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JEE Mains 1-Feb-2023 Shift 2 Solved Paper
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© examsnet.com
Question : 74
Total: 90
Let
α
x
=
exp
(
x
β
y
γ
)
be the solution of the differential equation
2
x
2
y
d
y
−
(
1
−
x
y
2
)
dx
=
0
,
x
>
0
,
y
(
2
)
=
√
log
e
2
. Then
α
+
β
−
γ
equals :
[1-Feb-2023 Shift 2]
1
−
1
0
3
Validate
Solution:
α
x
=
e
x
8
⋅
y
y
2
x
2
y
d
y
d
x
=
1
−
x
⋅
y
2
y
2
=
t
x
2
d
t
d
x
=
1
−
x
t
d
t
d
x
+
t
x
=
1
x
2
I.F.
=
e
t
n
x
=
x
t
(
x
)
=
∫
1
x
2
⋅
x
dx
y
2
⋅
x
=
ℓ
nx
+
C
∴
2
.
ln
2
=
ln
2
+
C
∴
C
=
ℓ
n
2
Hence,
xy
2
=
ℓ
n
2
x
∴
2
x
=
e
x
⋅
y
2
Hence
α
=
2
,
β
=
1
,
γ
=
2
© examsnet.com
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