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Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29 in terms of the fundamental matrix rt e1 0 Φ(t) = exp (D t) = . . . 0 0 ··· ··· .. . ··· er2 t . . . 0 0 0 . . . ern t The desired solution is x(t) = T y(t). . Example. Consider the system d x = A x + g(t) dt in terms of ￿ −2 A= 1 ￿ 1 , −2 ￿ −t ￿ 2e g(t) = . 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 ￿￿ ￿￿ ￿ ￿ ￿ ￿ 1 1 11 −3 0 (1) (2) ξ= , ξ= =⇒ T= , D= . −1 1 −1 1 0 −1 There are two methods one may choose from: #1. Find the general solution to the system y￿ = D y + h(t) in terms of ￿ −3t ￿ ￿ ￿ 1 2 e−t − 3 t e 0 exp (D t) = and h(t) = T−1 g(t) = ; 0 e−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 d x = P(t) x + g(t). dt Recall that if P(t) = A is a constant n × n matrix, then the general solution to this system is in the form ￿t x(t) = Ψ(t) c + Ψ(t) Ψ(τ )−1 g(τ ) dτ ; ￿ ￿￿ ￿ ￿ ￿￿ ￿ homogeneous solution 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 ￿t d Q(t) = 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. dt ...
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