1
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
Let f : [0,1] $$\to$$ R be such that f(xy) = f(x).f(y), for all x, y $$\in$$ [0, 1], and f(0) $$\ne$$ 0. If y = y(x) satiesfies the differential equation, $${{dy} \over {dx}}$$ = f(x) with y(0) = 1, then y$$\left( {{1 \over 4}} \right)$$ + y$$\left( {{3 \over 4}} \right)$$ is equal to :
A
3
B
4
C
2
D
5
2
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
If y = y(x) is the solution of the differential equation,

x$$dy \over dx$$ + 2y = x2, satisfying y(1) = 1, then y($$1\over2$$) is equal to :
A
$${{7} \over {64}}$$
B
$${{49} \over {16}}$$
C
$${{1} \over {4}}$$
D
$${{13} \over {16}}$$
3
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
4
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
$$- {8 \over 9}{\pi ^2}$$
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