b.
The set of ODEs is
dx
dt
=
t
2
u
du
dt
=
−
u
The solution of the first ODE requires the yet unknown
u
thus we tackle the second ODE
du
u
=
−
dt
Now integrate this
ln
u
(
x, t
) =
−
t
+
K
or
u
(
x, t
) =
C e
−
t
At
t
= 0 the solution is
u
(
x
0
,
0) =
f
(
x
0
) =
C
Thus
u
(
x, t
) =
f
(
x
0
)
e
−
t
Now substitute this solution in the characteristic equation (first ODE)
dx
dt
=
t
2
f
(
x
0
)
e
−
t
or
dx
=
f
(
x
0
)
t
2
e
−
t
dt
Integrate and continue as in part a of this problem
x
=
f
(
x
0
)
−
t
2
e
−
t
−
2
t e
−
t
−
2
e
−
t
+
C
For
t
= 0 we get
x
0
=
f
(
x
0
) [
−
2 +
C
]
Thus
C f
(
x
0
) =
x
0
+ 2
f
(
x
0
)
and the characteristics are
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 Fall '08
 BELL,D
 Ode, Characteristic polynomial, Substitute good, Complex number, Characteristic

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