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JEE Mains 11-Apr-2023 Shift 1 Solved Paper
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© examsnet.com
Question : 18
Total: 90
Let
y
=
y
(
x
)
be a solution curve of the differential equation.
(
1
−
x
2
y
2
)
dx
=
y
dx
+
x
d
y
. If the line
x
=
1
intersects the curve
y
=
y
(
x
)
at
y
=
2
and the line
x
=
2
intersects the curve
y
=
y
(
x
)
at
y
=
α
, then a value of
α
is :
[11-Apr-2023 shift 1]
1
+
3
e
2
2
(
3
e
2
−
1
)
1
−
3
e
2
2
(
3
e
2
+
1
)
3
e
2
2
(
3
e
2
−
1
)
3
e
2
2
(
3
e
2
+
1
)
Validate
Solution:
(
1
−
x
2
y
2
)
dx
=
y
dx
+
x
d
y
,
y
(
1
)
=
2
y
(
2
)
=
∞
d
x
=
d
(
x
y
)
1
−
(
x
y
)
2
∫
dx
=
∫
d
(
x
y
)
1
−
(
x
y
)
2
x
=
1
2
ln
|
1
+
x
y
1
−
x
y
|
+
C
Put
x
=
1
and
y
=
2
:
1
=
1
2
ln
|
1
+
2
1
−
2
|
+
C
C
=
1
−
1
2
ln
3
Now put
x
=
2
:
2
=
1
2
ln
|
1
+
2
α
1
−
2
α
|
+
1
−
1
2
ln
3
1
+
1
2
ln
3
=
1
2
|
1
+
2
α
1
−
2
α
|
2
+
ln
3
=
|
1
+
2
α
1
−
2
α
|
|
1
+
2
α
1
−
2
α
|
=
3
e
2
1
+
2
α
1
−
2
α
=
3
e
2
,
−
3
e
2
1
+
2
α
1
−
2
α
=
3
e
2
⇒
α
=
3
e
2
−
1
2
(
3
e
2
+
1
)
And
1
+
2
α
1
−
2
α
=
−
3
e
2
⇒
α
=
3
e
2
+
1
2
(
3
e
2
−
1
)
© examsnet.com
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