Lecture Notes in Differential Equations - Bruce E. Shapiro

313
y = 1 [( ) ( ) ] ( )
4 1
e 2t 1 −1
+
e −3t 6
5 4 1 −4 4 5
= 1 [( ) ( ]
29
e 2t 1
+ e
5 29 −4) −3t
= 29 ( ) 1
5 e2t + 1 ( ) 1
1 5 e−3t −4
(31.87)
(31.88)
(31.89)
We observe in passing that the last line has the form y = ax 1 e λ1t +bx 2 e λ2t .
Properties of Solutions
Theorem 31.6. y = e λt v is a solution of the system y ′ = Ay if and only
if v is an eigenvector of A with eigenvalue λ.
Proof. Suppose that {λ, v} are an eigenvalue/eigenvector pair for the matrix
A. Then
Av = λv (31.90)
and
d (
e λt v ) = λe λt v = e λt Av = A ( e λt v ) (31.91)
dt
Hence y = e λt v is a solution of y ′ = Ay.
To prove the converse, suppose that y = e λt v for some number λ ∈ C and
some vector v, is a solution of y ′ = Ay. Then y = e λt vmust satisfy the
differential equation, so
λe λt v = y ′ = Ay = Ae λt v (31.92)
Dividing by the common scalar factor of e λt , which can never be zero, gives
Av = λv. Hence λ is an eigenvalue of A with eigenvector v.
Theorem 31.7. The matrix e At is a fundamental matrix of y ′ = Ay.
Proof.
Let W = e At . From the previous theorem,
W ′ = d dt eAt = Ae At = AW (31.93)