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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, MARCH 13 Higher Order Linear Equations Fundamental Set of Solutions. Consider the initial value problem n ￿ j =0 Pn−j (t) y (j ) = G(t) where (j −1) y (j −1) (t0 ) = y0 i = 1, 2, . . . , n. , Say that we can find (n + 1) functions y1 = y1 (t), . . . , yn (t) and Y = Y (t) such that (1) Each yi = yi (t) is a solution to the homogeneous equation n ￿ j =0 (j ) Pn−j (t) yi =0 for i = 1, 2, . . . , n. (2) The n × n determinant is nonzero: ￿ ￿ y1 (t) y2 (t) ··· ￿ ￿ (1) (1) ￿ y (t) y2 (t) ··· ￿1 ￿ ￿ ￿ W y1 , y2 , . . . , yn (t) = ￿ . . .. . . ￿ . . . ￿ ￿ (n−1) (n−1) ￿y (t) y2 (t) · · · 1 ￿ yn (t) ￿ ￿ ￿ (1) yn (t) ￿ ￿ ￿ ￿ (j −1) ￿ ￿ ￿ = ￿y ￿ ￿= 0. ￿ i . . ￿ . ￿ ￿ (n−1) yn (t)￿ (3) Y = Y (t) is a solution to the nonhomogeneous equation n ￿ j =0 Pn−j (t) Y (j ) = G(t). Then there exist constants ci such that the solution to the differential equation is in the form ￿n ￿ ￿ y (t) = ci yi (t) + Y (t). i=1 For this reason, we call {y1 , y2 , . . . , yn } a fundamental set of solutions. We explain why. We know that u(t) = y (t) − Y (t) is a solution to the homogeneous equation by (3), so it suffices to find constants ci such that y (t) − Y (t) = u(t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t). is the solution to the initial value problem. We may differentiate this expression then evaluate at t = t0 : u(t0 ) = c1 y1 (t0 ) + c2 y2 (t0 ) u(1) (t0 ) = (1) c1 y1 (t0 ) + (1) c2 y2 (t0 ) u(2) (t0 ) = c1 y1 (t0 ) (2) + c2 y2 (t0 ) (n−1) (t0 ) + c2 y2 (2) . . . u(n−1) (t0 ) = c1 y1 (n−1) 1 + ··· + cn yn (t0 ) + cn yn (t0 ) + ··· + cn yn (t0 ) + ··· (t0 ) + · · · (1) (2) (n−1) + cn yn (t0 ) ...
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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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