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Mathematics
Problem
The general solution of
y
′
′
−
y
=
0
y''-y=0
y
′′
−
y
=
0
is
y
=
C
1
e
x
+
C
2
e
−
x
y=C_1e^{x}+C_2e^{-x}
y
=
C
1
e
x
+
C
2
e
−
x
. Find the particular solution satisfying
y
(
0
)
=
3
y(0)=3
y
(
0
)
=
3
and having a horizontal tangent at
x
=
0
x=0
x
=
0
. a)
y
=
2
e
x
+
e
−
x
y=2e^{x}+e^{-x}
y
=
2
e
x
+
e
−
x
b)
y
=
3
2
(
e
x
+
e
−
x
)
y=\dfrac{3}{2}\left(e^{x}+e^{-x}\right)
y
=
2
3
(
e
x
+
e
−
x
)
c)
y
=
3
2
(
e
x
−
e
−
x
)
y=\dfrac{3}{2}\left(e^{x}-e^{-x}\right)
y
=
2
3
(
e
x
−
e
−
x
)
d)
y
=
3
e
x
y=3e^{x}
y
=
3
e
x
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