Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29 • By the Product Rule for Matrices d 2
dQ dQ
Q =Q·
+
· Q = QP + PQ = 2QP
dt
dt
dt =⇒ 3 d k
Q = k Qk−1 P ;
dt so that we have the formula
∞
∞
d
1 d k 1
exp Q(t) =
Q=
k Qk−1 P = exp Q(t) P(t).
dt
k ! dt
k!
k=0 k=0 We attempt to solve the diﬀerential equation as before by using integrating factors:
d
x = P(t) x + g(t)
dt d
x − P(t) x = g(t)
dt
d
exp −Q(t) ·
x + exp −Q(t) −P(t) · x = exp −Q(t) g(t)
dt
d
exp −Q(t) · x = exp −Q(t) g(t)
dt
d
−1
Ψ(t) · x = Ψ(t)−1 g(t)
dt
t
−1
Ψ(t) · x(t) = c +
Ψ(τ )−1 g(τ ) dτ
t
x(t) = Ψ(t) c + Ψ(t)
Ψ(τ )−1 g(τ ) dτ for some constant vector c. Variation of Parameters. Consider the system of diﬀerential equations
d
x = P(t) x + g(t).
dt
Say that we can ﬁnd
i. a nonsingular matrix Ψ satisfying
d
Ψ = P(t) Ψ;
dt
ii. a particular solution x(p) (t) satisfying
d (p)
x = P(t) x(p) + g(t).
dt
It is easy to see that the general solution to the system is in the form x(t) = x(c) (t) + x(p) (t), where
x(c) (t) = Ψ(t) c is the general solution to the associated homogeneous equation. We have seen how to
compute a fundamental matrix
t
Ψ(t) = exp
P(τ ) dτ
under certain conditions on the coeﬃcient matrix P(t). We discuss how to ﬁnd such a particular solution
x(p) (t) just with the knowledge of Ψ(t). We make a guess that this solution is in the form x(p) = Ψ(t) u(t)
for some function u(t) to be found. This is known as the method of Variation of Parameters.
Indeed, by the Product Rule, we have
d (p)
d
d
x =Ψ·
u+
Ψ · u.
dt
dt
dt ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.
 Spring '09
 EdrayGoins
 Product Rule, Matrices

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