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lecture_39 (dragged) 2 - MA 36600 LECTURE NOTES WEDNESDAY...

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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.

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