lecture_23 (dragged) 2 - MA 36600 LECTURE NOTES: WEDNESDAY,...

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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 11 we have L[y ] = L ￿ m ￿ ￿ ci yi = i=1 m ￿ ci L[yi ] = i=1 m ￿ i=1 3 ci · 0 = 0. Hence y = y (t) is also a solution to the homogeneous problem. As a consequence, we show that the general solution to L[y ] = G(t) is y (t) = u(t) + Y (t), where L[u] = 0 and L[Y ] = G(t). To see why, let Y = Y (t) denote a particular solution, and y = y (t) be any solution to L[y ] = G(t). If we define u(t) = y (t) − Y (t), then L[u] = L[y ] − L[Y ] = G(t) − G(t) = 0. It suffices then to find the general solution to homogeneous equation L[y ] = 0. Linear Independence. We discuss when we have the general solution to L[y ] = 0. Consider the initial value problem n ￿ (j −1) Pn−j (t) y (j ) = G(t) where y (j −1) (t0 ) = y0 , i = 1, 2, . . . , n. i=1 Say that {y1 , y2 , . . . , ym } is a set of functions satisfying L[yi ] = 0, and consider the linear combination y (t) = c1 y1 (t) + c2 y2 (t) + · · · + cm ym (t) We know that L[y ] = 0. In order for y = y (t) to be the general solution, we need to find constants ci such (j −1) that y (j −1) (t0 ) = y0 . We may differentiate this expression then evaluate at t = t0 : y (t0 ) = c1 y1 (t0 ) + c2 y2 (t0 ) y (1) (t0 ) = (1) c1 y1 (t0 ) + (1) c2 y2 (t0 ) y (2) (t0 ) = c1 y1 (t0 ) (2) + c2 y2 (t0 ) (n−1) (t0 ) + c2 y2 + ··· We write this using the product of matrices: y0 y1 (t0 ) (1) (1) y0 y1 (t0 ) (2) (2) y 0 = y1 (t0 ) . . . . . . (n−1) y0 (n−1) y1 y2 (t0 ) (1) (2) (n−1) (t0 ) y2 cm ym (t0 ) .. (1) (2) (n−1) + cm ym c1 . (t0 ) · · · (t0 ) (1) ym (t0 ) c2 (2) ym (t0 ) c3 . . . . . . . (n−1) ym (t0 ) cm ··· . . . + ym (t0 ) ··· y2 (t0 ) cm ym (t0 ) (t0 ) + · · · ··· y2 (t0 ) + + ··· (2) (n−1) cm ym (t0 ) + ··· . . . y (n−1) (t0 ) = c1 y1 + (j −1) The matrix in the middle is an m × n matrix. In order for us to find the ci in terms of the y0 the m × n matrix to be invertible. Hence we must have m = n and determinant ￿ ￿ ￿ y1 (t0 ) y2 (t0 ) ··· ym (t0 ) ￿ ￿ ￿ ￿ (1) ￿ (1) (1) ￿ y1 (t0 ) y2 (t0 ) ··· ym (t0 ) ￿ ￿ ￿ ￿ (2) ￿ ￿ ￿ (2) (2) ￿ y (t0 ) y2 (t0 ) ··· ym (t0 ) ￿ ￿= 0. W y1 , y2 , . . . , yn (t0 ) = ￿ 1 ￿ ￿ ￿ ￿ ￿ . . . .. . . . ￿ ￿ . . . . ￿ ￿ ￿ (n−1) ￿ (n−1) (n−1) ￿y (t ) y (t ) · · · ym (t )￿ 1 0 2 0 0 , we need ...
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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 University-West Lafayette.

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