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# Ordinary Differential Equations MA 26600

Purdue

GPA 3.97

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This 8 page Class Notes was uploaded by Dorothea Bode on Saturday September 19, 2015. The Class Notes belongs to MA 26600 at Purdue University taught by Haijun Yu in Fall. Since its upload, it has received 29 views. For similar materials see /class/208140/ma-26600-purdue-university in Mathematics (M) at Purdue University.

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Date Created: 09/19/15

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 F0 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 12 yC In Wk 1197 7 is transformed intro a system of n first order equations Ii I27 05 I37 17 1 1m IFt7117127m71n Practice Problems Section 71 5 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 Homogeneous Linear Systems with Constant Coef cients 17 A17 1 By assuming a solution is of the form 17 5equot where 5 is nonzero constant vecter and plugging it in 1 one obtains A n X n real constant matrix As A rsgte 0 which is equivalent to Aer1gtso lt2 To ensure a nonzero solution 5 exists the coef cient matrix A 7 TI must be singular matrix which means detAirl0 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 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 first case i suppose the n eigenvalues are n Tn the associated eigenvectors are 51 EM respec tively then 171 51 er 17 n EM e are n linearly indepenent Wronskian Wm1 14ml 0 solutions In this case the general solution of linear system 1 is given by 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 sinmt 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 17 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 A i T gt77 E 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 n gt 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 m i 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 r2 r 39 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 A Brief Review Chapter 1 Introduction Chapter 2 First Order Differential Equations 1 Chapter 1 Introduction There is nothing solid in Chapter 1 However sometimes when you can7t gure out how to solve an equa tion or system with methods learned in other chapters you may try 1 Classi cation 9 first order linear the most important one integrating factor or formula in Sec21 ET first order nonlinear the only nonlinear one a separable homogenous or exact n high order linear methods in Chapter 34 or Chapter 6 Laplace transform F linear system a Chapter 7 2 sketch the direction field of a given equation see which answer solution fits the direction field 3 directly verify a solution by plugging it in the equation 2 Chapter 2 First Order Differential Equations 1 split the content of this chapter into three catolog see the subsection below Note that Section 27 Numerical Approximation Euler7s Method is not required Remember the formulae 1 list below at least do two practice problems for each topic or go over corresponding homework If you can solve a certain differential equation also depend on if you are familar with the basic inte grals You can find some review material on integrals on my course webpage the bottom of Update List 21 How to solve a rst order equation 0 First Order Linear Equation Section 21 d inM 90 1 Step 1 Calculate the integrating factor mt exp plttgtdt lt2 1 ytitgtdtC 3 ltgt Ml 00 ltgt Eqns 2 and 3 give an explicit formula of the solution to equation 1 which is the MOST IMPORTANT one in this course Probably you will use it again in other math or physics courses Practice Problems Section 21 1 15 16 Step 2 Calculate the solution Separable Equation Section 22 dz Cy Mo Ny 0 or Mo dz Ny dy 0 or The equation can be solved by first separating terms and then take the direct integration SECTION 2 Practice Problems Section 22 1 14 Homogeneous Equation Section 22 d digfr7yyyI 4 Step 1 Introduce a new depenent variable 1 so that v yz or yz 1 which lead to dy 7 dzv 7 d1 3 71 5 Step 2 In equation 4 replace yx by v by v x37 we get 1 1117 91 or dv IE 7 W 7 v lt6 This is a separable equation can be solved by moving all the 1 terms to the left side and all the 1 terms to the rightnote that you can7t put the dv or d1 terms in denominator and then taking the integration ie dv dz 7 7 5 91 7 v 1 Practice Problems Section 22 32 36 Exact Equation See Example 2 of section 26 if you don7t like the general formulation A first order equation MI7yNI7yy 0 7 is an exact equation if and only if Myr7yNxI7y 8 Procedure to solve an equation of form 7 Step 1 check if 7 is an exact equation by verify 8 if true then there is a functino z y such that ML y MW 9 lt1gtyr7yNr7y Step 2 Take integration of Mz y with respect to I holding y constant we get gtz7yQz7yhy7 whereQz7yMz7ydr lt9 Step 3 Take partial derivative of z y with respect to y and let it equal to Nx y ltIgtyx7 y am y h y NW y then solving for h y we have My NW y 7 Qym y Step 4 The right side of equatin 10 should be a function of y only taking the integration on the both side we get an expression of hy plugin the value of hy in equation 9 and eventu ally an implicit solution is given by ay QI7yhy 0 Practice Problems Section 26 3 7 13 10 CHAPTER 2 FIRST ORDER DIFFERENTIAL EQUATIONS 22 The Existence and Uniqueness Theorem 1 Knowing how to determine the interval in which the solution is certain to exist Section 24 Note that you must convert the equation in following form the coeP cent of y term is 1 y pty 90 Practice Problems Section 24 1 3 2 That7s all 2 3 Applications 1 Salt water mixing in a tank know how to build a governing equation Section 2 example 1 rate in 7 rate out Practice Problems Section 23 2 3 2 Population dynamicsSecti0n25 How to nd the equilibrium solutions of an autonomous equa tionzeros of f y dy E i f Determine the stability of the equilibrium solutions by drawing some arrows on phase line There are several kind of stability stable including asymptotically stable unstable including semistable see Problem 9 section 25 Practice Problems Section 25 3 9 A Brief Review Chapter 6 The Laplace Transform The Laplace transform is very useful for solving linear differential equations with discontinuous or impulsive forcing terms A table of elementary Laplace transforms will be provided in the final you needn7t memorize the formulae in that table however you should be familar with them and know how to use them Actually remember them are more helpful 1 Know how to rewrite a piecewise continuous function in terms of unit step function uct For a two step function 7 f1t 0lttlt1 m f Mt tgt1 where f1t f2t can by any given function We can rewrite it as f 1 u1f1t u1 f2t f1tu1f2t f1t For a three step function f1t 0lttlt1 ft f2t1lttlt2 f3t7 tgt2 where f1t f2t f3t can be any given function We can rewrite function ft as fa 1 u1f1tu1 7 W 16205 W 16305 f1tu1f2t f1tu2fst 1623 Practice Problems Section 63 1 2 11 2 Be familar with partial fractions You need do partial fraction expansion to find the inverse trans form of a fraction function Practice Problems Section 62 2 4 8 3 Solve an initial value problem with piecewise continuous forcing term Section 64 Example 2 y4y9tv 1 ND 07 MO 07 where 0 0lttlt5 gt t755 5lttlt10 1 t Step 1 calculate the Laplace transform of nonhomogeneous term gt first rewrite gt in terms of unit step function W 1u50u57u10t55u101 7 75 u 7710 10 t 7 5 then the Laplace transform of gt is 7 fl l fl 755il710si7 i 68 4w 5mm 5 5 umltt 10gt7 5e 82 5e 82 i 5 82 Step 2 Take the Laplace transform on the both side of the first equation of 1 using the formula y t 3238 i 8 90 MO we get 82 4 Y8 8 90 MO CS plugging in the intial value7 we get 32 4Ys Cs solve it for ls7 CS 7 6755 767105 7 3247 Hs7 whereHs7 1 Y Step 3 The solution is given by the inverse transform of Ysi Suppose ht is the inverse trans form of Hs7 then the solution is W 7 1YS 7 glus 7 5 7 u10tht7 10 2 To find ht7 we need we take the partial fraction expansion of Hs7 which is l l l l H3gt7m71lt7mgt l l ht 7 1t 7 g s1n2t Other Practice Problems Section 64 6 7 Solve an initial value problem with impulsive forcing term Section 65 example 1 2y y 2y75t75 yltogto7 yltogtoi 3 Practice Problems Section 65 1 i The convolution integral Know the definition of convolution t t f9 ft7TyTdT7 fTyt7Tdn and the Laplace transform of f 6 9 H1 9 7F8 CSL 1F8GS 7 f y where F8 7 ft7 08 7 9t Practice Problems Section 66 4 9

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