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Mathematics
Problem
Solve the initial–value problem
d
y
d
x
=
(
1
+
y
2
)
e
x
,
y
(
0
)
=
0
\dfrac{dy}{dx}=(1+y^{2})e^{x},\quad y(0)=0
d
x
d
y
=
(
1
+
y
2
)
e
x
,
y
(
0
)
=
0
. Which of the following gives
y
y
y
as an explicit function of
x
x
x
? a)
y
=
tan
(
e
x
)
y=\tan(e^{x})
y
=
tan
(
e
x
)
b)
y
=
tan
(
e
x
−
1
)
y=\tan(e^{x}-1)
y
=
tan
(
e
x
−
1
)
c)
y
=
arctan
(
e
x
−
1
)
y=\arctan(e^{x}-1)
y
=
arctan
(
e
x
−
1
)
d)
y
=
tan
(
e
x
)
−
tan
1
y=\tan(e^{x})-\tan 1
y
=
tan
(
e
x
)
−
tan
1
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