285lect31 - 1 1.1 Lecture 32 Discussion of general...

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1 Lecture 32 1.1 Discussion of general non-homogeneous problem Consider ODE x 00 + ! 2 0 x = f ( t ) that models the behavior of a mass-spring system with natural circular fre- quency ! 0 , under in&uence of an external force of magnitude f ( t ) per unit of mass. If f ( t ) is a simple harmonic force, f ( t ) = A cos !t; ! 6 = ! 0 , then, as we already know, a particular solution of the equation x 00 + ! 2 0 x = A cos !t is x p ( t ) = A ! 2 0 & ! 2 cos !t: Now consider the external force f ( t ) of the following more general form: f ( t ) = N X n =1 A n cos ! n t , where ! n 6 = ! 0 ; for every n . The particular solution is obtained by superposition: x p ( t ) = N X n =1 A n ! 2 0 & ! 2 n cos ! n t: A similar formula can be obtained for the case when f ( t ) = N X n =1 B n sin ! n t , where ! n 6 = ! 0 ; for every n . It can be proved that any continuously di/erentiable function can be represented in the form: f ( t ) = a 0 2 + 1 X n =0 & a n cos &nt L + b n sin &nt L ± ; 0 < t < L: Then, use superposition of solutions to construct x p ( t ) . 1
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1.2 tions Now we consider several more examples. Example 1 f ( t ) = cos 2 &t: cos 2 &t = cos (2 &t + 2 & ) = cos 2 & ( t + 1) : It evident that cos 2 &t 6 = cos 2 & ( t + p ) , for any 0 < p < 1 : So, f ( t ) = cos 2 &t is a periodic function with period = 1 . Example 2 f ( t ) = cos &t 3 : cos &t 3 = cos & &t 3 + 2 & ± = cos & 3 ( t + 6) : It evident that cos 2 &t 6 = cos 2 & ( t + p ) , for any 0 < p < 6 : So, f ( t ) = cos &t 3 is a periodic function with period = 6 .
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This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285lect31 - 1 1.1 Lecture 32 Discussion of general...

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