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Byju's Answer
Standard XII
Mathematics
Derivative of Standard Functions
The solution ...
Question
The solution of
x
2
d
y
−
y
2
d
x
−
x
y
2
(
x
−
y
)
d
y
=
0
A
ln
∣
∣
∣
x
−
y
x
y
∣
∣
∣
=
y
2
2
−
c
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B
ln
∣
∣
∣
x
y
x
−
y
∣
∣
∣
=
x
2
2
+
c
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C
ln
∣
∣
∣
x
−
y
x
y
∣
∣
∣
=
x
2
2
+
c
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D
ln
∣
∣
∣
x
−
y
x
y
∣
∣
∣
=
x
+
c
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Solution
The correct option is
A
ln
∣
∣
∣
x
−
y
x
y
∣
∣
∣
=
y
2
2
−
c
x
2
d
y
−
y
2
d
x
=
x
y
(
x
−
y
)
(
y
d
y
)
let
d
y
d
x
=
y
′
x
2
y
′
−
y
2
=
x
y
(
x
−
y
)
(
y
y
′
)
it appears that we need and subtract the terms to manage beause u can see on the RHS iyt is obtained by some
ln
function resulting with terms
x
y
,
(
x
−
y
)
x
2
y
′
−
y
2
+
x
y
y
′
−
x
y
y
′
+
x
y
−
x
y
=
x
y
y
′
−
x
y
−
y
2
+
x
y
−
x
y
y
′
+
x
2
y
′
=
x
y
(
y
′
−
1
)
−
(
y
−
x
)
(
y
+
x
y
′
)
∫
[
x
y
x
−
y
×
x
y
(
1
−
y
′
)
−
(
x
−
y
)
(
y
+
x
y
′
)
(
x
y
)
2
]
d
x
=
∫
y
y
′
d
x
ln
∣
∣
∣
x
−
y
x
y
∣
∣
∣
=
y
2
2
−
c
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0
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