Lecture Notes in Differential Equations - Bruce E. Shapiro

306 LESSON 31. LINEAR SYSTEMS
and that if c = 0 but b ≠ 0,
u ′′ − (a + d)u ′ + adu = 0 (31.25)
In every case in which one of the equations is second order, the characteristic
equation of the differential equation is identical to the characteristic
equation of the matrix, and hence the solutions are linear combinations of
e λ1t and e λ2t , if the eigenvalues are distinct, or are linear combinations of
e λt and te λt if the eigenvalue is repeated. Furthermore, if one (or both) the
equations turn out to be first order, the solution to that equation is still
e λt where λ is one of the eigenvalues of A.
Theorem 31.2. Let
( ) a b
A =
c d
(31.26)
Then the solution of y ′ = Ay is given be one of the following four cases.
1. If b = c = 0, then the eigenvalues of A are a and d, and
}
y 1 = C 1 e at
y 2 = C 2 e dt
(31.27)
2. If b ≠ 0 but c = 0, then the eigenvalues of A are a and d, and
{
C 1 e at + C 2 e dt a ≠ d
y 1 =
(C 1 + C 2 t)e at a = d
⎫
⎪⎬
⎪ ⎭
(31.28)
y 2 = C 3 e dt
3. If b = 0 but c ≠ 0, then the eigenvalues of A are a and d, and
y 1 = C 1 e dt
⎫
{
⎪⎬
C 2 e at + C 3 e dt a ≠ d
(31.29)
y 2 =
⎪⎭
(C 2 + C 3 t)e at a = d
4. If b ≠ 0 and c ≠ 0, then the eigenvalues of A are
λ = 1 [
T ± √ ]
T
2
2 − 4∆
(31.30)
and the solutions are
(a) If λ 1 = λ 2 = λ
y i = (C i1 + C i2 t)e λt (31.31)