Mathematical Methods for Physicists: A concise introduction - Site Map

ORDINARY DIFFERENTIAL EQUATIONS
General solutions of the second-order equations
Suppose that we can ®nd one solution, y p …t† say, of Eq. (2.15):
…D 2 ‡ aD ‡ b†y p …t† ˆf …t†:
…2:15a†
Then on de®ning
y c …t† ˆy…t†y p …t†
we ®nd by subtracting Eq. (2.15a) from Eq. (2.15) that
…D 2 ‡ aD ‡ b†y c …t† ˆ0:
That is, y c …t† satis®es the corresponding homogeneous equation (2.16), and it is
known as the complementary function y c …t† of non-homogeneous equation (2.15).
while the solution y p …t† is called a particular integral of Eq. (2.15). Thus, the
general solution of Eq. (2.15) is given by
y…t† ˆAy c …t†‡By p …t†:
…2:17†
Finding the complementary function
Clearly the complementary function is independent of f …t†, and hence has nothing
to do with the behavior of the system in response to the external applied in¯uence.
What it does represent is the free motion of the system. Thus, for example, even
without external forces applied, a spring can oscillate, because of any initial
displacement and/or velocity. Similarly, had a capacitor already been charged
at t ˆ 0, the circuit would subsequently display current oscillations even if there
is no applied voltage.
In order to solve Eq. (2.16) for y c …t†, we ®rst consider the linear ®rst-order
equation
a dy ‡ by ˆ 0:
dt
Separating the variables and integrating, we obtain
y ˆ Ae bt=a ;
where A is an arbitrary constant of integration. This solution suggests that Eq.
(2.16) might be satis®ed by an expression of the type
y ˆ e pt ;
where p is a constant. Putting this into Eq. (2.16), we have
e pt …p 2 ‡ ap ‡ b† ˆ0:
Therefore y ˆ e pt is a solution of Eq. (2.16) if
p 2 ‡ ap ‡ b ˆ 0:
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