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The solution of
(
1
+
x
2
)
d
y
d
x
+
2
x
y
−
4
x
2
=
0
is :
(
a
)
3
x
(
1
+
y
2
)
=
4
y
3
+
c
(
b
)
3
y
(
1
+
x
2
)
=
4
x
3
+
c
(
c
)
3
x
(
1
−
y
2
)
=
4
y
3
+
c
(
d
)
3
y
(
1
+
y
2
)
=
4
x
3
+
c
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