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JEE Main 6 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 18
Total: 89
Let
y
=
y
(
x
)
be the solution of the differential equation
(
2
x
log
e
x
)
d
y
d
x
+
2
y
=
3
x
log
e
x
,
x
>
0
and
y
(
e
−
1
)
=
0
. Then,
y
(
e
)
is equal to
[6 Apr 2024 Shift 1]
−
3
2
e
−
2
3
e
−
3
e
−
2
e
Validate
Solution:
👈: Video Solution
d
y
d
x
+
y
x
ln
x
=
3
2
x
2
∴
I.F.
=
e
∫
1
x
ln
x
dx
=
e
ln
(
ln
(
x
)
)
=
ln
x
∴
y
ln
x
=
∫
3
ln
x
2
x
2
dx
=
3
ln
x
2
∫
x
−
2
dx
−
∫
(
3
2
x
⋅
∫
x
−
2
dx
)
dx
=
3
ln
x
2
(
−
1
x
)
−
∫
3
2
x
(
−
1
x
)
dx
y
⋅
ln
x
=
−
3
ln
x
2
x
−
3
2
x
+
C
∵
y
(
e
−
1
)
=
0
∴
0
(
−
1
)
=
3
e
2
−
3
e
2
+
C
⇒
C
=
0
∴
y
=
−
3
ln
x
2
x
−
3
2
x
∴
y
(
e
)
=
−
3
2
e
−
3
2
e
=
−
3
e
© examsnet.com
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