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b. The set of ODEs is
dx
dt
=
t
2
u
du
dt
=
−
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
)=
−
t
+
K
or
u
(
x, t
Ce
−
t
At
t
=0thesolutionis
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 (Frst ODE)
dx
dt
=
t
2
f
(
x
0
)
e
−
t
or
Z
dx
=
f
(
x
0
)
Z
t
2
e
−
t
dt
Integrate and continue as in part a of this problem
x
=
f
(
x
0
)
h
−
t
2
e
−
t
−
2
te
−
t
−
2
e
−
t
+
C
i
±or
t
=0weget
x
0
=
f
(
x
0
)[
−
2+
C
]
Thus
Cf
(
x
0
x
0
+2
f
(
x
0
)
and the characteristics are
x
=
f
(
x
0
)
h
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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