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JEE Main 1-Sep-2021 Shift 2 Solved Paper
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© examsnet.com
Question : 67
Total: 90
If
y
=
y
(
x
)
is the solution curve of the differential equation
x
2
d
y
+
(
y
−
1
x
)
d
x
=
0
;
x
>
0
and
y
(
1
)
=
1
, then
y
(
1
2
)
is equal to
[1 Sep 2021 Shift 2]
3
2
−
1
√
e
3
+
1
√
e
3
+
e
3
−
e
Validate
Solution:
x
2
d
y
+
(
y
−
1
x
)
d
x
=
0
⇒
x
2
d
y
+
y
d
x
=
d
x
x
⇒
d
y
d
x
+
y
x
2
=
1
x
3
This in linear differential equations
IF
=
e
∫
1
x
2
d
x
=
e
−
1
x
⇒
y
.
e
−
1
∕
x
=
∫
e
−
1
∕
x
.
1
x
3
d
x
+
c
Let
−
1
x
=
t
⇒
1
x
2
d
x
=
d
t
⇒
y
.
e
−
1
∕
x
=
∫
−
t
e
t
d
t
+
c
=
−
(
t
e
t
−
e
t
)
+
c
⇒
y
e
−
1
∕
x
=
1
x
e
−
1
x
+
e
−
1
x
+
c
Put
x
=
1
,
y
=
1
(1)
e
−
1
=
e
−
1
1
+
e
−
1
+
c
⇒
c
=
−
e
−
1
Solution
y
.
e
−
1
∕
x
=
1
x
e
−
1
x
+
e
−
1
x
−
e
−
1
⇒
y
=
1
x
+
1
−
e
1
∕
x
e
Put
x
=
1
2
y
=
2
+
1
−
e
2
e
⇒
y
=
3
−
e
© examsnet.com
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