solve (x^2-y)dx+(y^2-x)y=0 if it passes through the origin
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Answer:
(
x
2
−
y
)
d
x
+
(
y
2
−
x
)
d
y
=
0
Rewrite:
x
2
d
x
−
y
d
x
+
y
2
d
y
−
x
d
y
=
0
Rearrange the terms:
x
2
d
x
+
y
2
d
y
−
(
x
d
y
+
y
d
x
)
=
0
Since we know:
(
f
.
g
)
′
=
f
′
g
+
f
g
′
, in this problem
f
=
x
,
g
=
y
Hence,
x
d
y
+
y
d
x
=
d
(
x
y
)
x
2
d
x
+
y
2
d
y
−
d
(
x
y
)
=
0
Integrating both sides:
∫
x
2
d
x
+
∫
y
2
d
y
−
∫
d
(
x
y
)
=
C
Applying the common integrals:
x
3
3
+
y
3
3
−
x
y
=
C
It is given that the given equation passes through the origin. So, plug in
x
=
0
,
y
=
0
to get the value of the constant.
C
=
0
Hence,
x
3
3
+
y
3
3
−
x
y
=
0
Therefore, the general solution of
(
x
2
−
y
)
d
x
+
(
y
2
−
x
)
d
y
=
0
is:
x
3
3
+
y
3
3
=
x
y
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