Mathematical Methods for Physicists: A concise introduction - Site Map

SECOND-ORDER EQUATIONS WITH CONSTANT COEFFICIENTS
Now
2 ‡ a ‡ b ˆ 0; and a ‡ 2 ˆ 0
so that
v 00 ˆ 0:
Hence, integrating gives
v ˆ At ‡ B;
where A and B are arbitrary constants, and the general solution of Eq. (2.16) is
y ˆ…At ‡ B†e t
…2:21†
Example 2.11
Solve the equation (D 2 4D ‡ 4†y ˆ 0 given that y ˆ 1andDy ˆ 3 when t ˆ 0:
Solution: The auxiliary equation is p 2 4p ‡ 4 ˆ…p 2† 2 ˆ 0 which has one
root p ˆ 2. The general solution is therefore, from Eq. (2.21)
y ˆ…At ‡ B†e 2t :
Since y ˆ 1 when t ˆ 0, we have B ˆ 1. Now
y 0 ˆ 2…At ‡ B†e 2t ‡ Ae 2t
and since Dy ˆ 3 when t ˆ 0,
3 ˆ 2B ‡ A:
Hence A ˆ 1 and the solution is
y ˆ…t ‡ 1†e 2t :
Finding the particular integral
The particular integral is a solution of Eq. (2.15) that takes the term f …t† on the
right hand side into account. The complementary function is transient in nature,
so from a physical point of view, the particular integral will usually dominate the
response of the system at large times.
The method of determining the particular integral is to guess a suitable functional
form containing arbitrary constants, and then to choose the constants to
ensure it is indeed the solution. If our guess is incorrect, then no values of these
constants will satisfy the di€erential equation, and so we have to try a di€erent
form. Clearly this procedure could take a long time; fortunately, there are some
guiding rules on what to try for the common examples of f (t):
77