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JEE Main 11 Jan 2019 Shift 1 Solved Paper
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© examsnet.com
Question : 53
Total: 90
If y(x) is the solution of the differential equation
d
y
d
x
+
(
2
x
+
1
x
)
y =
e
−
2
x
, x > 0
where y (1) =
1
2
e
−
2
, then :
y(x) is decreasing in (0, 1)
y(x) is decreasing in
(
1
2
,
1
)
y
(
l
o
g
e
2
)
=
l
o
g
e
2
4
y
(
l
o
g
e
2
)
=
l
o
g
e
4
Validate
Solution:
d
y
d
x
+
(
2
x
+
1
x
)
y =
e
−
2
x
I.F. =
e
∫
(
2
x
+
1
x
)
d
x
=
e
∫
(
2
+
1
x
)
d
x
=
e
2
x
+
l
n
x
=
e
2
x
. x
So ,
y
(
x
e
2
x
)
=
∫
e
−
2
x
.
x
e
2
x
+ C
⇒
x
y
e
x
= ∫ x dx + C
⇒
2
x
y
e
2
x
=
x
2
+ 2C
It passes through
(
1
,
1
2
e
−
2
)
we get C = 0
y =
x
e
−
2
x
2
⇒
d
y
d
x
=
1
2
e
−
2
x
(- 2x + 1)
⇒ f (x) is decreasing in
(
1
2
,
1
)
y
(
l
o
g
e
2
)
=
(
l
o
g
e
2
)
e
−
2
(
l
o
g
e
2
)
2
=
1
2
l
o
g
e
2
© examsnet.com
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