- {4 \over 9}{\pi ^2}$$
B
$${4 \over {9\sqrt 3 }}{\pi ^2}$$
C
$$- {8 \over {9\sqrt 3 }}{\pi ^2}$$
D
$$- {8 \over 9}{\pi ^2}$$
3
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
The curve satifying the differeial equation, (x2 $$-$$ y2) dx + 2xydy = 0 and passing through the point (1, 1) is :
A
B
a hyperbola.
C
an ellipse.
D
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + 2y = f\left( x \right),$$

where $$f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$$

If y(0) = 0, then $$y\left( {{3 \over 2}} \right)$$ is :
A
$${{{e^2} + 1} \over {2{e^4}}}$$
B
$${1 \over {2e}}$$
C
$${{{e^2} - 1} \over {{e^3}}}$$
D
$${{{e^2} - 1} \over {2{e^3}}}$$
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