Mathematical Methods for Physicists: A concise introduction - Site Map

ORDINARY DIFFERENTIAL EQUATIONS
and the general solution can be written as
y ˆ Ae p 1t ‡ Be p 2t
ˆ e rt ‰…A ‡ B† cos st ‡ i…A B† sin stŠ
ˆ e rt ‰A 0 cos st ‡ B 0 sin stŠ
…2:20†
with A 0 ˆ A ‡ B; B 0 ˆ i…A B†:
The solution (2.20) may be expressed in a slightly di€erent and often more
useful form by writing B 0 =A 0 ˆ tan . Then
y ˆ…A 2 0 ‡ B 2 0† 1=2 e rt …cos cos st ‡ sin sin st† ˆCe rt cos…st †;
where C and are arbitrary constants.
…2:20a†
Example 2.10
Solve the equation …D 2 ‡ 4D ‡ 13†y ˆ 0, given that y ˆ 1 and y 0 ˆ 2 when t ˆ 0.
Solution: The auxiliary equation is p 2 ‡ 4p ‡ 13 ˆ 0, and hence p ˆ2 3i.
The general solution is therefore, from Eq. (2.20),
y ˆ e 2t …A 0 cos 3t ‡ B 0 sin 3t†:
Since y ˆ l when t ˆ 0, we have A 0 ˆ 1. Now
y 0 ˆ2e 2t …A 0 cos 3t ‡ B 0 sin 3t†‡3e 2t …A 0 sin 3t ‡ B 0 cos 3t†
and since y 0 ˆ 2 when t ˆ 0, we have 2 ˆ2A 0 ‡ 3B 0 . Hence B 0 ˆ 4=3, and the
solution is
3y ˆ e 2t …3 cos 3t ‡ 4 sin 3t†:
(iii) Coincident roots
When a 2 ˆ 4b, the auxiliary equation yields only one value for p, namely
p ˆ ˆa=2, and hence the solution y ˆ Ae t . This is not the general solution
as it does not contain the necessary two arbitrary constants. In order to obtain the
general solution we proceed as follows. Assume that y ˆ ve t , where v is a function
of t to be determined. Then
y 0 ˆ v 0 e t ‡ ve t ; y 00 ˆ v 00 e t ‡ 2v 0 e t ‡ 2 ve t :
Substituting for y; y 0 , and y 00 in the di€erential equation we have
and hence
e t ‰v 00 ‡ 2v 0 ‡ 2 v ‡ a…v 0 ‡ v†‡bvŠ ˆ0
v 00 ‡ v 0 …a ‡ 2†‡v… 2 ‡ a ‡ b† ˆ0:
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