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Byju's Answer
Standard XII
Mathematics
Solving Linear Differential Equations of First Order
The solution ...
Question
The solution of the differential equation:
(
x
2
−
1
)
d
y
d
x
+
2
x
y
=
1
(
x
2
−
1
)
A
y
(
x
2
−
1
)
=
1
2
log
∣
∣
∣
x
−
1
x
+
1
∣
∣
∣
+
C
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B
y
(
x
2
+
1
)
=
1
2
log
∣
∣
∣
x
−
1
x
+
1
∣
∣
∣
−
C
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C
y
(
x
2
−
1
)
=
5
2
log
∣
∣
∣
x
−
1
x
+
1
∣
∣
∣
+
C
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D
None of these
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Solution
The correct option is
A
y
(
x
2
−
1
)
=
1
2
log
∣
∣
∣
x
−
1
x
+
1
∣
∣
∣
+
C
(
x
2
−
1
)
d
y
d
x
+
2
x
y
=
1
x
2
−
1
⇒
d
d
x
[
y
(
x
2
−
1
)
]
=
1
x
2
−
1
⇒
y
(
x
2
−
1
)
=
∫
d
x
x
2
−
1
2
⇒
y
(
x
2
−
1
)
=
1
2
ln
∣
∣
∣
x
−
1
x
+
1
∣
∣
∣
+
c
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is equal to