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# 208 Class Note for MA 26600 with Professor Yu at Purdue

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COURSE
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KARMA
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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 19 views.

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Date Created: 02/06/15
A Brief Review Chapter 7 System of First Order Linear Equations Following is a brief review or check list of Chapter 7 The third part homogenous linear system with constant coef cients is the most important part The last quiz on Thursday April 23 will cover the first three parts 1 Know how to transform an equation of high order into a system of first order equations Section 71 example 1 An arbitrary nth order equation a y Ft y y y 1 by introducing n new dependent variables 1197 I2y 7 iiii H7 Iny 1 is transformed intro a system of n first order equations 11127 12137 17 1 1m IFt7117127m71n Practice Problems Section 71 5 2 Basic skills related to matrix operation and linear algebra i Row reduction Gaussian elimination used to find inverse of a matrix or solve linear system Section 72 example 2 Section 73 example 13 ii Know how to find eigenvalues and eigenvectors of a matrix Section 73 example 4 Practice Problems Section 73 example 1 example 2 example 4 3 Homogeneous Linear Systems with Constant Coef cients 17 Azt A n X n real constant matrix 1 By assuming a solution is of the form 17 5equot where 5 is nonzero constant vecter and plugging it in 1 one obtains As A rsgte 0 which is equivalent to A7T150 2 To ensure a nonzero solution 5 exists the coef cient matrix A 7 TI must be singular matrix which means detA 7 T I 0 3 From the above analysis we see that 7 is an eigenvalue of A and 5 is the associated eigenvector Provided A is an n X n real constant matrix detA 7 7 I is a polynomial of degree n It have n zeros which are eigenvalues of A these eigenvalues might be i real and different from each other ii some ones are complex conjugates iii real repeated we only consider real repeated zeros here for repeated complex zeros a little bit complex For the rst case i suppose the n eigenvalues are n Tn the associated eigenvectors are 51 EM respec tively then 171 51 er 17W EM e are n linearly indepenent Wronskian Wm1 14ml 0 solutions In this case the general solution of linear system 1 is given by 7 t 5151 erlt Cg 52 e 07500 er 4 Practice Problems Section 75 1 2 16 For the second case ii we can verify that the complex eigenvalues always appear in complex conjugate pairs Suppose r1 A 239 M is one complex eigenvalue and 51 a 239 b is the associated eigenvector then r A 7 239 M is another eigenvalue and 52 a 7 239 b is the eigenvector associated to T2 and 131 51 E7117 132 52 em are two solutions of linear system 1 and they are complex solutions However the real part and imaginary part of 171 or 172 since 171 and 172 are complex conjugates give us two real solu tions namely ut eMa cosbt 7 b sinbt vt eMa sinbt b cosbt These two solutions together with other solutions form a fundamental set of solutions Practice Problems Section 76 1 2 10 For the third case iii Their might be two subcases First if the repeated eigenvalue r with algebraic multiplicity s have 3 associated linear indepen dent eigenvectors suppose they are 51 55 then 51 equot 55 e are 3 linearly independent solutions together with other solutions form a fundamental set all the solution can be given as linear combination of the fundamental set However for some matrix excluding real symmetric matrices we may not find 3 linear inde pent eigenvectors associated to one eigenvalue with algebraic multiplicity 3 Consider that n r2 r is a double eigenvalue with multiplicity 2 if there is only one linear independent eigenvector associated to r denoted by 5 then we have one solution of form 171 5equot The other solution related to r is given by m2t 5 t e 17 equot where 17 is the so called generalized eigenvector associated to T which satisfies equation ArT177E 5 Solution 171 172 are linearly independent together with other solutions could form a fundamental set of solutions Practice Problems Section 78 3 8 Phase Portrait For a system has more than 2 dependent variables its hard to plot the solutions For a system with 2 dependent variables we can draw a phase portrait of the solutions Depend on the eigen values of coef cients matrix we may have several different phase portraits a real different eigenvalues 7 17 2 if n lt 0r2 lt 0 the origin is a stable node Sec75 example 2 Figure 753 754 if r1gt 0r2 gt 0 the origin is an unstable node if n and T have different sign r1 r2 lt 0 then the origin is a saddle point Sec75 example 1 Figure 751 and 752 saddle point are always unstable b complex comjugate eigenvalues r1 A m r A 7 iii if A gt 0 the trajectories of the solutions are spiral lines the origin is an unstable spiral point if A lt 0 the origin is a stable spiral point Sec7i6 examaple 1 Figure 761 762 if A 0 the trajectories are closed curves about the origin the origin is a center cen ters are always stable but not asymptotically stable c real repeated eigenvalue r1 7 2 r if r gt 0 the origin is an unstable improper node Sec78 example 2 Figure 781 782 if r lt 0 the origin is a stable improper node 4 Nonhomogeneous Linear System There are four methods to solve a nonhomogenous linear system WPtwyt 6 or nonhomogenous linear equation with constant coef cients m Amyt 7 They are a Diagonalization Sec 79 example 1 b Undetermined CoeP ents Sec7i9 example 2 c Variation of Parameters Sec7i9 example 3 d Laplace TransformSec7i9 example 4 Each of those methods has some advantages and disadvantages You needn7t know all those methods be pro cient in using one or two methods is enough Practice Problems Section 79 1

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