Lecture Notes in Differential Equations - Bruce E. Shapiro

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is
where
λ 2 − T λ + ∆ = 0 (31.14)
}
T = a + d = trace(A)
∆ = ad − bc = det(A)
(31.15)
The roots λ 1 and λ 2 of (31.14) are the eigenvalues of A; we will make use
of this fact shortly.
By rearranging the system (31.12) it is possible to separate the two first
order equations into equivalent second order equations with the variables
separated.
If both b = 0 and c = 0, we have two completely independent first order
equations,
u ′ = au, v ′ = dv (31.16)
If b ≠ 0, we can solve the first of equations (31.12) for v,
v = 1 b [u′ − au] (31.17)
Substituting (31.17) into both sides of the second of equations (31.12),
Rearranging,
1
b [u′′ − au ′ ] = v ′ = cu + dv = cu + d b [u′ − au] (31.18)
u ′′ − (a + d)u ′ + (da − bc)u = 0 (31.19)
If b = 0 and c ≠ 0 the first of equations (31.12) becomes
and the second one can be solved for u,
Substituting (31.21) into (31.20),
Rearranging,
u ′ = au (31.20)
u = 1 c [v′ − dv] (31.21)
1
c [v′′ − du ′ ] = u ′ = au = a u [v′ − dv] (31.22)
By a similar process we find that if c ≠ 0
v ′′ − (a + d)v ′ + adv = 0 (31.23)
v ′′ − (a + d)v ′ + (ad − bc)v = 0 (31.24)