MA  Linear Algebra and Dierential Equations Fall  Purdue Quiz   Underdamping For which values of the parameter will the equation  y
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MA Linear Algebra and Dierential Equations Fall Purdue Quiz Underdamping For which values of the parameter will the equation y

1 have all its solutions tending to zero as tends to in64257nity in a manner similar to that depicted in Fig1 872215 87221 872205 05 15 y e 8722t cos t Underdamping Figure 1 Underdamping Solution The auxiliary polynomial of equation 11 is 945r 1 0

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MA Linear Algebra and Dierential Equations Fall Purdue Quiz Underdamping For which values of the parameter will the equation y




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Presentation on theme: "MA Linear Algebra and Dierential Equations Fall Purdue Quiz Underdamping For which values of the parameter will the equation y"— Presentation transcript:


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MA 262: Linear Algebra and Differential Equations Fall 2011, Purdue Quiz 12 1 Underdamping For which values of the parameter will the equation 00 αy = 0 (1.1) have all its solutions tending to zero as tends to infinity in a manner similar to that depicted in Fig.1? −1.5 −1 −0.5 0.5 1.5 y = e −t cos( t+ Underdamping Figure 1: Underdamping Solution: The auxiliary polynomial of equation (1.1) is αr +1 = 0 (1.2) In order to get a graph like Fig.1, the general solution to the equatio n (1.1) should be in this form: ) = cos( βt )+

sin( βt (1.3) It follows that there are two complex roots for the polynomial (1.2) , which means < α < Besides, the real part of the complex root should be positive, which means α > Now we know that the (0 2). Copy right reserved by Yingwei Wang
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MA 262: Linear Algebra and Differential Equations Fall 2011, Purdue 2 Variation-of-parameter method Find a particular solution for 00 +2 (2.1) Solution: 1. Find the general solution to the homogeneous equation 00 +2 == 0. It is easy to know that 00 +2 == 0 +2 +1 = 0 2. Do the parameter variation to find a

particular solution to the non homogeneous equation (2.1). Suppose )e , then we can get (2.2) where = e , y , F From (2.2), we can get 1 1 x, = ln x, Now we find a particular solution of (2.1) as )e ln Copy right reserved by Yingwei Wang