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BITSAT 2020 Solved Question Paper
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© examsnet.com
Question : 139
Total: 150
The solutions of
(
x
+
y
+
1
)
d
y
=
d
x
are
x
+
y
+
2
=
C
e
y
x
+
y
+
4
=
C
log
y
log
(
x
+
y
+
2
)
=
C
y
log
(
x
+
y
+
2
)
=
C
−
y
Validate
Solution:
Putting
x
+
y
+
1
=
u
,
we have
d
u
=
d
x
+
d
y
and the given equation reduces to
u
(
d
u
−
d
x
)
=
d
x
⇒
u
d
u
u
+
1
=
d
x
⇒
u
−
log
(
u
+
1
)
=
x
⇒
log
(
x
+
y
+
2
)
=
y
+
constant
⇒
x
+
y
+
2
=
C
e
y
© examsnet.com
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