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lecture_27 (dragged) 1 - 2 MA 36600 LECTURE NOTES WEDNESDAY...

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Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 1 We will choose the functions ui (t) such that n ￿ dui i=1 dt (j −1) yi 0 (t) = G(t)/P (t) for j = n. 0 Hence we have the expression n ￿ dm Y Pn−m (t) m = dt m=0 = ￿n ￿ ￿ n ￿ for j = 1, 2, . . . , (n − 1); d m yi ui (t) Pn−m (t) m dt m=0 i=1 ￿n ￿ ￿ ui (t) [0] i=1 + ￿￿ ￿ + n m ￿￿ m=0 j =1 0 + · · · + 0 + P0 (t) ￿ = G(t). Pn−m (t) n ￿ dui i=1 (n−1) yi dt ￿ ￿(m−j ) n ￿ dui i=1 (j −1) yi (t) dt ￿￿ (t) Hence we must solve the following system of equations: du1 dt du1 (1) y1 (t) dt y1 (t) du1 dt du1 (n−1) y1 (t) dt (n−2) y1 (t) du2 dt du2 (1) y2 (t) dt + ··· + + ··· + du2 dt du2 (n−1) + y2 (t) dt + ··· + yn + y2 (t) + (n−2) + y2 (t) + ··· dun dt dun (1) yn (t) dt = 0 = 0 dun dt dun (n−1) + yn (t) dt = 0 = G(t) P0 (t) yn (t) (n−2) (t) . . . Using Cramer’s Rule, we can solve for the functions ui (t): ￿ ￿ y1 (t) ￿ ￿ (1) ￿ y (t) ￿1 ￿ . dui 1 G(t) ￿ . ￿ ￿ . =￿ ￿ dt W y1 , y2 , . . . , yn (t) P0 (t) ￿ ￿y (n−2) (t) ￿1 ￿ ￿y (n−1) (t) ￿1 ￿ ··· ··· .. . ··· ··· yi−1 (t) 0 yi+1 (t) (1) 0 yi+1 (t) . . . . . . . . . yi−1 (t) (1) (n−2) ··· ··· .. . (n−2) yi−1 (t) 0 yi+1 (t) · · · (n−1) (n−1) yi−1 (t) 1 yi+1 (t) · · · ￿￿ Wi (t) Upon integrating, we see that Y (t) = u1 (t) y1 (t) + u2 (t) y2 (t) + · · · + un (t) yn (t) = ￿ t G(τ ) K (τ , t) dτ P0 (τ ) ￿ yn (t) ￿ ￿ ￿ (1) yn (t) ￿ ￿ ￿ ￿ . . ￿. . ￿ ￿ (n−2) yn (t)￿ ￿ ￿ (n−1) yn (t)￿ ￿ ￿ ...
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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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