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Unformatted text preview: 6. The set of ODEs is du
dx
= t2 u
=5
dt
dt
The solution of the ﬁrst ODE requires the yet unknown u thus we tackle the second ODE
du = 5 dt
Now integrate this
u(x, t) = 5t + K
At t = 0 the solution is
u(x0 , 0) = f (x0 ) = x0 = K
Thus
u(x, t) = x0 + 5t
Now substitute this solution in the characteristic equation (ﬁrst ODE)
dx
= 5t3 + x0 t2
dt
Integrate
x= 1
54
t + t3 x0 + C
4
3 For t = 0 we get
x0 = 0 + 0 + C
Thus
C = x0
and the characteristics are
x= 13
t +1
3 54
t+
4 Solve this for x0
x0 = x−
1+ The solution is then given by u(x, t) = 5t + 54
t
4
13
t
3 x−
1+ 256 54
t
4
13
t
3 x0 ...
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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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