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lecture_18 (dragged) 2 - MA 36600 LECTURE NOTES: FRIDAY,...

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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 3 Then the general solution to the nonhomogeneous equation is the function y (t) = c1 y1 (t) + c2 y2 (t) + Y (t) for some constants c1 and c2 . We remark that typically finding y1 and y2 is easy: when the coefficients a(t), b(t), and c(t) are constant functions, one can compute y1 and y2 by considering roots of the characteristic equation a r2 + b r + c = 0. We will spend the next couple of lectures discussing how to find a solution Y . Note that once one solution Y is found, we can compute all solutions y . We give a proof of the claim above. Say that y = y (t) is any solution to the nonhomogeneous equation, and let Y = Y (t) be as in assumption (iii). Denote the difference u(t) = y (t) − Y (t). We see that u = u(t) is a solution to the homogeneous equation: ￿ ￿ ￿ ￿ ￿ ￿ a(t) u￿￿ + b(t) u￿ + c(t) u = a(t) y ￿￿ − Y ￿￿ + b(t) y ￿ − Y ￿ + c(t) y − Y ￿ ￿￿ ￿ ￿ ￿￿ ￿ = a(t) y ￿￿ + b(t) y ￿ + c(t) y − a(t) Y ￿￿ + b(t) Y ￿ + c(t) Y = f (t) − f (t) = 0. By assumption (i), both y1 and y2 are also solutions to a(t) u￿￿ + b(t) u￿ + c(t) u = 0; and by assumption (ii) these functions are linearly independent. Hence {y1 , y2 } is a fundamental set of solutions for the homogeneous equation. That means there exist constants c1 and c2 such that u(t) = c1 y1 (t) + c2 y2 (t) =⇒ y (t) = u(t) + Y (t) = c1 y1 (t) + c2 y2 (t) + Y (t). Example. Consider the differential equation y ￿￿ − 3 y ￿ − 4 y = 3 e2t . We will find the general solution of this equation. To do so, we compute functions y1 = y1 (t), y2 = y2 (t), and Y = Y (t) as above. First we find the functions y1 and y2 as solutions to the homogeneous equation We have the characteristic polynomial Hence we have the roots y ￿￿ − 3 y ￿ − 4 y = 0. r2 − 3 r − 4 = 0 =⇒ (r + 1) (r − 4) = 0 ￿ =⇒ ￿ r1 r2 = −1 = 4 y1 (t) = e−t y2 (t) = e4t Since r1 ￿￿= r2 we see that the functions y1 and y2 are linearly independent. That is, the Wronskian ￿ W y1 , y2 (t) ￿= 0. Second we seek a function Y = Y (t) as a solution to the nonhomogeneous equation Y ￿￿ − 3 Y ￿ − 4 Y = 3 e2t . It suffices just to find one solution Y = Y (t). To this end, we make a guess that such a solution is in the form Y (t) = A eαt for some constants A and α. We have the derivatives Y = A eαt ￿ ￿ ￿ Y = α A eαt =⇒ Y ￿￿ − 3 Y ￿ − 4 Y = α2 − 3 α − 4 A eαt = 3 e2t . Y ￿￿ = α2 A eαt ￿ ￿ We now choose α = 2, so that α2 − 3 α − 4 A eαt = −6 A e2t . Similarly, we choose A = −1/2. This means that one solution to the nonhomogeneous equation is the function 1 Y (t) = − e2t . 2 Hence the general solution to the nonhomogeneous differential equation y ￿￿ − 3 y ￿ − 4 y = 3 e2t ...
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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.

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