lecture_27 (dragged) 1

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

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2 MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 1 We will choose the functions u i ( t ) such that n i =1 du i dt y ( j 1) i ( t ) = 0 for j = 1 , 2 , . . . , ( n 1); G ( t ) /P 0 ( t ) for j = n . Hence we have the expression n m =0 P n m ( t ) d m Y dt m = n i =1 u i ( t ) n m =0 P n m ( t ) d m y i dt m + n m =0 m j =1 P n m ( t ) n i =1 du i dt y ( j 1) i ( t ) ( m j ) = n i =1 u i ( t ) [0] + 0 + · · · + 0 + P 0 ( t ) n i =1 du i dt y ( n 1) i ( t ) = G ( t ) . Hence we must solve the following system of equations: y 1 ( t ) du 1 dt + y 2 ( t ) du 2 dt + · · · + y n ( t ) du n dt = 0 y (1) 1 ( t ) du 1 dt + y (1) 2 ( t ) du 2 dt + · · · + y (1) n ( t ) du n dt = 0 . . . y ( n 2) 1 ( t ) du 1 dt + y ( n 2) 2 ( t ) du 2 dt + · · · + y ( n 2) n ( t ) du n dt = 0 y ( n 1) 1 ( t ) du 1 dt + y ( n 1) 2 ( t ) du 2 dt + · · · + y ( n 1) n ( t ) du n dt = G ( t ) P 0 ( t ) Using Cramer’s Rule, we can solve for the functions u i ( t ): du i dt = 1 W y 1 , y 2 , . . . , y n ( t ) G ( t ) P 0 ( t ) y 1 ( t ) · · · y i 1 ( t ) 0 y i +1 ( t ) · · · y n ( t ) y (1) 1 ( t ) · · · y (1) i 1 ( t ) 0 y (1) i +1 ( t ) · · · y (1) n ( t ) . . . . . . . . . . . . . .
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