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Byju's Answer
Standard XII
Mathematics
Variable Separable Method
The solution ...
Question
The solution of
(
1
+
e
x
)
y
d
y
=
e
x
d
x
is:
A
y
2
=
log
c
(
e
x
+
1
)
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B
y
2
2
=
log
c
e
x
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C
y
2
2
=
log
c
(
e
x
+
1
)
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D
2
y
=
log
c
(
e
x
+
1
)
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Solution
The correct option is
D
y
2
2
=
log
c
(
e
x
+
1
)
(
1
+
e
x
)
y
d
y
=
e
x
d
x
∫
y
d
y
=
∫
(
e
x
1
+
e
x
)
d
x
+
c
⇒
y
2
2
=
log
(
1
+
e
x
)
c
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0
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