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Byju's Answer
Standard XII
Mathematics
Solving Linear Differential Equations of First Order
The solution ...
Question
The solution of
d
y
d
x
=
e
x
−
y
+
e
2
l
o
g
x
−
y
A
e
y
=
e
x
+
x
2
3
+
c
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B
e
y
=
e
x
+
x
3
3
+
c
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C
e
y
=
e
x
+
l
o
g
x
+
c
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D
y
=
e
3
x
+
y
3
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Solution
The correct option is
B
e
y
=
e
x
+
x
3
3
+
c
d
y
d
x
=
e
x
−
y
+
e
2
l
o
g
x
−
y
⇒
e
y
d
y
=
(
e
x
+
x
2
)
d
x
⇒
e
y
=
e
x
+
x
3
/
3
+
c
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Q.
The solution of the equation
d
y
d
x
−
e
x
−
y
+
x
2
e
−
y
is
[MP PET 2004]