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Byju's Answer
Standard XII
Mathematics
Particular Solution of a Differential Equation
The solution ...
Question
The solution of
y
d
x
−
x
d
y
+
l
o
g
x
d
x
=
0
is:
A
(
x
+
1
)
+
log
x
=
c
y
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B
y
+
1
+
log
x
=
c
x
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C
(
y
+
1
)
2
2
+
log
x
=
c
y
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D
(
y
+
1
)
2
=
log
c
x
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Solution
The correct option is
A
y
+
1
+
log
x
=
c
x
y
d
x
=
x
d
y
−
l
o
g
x
d
x
y
=
x
d
y
d
x
−
l
o
g
x
⇒
l
o
g
x
x
=
d
y
d
x
−
1
/
x
⇒
l
o
g
x
x
2
=
d
d
x
(
y
/
x
)
∫
l
o
g
x
x
2
d
x
=
∫
d
(
y
/
x
)
+
c
1
∫
l
o
g
x
x
2
d
x
=
y
/
x
+
c
1
l
e
t
l
o
g
x
=
t
⇒
1
/
x
d
x
=
d
t
x
=
e
t
1
/
x
=
e
−
t
⇒
∫
t
e
−
t
d
t
=
−
t
e
−
t
−
e
−
t
=
−
(
1
+
t
e
t
)
=
−
(
1
+
l
o
g
x
)
x
⇒
−
(
1
+
l
o
g
x
x
)
=
y
/
x
+
c
1
⇒
1
+
l
o
g
x
+
y
+
c
1
x
=
0
p
u
t
c
1
=
−
c
⇒
1
+
l
o
g
x
+
y
=
c
x
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