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Joint Entrance Examination
Mathematics
Differential Equations
Question 298
Question 298 of 490
Medium
MCQ
Let
y
(
x
)
y(x)
y
(
x
)
solve
(
x
log
x
)
d
y
d
x
+
y
=
2
x
log
x
(x\log x)\dfrac{dy}{dx}+y=2x\log x
(
x
lo
g
x
)
d
x
d
y
+
y
=
2
x
lo
g
x
,
x
≥
1
x\ge1
x
≥
1
. Then
y
(
e
)
y(e)
y
(
e
)
equals:
A
2
B
2
e
2e
2
e
C
e
e
e
D
0
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Solutions · 1
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Official solution
I.F.
=
log
x
=\log x
=
lo
g
x
.
y
log
x
=
2
[
x
log
x
−
x
]
+
c
y\log x=2[x\log x-x]+c
y
lo
g
x
=
2
[
x
lo
g
x
−
x
]
+
c
. Using
x
=
1
x=1
x
=
1
:
0
=
−
2
+
c
⟹
c
=
2
0=-2+c \implies c=2
0
=
−
2
+
c
⟹
c
=
2
. At
x
=
e
x=e
x
=
e
:
y
=
2
(
0
)
+
2
=
2
y=2(0)+2 = 2
y
=
2
(
0
)
+
2
=
2
.
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