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Mathematics
Mathematics
Problem
Let
y
=
y
(
x
)
y = y(x)
y
=
y
(
x
)
be the solution of the differential equation
cos
x
(
log
e
(
cos
x
)
)
2
d
y
+
(
sin
x
−
3
y
sin
x
log
e
(
cos
x
)
)
d
x
=
0
\cos x (\log_e(\cos x))^2 dy + (\sin x - 3y \sin x \log_e(\cos x)) dx = 0
cos
x
(
lo
g
e
(
cos
x
)
)
2
d
y
+
(
sin
x
−
3
y
sin
x
lo
g
e
(
cos
x
))
d
x
=
0
,
x
∈
(
0
,
π
2
)
x \in \left(0, \frac{\pi}{2}\right)
x
∈
(
0
,
2
π
)
. If
y
(
π
4
)
=
−
1
log
e
2
y\left(\frac{\pi}{4}\right) = \frac{-1}{\log_e 2}
y
(
4
π
)
=
l
o
g
e
2
−
1
, then
y
(
π
6
)
y\left(\frac{\pi}{6}\right)
y
(
6
π
)
is: (a)
2
log
e
(
3
)
−
log
e
(
4
)
\frac{2}{\log_e(3) - \log_e(4)}
l
o
g
e
(
3
)
−
l
o
g
e
(
4
)
2
(b)
1
log
e
(
4
)
−
log
e
(
3
)
\frac{1}{\log_e(4) - \log_e(3)}
l
o
g
e
(
4
)
−
l
o
g
e
(
3
)
1
(c)
−
1
log
e
(
4
)
-\frac{1}{\log_e(4)}
−
l
o
g
e
(
4
)
1
(d)
1
log
e
(
3
)
−
log
e
(
4
)
\frac{1}{\log_e(3) - \log_e(4)}
l
o
g
e
(
3
)
−
l
o
g
e
(
4
)
1
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