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Mathematics
Mathematics
Problem
If
y
=
y
(
x
)
y = y(x)
y
=
y
(
x
)
is the solution of the differential equation,
d
y
d
x
+
2
y
tan
x
=
sin
x
\frac{dy}{dx} + 2y \tan x = \sin x
d
x
d
y
+
2
y
tan
x
=
sin
x
,
y
(
π
3
)
=
0
y\left(\frac{\pi}{3}\right) = 0
y
(
3
π
)
=
0
, then the maximum value of the function
y
(
x
)
y(x)
y
(
x
)
over
R
\mathbb{R}
R
is equal to : [JEE MAINS - ONLINE - 2021] (1) 8 (2)
1
2
\frac{1}{2}
2
1
(3)
−
15
4
-\frac{15}{4}
−
4
15
(4)
1
8
\frac{1}{8}
8
1
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