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Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29 in terms of the fundamental matrix rt
0 Φ(t) = exp (D t) = .
0 0 ···
··· er2 t
ern t The desired solution is x(t) = T y(t). . Example. Consider the system
x = A x + g(t)
dt in terms of
3t We outline two methods to ﬁnding the general solution to this system.
First we compute the matrices T and D. It is easy to check that A has eigenvalues r1 = −3 and r2 = −1;
and corresponding eigenvectors
There are two methods one may choose from: #1. Find the general solution to the system y = D y + h(t) in terms of
1 2 e−t − 3 t
exp (D t) =
h(t) = T−1 g(t) =
2 2 e−t + 3 t
then compute the product x(t) = T y(t).
#2. Compute the solution x(t) directly from the fundamental matrix
1 e−t + e−3t e−t − e−3t
exp (A t) = T exp (D t) T−1 =
2 e−t − e−3t e−t + e−3t
Nonconstant Coeﬃcients. We conclude our discussion of the general system of ﬁrst order linear equations
in the form
x = P(t) x + g(t).
Recall that if P(t) = A is a constant n × n matrix, then the general solution to this system is in the form
Ψ(τ )−1 g(τ ) dτ ;
particular solution in terms of a constant vector c and the fundamental matrix
Ψ(t) = exp (A t) =⇒ d
Ψ = A Ψ.
dt Today we discuss how to ﬁnd the solution if P(t) is not a constant matrix.
Denote the matrix
P(τ ) dτ
Q = P(t).
dt We will show that if P and Q commute i.e., if P Q = Q P, then a fundamental matrix is
Ψ = exp Q(t) First, observe two facts:
• Ψ(t)−1 = exp −Q(t) . =⇒ d
Ψ = P(t) Ψ.
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