5. a. The set of ODEs is
dx
dt
=
−
u
2
du
dt
=3
u
The solution of the Frst ODE requires the yet unknown
u
thus we tackle the second ODE
du
u
dt
Now integrate this
ln
u
(
x, t
)=3
t
+
K
or
u
(
x, t
)=
Ce
3
t
At
t
=0thesolutionis
u
(
x
0
,
0) =
f
(
x
0
C
Thus
u
(
x, t
f
(
x
0
)
e
3
t
Now substitute this solution in the characteristic equation (Frst ODE)
dx
dt
=
−
f
(
x
0
)
e
3
t
2
=
−
(
f
(
x
0
))
2
e
6
t
Integrating
x
=
−
(
f
(
x
0
))
2
Z
e
6
t
dt
=
−
1
6
(
f
(
x
0
))
2
e
6
t
+
K
±or
t
=0weget
x
0
=
−
1
6
(
f
(
x
0
))
2
+
K
Thus
K
=
x
0
+
1
6
(
f
(
x
0
))
2
and the characteristics are
x
=
−
1
6
(
f
(
x
0
))
2
e
6
t
+
x
0
+
1
6
(
f
(
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 Fall '08
 BELL,D
 Ode, Characteristic polynomial, Substitute good, Complex number, Characteristic

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