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Unformatted text preview: t by partial integration
(since it contains derivatives with respect to each variable). So, we’ll use the
method of characteristics instead. To ﬁnd the “base characteristics”, we need
to solve the DE
That’s easy: y = x/2+C . Expressing this in implicit form, we have x 2y = C2 ,
(these are the characteristics) which tells us that we should let one of our new
variables be x 2y . So, let’s let
⇠=x 2y, ⌘ = y. With this choice, we have
u x = u ⇠ ⇠ x + u ⌘ ⌘x = u ⇠
u y = u ⇠ ⇠ y + u ⌘ ⌘y =
ˆ 2 u⇠ + u⌘
ˆ and this converts the original PDE into the form
2ˆ⇠ + ( 2u⇠ + u⌘ ) + u = 0,
ˆ This is an equation we can solve! We just need to observe how similar it is
to the ODE
= u. Since we know that the general solution to this is (by
inspection) u = Ce ⌘ , we know immediately that the solution to our PDE is
u = f (⇠ ) e
ˆ ⌘ where f is an arbitrary function of one variable.
Returning to the original variables, w...
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- Fall '14