1
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
2
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
$${{{e^2} - 1} \over {2{e^3}}}$$
3
JEE Main 2017 (Online) 9th April Morning Slot
+4
-1
If 2x = y$${^{{1 \over 5}}}$$ + y$${^{ - {1 \over 5}}}$$ and

(x2 $$-$$ 1) $${{{d^2}y} \over {d{x^2}}}$$ + $$\lambda$$x $${{dy} \over {dx}}$$ + ky = 0,

then $$\lambda$$ + k is equal to :
A
$$-$$ 23
B
$$-$$ 24
C
26
D
$$-$$ 26
4
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
The curve satisfying the differential equation, ydx $$-$$(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :
A
$$\left( {{1 \over 4}, - {1 \over 2}} \right)$$
B
$$\left( { - {1 \over 3},{1 \over 3}} \right)$$
C
$$\left( {{1 \over 3}, - {1 \over 3}} \right)$$
D
$$\left( {{1 \over 4}, {1 \over 2}} \right)$$
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