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Unformatted text preview: 4. a. The set of ODEs is
du
= e−3x
dt dx
=c
dt
The solution of the ﬁrst is x = x0 + ct
Substituting x in the second ODE
du
= e−3(x0 +ct)
dt
Now integrate
u(x, t) = K + e−3x0 1 −3ct
e
−3c At t = 0 we get
f (x0 ) = u(x0 , 0) = K + e−3x0 1
−3c Therefore the constant of integration K is
K = f (x0 ) + e−3x0 1
3c Substitute this K in the solution
u(x, t) = f (x0 ) + e−3x0 1
1 −3ct
− e−3x0
e
3c
3c Recall that x0 = x − ct thus
u(x, t) = f (x − ct) +
b. The set of ODEs is 1 −3x
1 −3(x−ct)
e
e
−
3c
3c dx
=t
dt du
=5
dt The solution of the ﬁrst is
x = x0 + 12
t
2 Now integrate the second ODE
u(x, t) = 5t + K
At t = 0 the solution is
u(x0 , 0) = f (x0 ) = K plug t = 0 in the solution u Thus when substituting for x0 in the solution
u(x, t) = 5t + f x − 251 12
t
2 ...
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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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