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Byju's Answer
Standard XII
Mathematics
Higher Order Equations
The solution ...
Question
The solution of
(
1
x
−
1
(
x
−
y
)
2
)
d
x
+
(
1
(
x
−
y
)
2
−
1
y
)
d
y
=
0
.
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Solution
∙
Given the difference equation is
{
1
x
−
1
(
x
−
y
)
2
}
d
x
+
{
1
(
x
−
y
)
2
−
1
y
}
d
y
=
0
on,
d
y
d
x
−
d
y
y
+
d
y
−
d
x
(
x
−
y
)
2
=
0
on,
d
x
x
−
d
y
y
+
d
(
y
−
x
)
(
y
−
x
)
2
=
0
Now, integrating both sides we have,
log
|
x
|
−
log
|
y
|
+
(
−
1
y
−
x
)
=
c
on,
log
|
x
|
−
log
|
y
|
=
1
(
y
−
x
)
+
c
where
′
c
′
is integrating constant.
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Similar questions
Q.
Solution of the differential equation
{
1
x
−
y
2
(
x
−
y
)
2
}
d
x
+
{
x
2
(
x
−
y
)
2
−
1
y
}
d
y
=
0
is
(where
c
is arbitrary constant).
Q.
Consider the differential equation
y
2
d
x
+
(
x
−
1
y
)
d
y
=
0.
If y (1) =1, then x is given by:
Q.
Find a particular solution of each of the following differential equations:
(i)
1
+
x
2
d
y
d
x
+
2
x
y
=
1
1
+
x
2
;
y
=
0
,
when
x
=
1
(ii) (x + y) dy + (x − y) dx = 0; y = 1 when x = 1
(iii) x
2
dy + (xy + y
2
) dx = 0; y = 1 when x = 1
Q.
STATEMENT -1 : Solution of
(
1
+
x
√
x
2
+
y
2
)
d
x
+
y
(
−
1
+
√
x
2
+
y
2
)
d
y
=
0
is
x
−
y
2
2
+
1
3
(
x
2
+
y
2
)
3
/
2
+
C
1
=
0
,
C
1
being arbitrary constant.
STATEMENT-2 : Solution of
x
d
y
−
y
d
x
=
√
x
2
−
y
2
d
x
is
s
i
n
−
1
(
y
x
)
=
l
n
x
+
C
2
,
C
2
being arbitrary constant.
Q.
Solve
(
1
+
y
2
)
+
(
x
−
e
tan
−
1
y
)
d
y
d
x
=
0
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