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Byju's Answer
Standard XII
Mathematics
General Solution of a Differential Equation
Verify that ...
Question
Verify that
x
y
=
log
y
+
c
is a solution of the differential equation
(
x
y
−
1
)
d
y
d
x
+
y
2
=
0
.
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Solution
Given
x
y
=
l
o
g
y
+
c
Differentiate on both sides
⇒
x
y
′
+
y
=
1
y
y
′
⇒
y
′
=
y
2
1
−
x
y
We will substitute
y
′
in given DE and we will verify
⇒
(
x
y
−
1
)
y
2
1
−
x
y
+
y
2
⇒
−
y
2
+
y
2
=
0
=
R
H
S
Given solution is correct for above differential equation
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0
Similar questions
Q.
Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
1.
y
=
e
x
+
1
:
y
′′
−
y
=
0
2.
y
=
x
2
+
2
x
+
C
:
y
′
−
2
x
−
2
=
0
3.
y
=
cos
x
+
C
:
y
′
+
sin
x
=
0
4.
y
=
√
1
+
x
2
:
y
′
=
x
y
1
+
x
2
5.
y
=
A
x
:
x
y
=
y
(
x
≠
0
)
6.
y
=
x
sin
x
:
x
y
=
y
+
x
√
x
2
y
2
(
x
≠
0
a
n
d
x
>
y
o
r
x
<
y
)
7.
x
y
=
log
y
+
C
:
y
′
=
y
2
1
−
x
y
(
x
y
≠
1
)
8.
y
−
cos
y
=
x
:
(
y
sin
y
+
cos
y
+
x
)
y
′
=
y
9.
x
+
y
=
tan
−
1
y
:
y
2
y
′
+
y
2
+
1
=
0
10.
y
=
√
a
2
−
x
2
x
ϵ
(
−
a
,
a
)
:
x
+
y
d
y
d
x
=
0
(
y
≠
0
)
Q.
Verify that xy = a e
x
+ b e
−x
+ x
2
is a solution of the differential equation
x
d
2
y
d
x
2
+
2
d
y
d
x
-
x
y
+
x
2
-
2
=
0
.
Q.
Solve
(
x
y
−
1
)
d
y
d
x
+
y
2
=
0
.
Q.
The solution of the differential equation
d
y
d
x
=
x
y
x
2
+
y
2
is -
Q.
The solution of the differential equation
e
−
x
(
y
+
1
)
d
y
+
(
cos
2
x
+
sin
2
x
)
y
d
x
=
0
subjected to the condition that
y
=
1
when
x
=
0
is
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