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Unformatted text preview: he solution we need is u(x, t) = 5
e
x3 3t . Comment: you can see that the initial condition was chosen carefully so that
the method would work. However, for this particular pde, we can make the
method work for many other initial conditions! Here’s how:
• Suppose we are given u (x, 0) = f (x). • We’ve shown that u(x, t) = C5 x e t is a solution to the PDE for any value
of C5 and any value of . By the principle of superposition, then, any sum
of these solutions will also be a solution. AMATH 350 Assignment #8 Solutions  Fall 2013 Page 7 • This means that we if we construct an inﬁnite series, by letting take on
the integer values 0, 1, 2, . . . , and letting C be an arbitrary constant for
each one of these values, the series will also represent a solution, as long
1
X
as it converges! So, we may write u (x, t) =
Cn xn ent .
n=0 • If we take this expression and set t = 0, we ﬁnd that we can match
any initial...
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This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.
 Fall '12
 DavidHamsworth

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