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by: Reyes Glover


Reyes Glover
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This 43 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 427K at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/181494/m-427k-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.

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Date Created: 09/06/15
M427K Handout Linear Systems Systems of Linear First Order ODEs lll Salman Butt April 13 2006 NOTES 1 Office hours this Monday are abbreviated to 10 1030am COMPLEX EIGENVALUES Last time7 we left off consider the linear system 4717132 32 1 m7lt41gt7 0 1 We found the matrix had complex eigenvalues 1 i 2i with eigenvector 17 17 72 We wrote our solution as 105 175ltH21gtt7 determined its real and imaginary parts and wrote down our general a 7 t cos2t t sin2t Mt 7615 lt 2sin2t gt 625 lt 72cos2t Let s determine our constants 0102 solution EXERCISE Find the solution to the linear system EXPONENTIAL OF A MATRIX Recall that when we have the rst order differential equation z am the solution was t z0e t We use this as inspiration to guess that a possible solution to the differential equation i Ai is f a05 But what is 5quot or more simply what is 5 4 We de ne it as follows Thus 5quot is merely So if want to plug this guess into our differential equation we need to compute the derivative of 5quot Writing our differential equation as i 7 Ai O we have So if we can compute eAt quickly we can nd our solution in no time And when can we compute eAt quickly Let s consider the matrices 32 32gt These examples hold true in general ifA is upper lower triangular or is diagonal7 determining t is easy Let s actually solve a differential equation using this technique 47 20 32 32 1 x7lt12gt7 0 1 5A REPEATED EIGENVALUES In this section7 we consider what happens when our homogeneous system i Ai has repeated eigenvalues r1 r2 As before we compute the eigenvector 17 which is now associated to both 71 and 72 This will give us one solution f1 175 But what about a second solution Well7 it turns out that the second solution will have the form E2 its equot where 17 is the eigenvector for r above and 7 satis es the equation A 7 r 1 17 Such a vector 73 is called a generalized eigenvector associated to r Finally we set the general solution to be t 01 1t cgi t M427K Handout Separable and Exact Equations Salman Butt January 267 2006 This handout describes the procedure used to solve two speci c types of rst order differential equations separable and exact equations SEPARABLE EQUATIONS Observe that we can write any differential equation 3 my lt1 Mzy 0 2 dm Now separable differential equations are of the speci c form Note the key fact here These equations are solved easily by integrating M and N We do this by rst writing our separable equation as 4 And now simply integrate both sides to get the solution An example illustrates this procedure well Consider the initial value problem 72751 l 92 y 7 240 1 5 Observe by the way that we made no assumption that our differential equation is linear Thus our M and N may be non linear but so long as it is a separable equation and we can integrate M and N we can nd a solution using this separation of variables procedure EXACT EQUATIONS We begin by writing a general rst order differential equation as Ma y Na 24 0 lt6 We say this equation is exact if and only if there is a function 11z y such that 1 Mm y and 112 Nzy where 1 37412 But by Theorem 261 in the text this happens if and only if My Nm In this case we have a method to nd the solution to our differential equation Namely we integrate M with respect to z to give us our desired function 11m y bearing in mind that M is a function of both z and y so inde nite integration gives us a constant which depends on y denoted by My 7 But 112 N thus 8 We can now isolate our gt y to get 9 Once we have our gt y we merely integrate it with respect to y to get our unknown function y and we then plug this into equation Observe that for such a function m y d 8 8 WW g 871511 May New 0 lt10 Thus 11m y c so our equation 7 becomes merely 11 Again an example is worth a thousand words Consider the differential equation liysinzcoszy 0 12 M427K C Handout Variation of Parameters Salman Butt September 28 2005 This paper describes the derivation of the formulas used to nd the particular solution to a second order non homogeneous equation with arbitrary coefficients That is we try to nd the particular solution to the differential equation 24 pty 6175 W 1 We rst assume that we already have a fundamental set of solutions to the associated homogeneous equation 24 pty 6175 0 2 That is we have functions y1t y2t that are linearly independent and are solutions to equation 2 The key idea is to guess that yp is a combination of y1 and yg where the coefficients of y1y2 are functions ie not just constants That is we guess 24p 11109341 112092 3 Now we want yp to solve our original differential equation So we want 4 pty qtyp 975 4 In order to plug into this equation we need to determine y and yg yp U1y1 Uzyzy Uiyl Ulyi 11292 11212 lviyl UzyZl Ulyi 11212 We now introduce a constraint 11341 U yz 0 5 This gives us two constraints on 121122 which will be necessary to determine 121122 The other constraint is equation Applying this constraint to yg the term in the brackets becomes 0 and we have 24 viz1 11212 6 From this equation we determine y Ulyl Ulyi v y vzyg 7 Plugging all this back into the left hand side of equation 4 we get 242 WM qtyp viz1 v1yf v zy z 11232 ptv1y 1 11222 qtv1y1 vzyz v1yl ptyl qty1 v2y pty qty2 Ulyl 1222 M427K Handout Advanced Numerical Methods Salman Butt April 21 2006 RUNGE KUTTA METHOD This method is given by the formula lts global truncation error is bounded by a constant multiple Of h4 and its local truncation error is on the order Of 715 MULTI STEP METHOD 1 ADAMS METHODS The Adams method is a multistep method We x a step size h Adams methods use a polynomial Of degree k denoted Pkt tO approximate gt t in the equation gtlttn1gt e gtlttngt gt lttgtdt We take the simple case with k 1 and set P1t At B We require P105 tmyn and P1tn1 ftn1yn1 Plugging in for P1 and solving for A and B7 we nd Plugging this into our integral equation7 integrating7 and plugging y s in for the gt7s we get the equation This is the second order Adams Bashforth formula It requires knowing yn and yn71 lts local truncation error is on the order Of hs Note that the rst order Adams Bashforth formula ie when k 0 is merely Euler s method Note also that the accuracy Of this method increases as you increase k Varying this formula slightly by forcing P1tn1 tO be equal tO ftn1yn1 we get a new formula This is the second order Adams Moulton formula Note this is an implicit equation with local truncation error on the order of hs We can increase the degree of the polynomial as in A B to get more accurate results Note that the rst order A M formula ie k 0 is just the backward Euler method Though it has the same LTE as the A B method A M is considerably more accurate for moderate orders though it is slower What method to use depends on the given problem These two methods can be used together as a predictor corrector method A B is the predictor while A M plays the role of the corrector We use A B to compute the rst approximation yn We then compute fn1 for A M and get a more accurate value of yn1 If the difference between these two values is too large we use A M again If we have to use A M more than once or twice our step size h is too large and must be decreased In order to compute the rst few terms needed for the multi step methods you can use a one step method to compute these values Another approach is to use a low order method with a very small step size to calculate some initial values and then increase the order and step size gradually until enough values have been computed to use the higher order method MULTI STEP METHOD 2 BACKWARD DIFFERENTIATION FORMULA In this method the idea is use a polynomial of degree k denoted Pkt to approximate gtt as opposed to gt t for the formula gt t fty We then differentiate P and set P tn ftn1yn1 to get an implicit formula for yn1 Taking the k 1 case we require P105 y t y and P1tn1yn1 ftn1yn1 as well as the condition on Pl Plugging all this in to our equation we get the formula Note this is just the backward Euler formula Increasing the degree of the polynomial yields more accurate backward differentiation formulas For example the second order formula is given by See the end of section 84 for a discussion of relative merits of R K and these multi step methods Here s a brief rundown 1 M427K Handout Fundamental Solutions of Linear Homogeneous Equations and Linear lndependence Salman Butt February 7 2006 This handout discusses 32 and 33 Of the text as well as providing some differential equations tO solve NOTES 1 During the last couple Of minutes Of class today I want you tO take out a sheet Of paper and let me know how would you rate the pace Of discussion session and what you are most confused by in the class so far 2 Your exam is next week on Wednesday February 15 I will review material in class on Tuesday February 13th It is in your best interest tO come tO discussion section on that day EXAMPLES Solve the following differential equations 2 emly7 240 0 iii4 754 uvduuivdv 0 2y y y 07901790 0 FUNDAMENTAL SOLUTIONS OF LINEAR HOMOGENEOUS EQUATIONS We introduce the following notation L gt lt1 pd q gt where j is a twice differentiable function and p q are continuous on an interval 04 lt t lt 3 Note L is a functional it takes as input a function and outputs another function SO we are studying differential equations Of the form My O7 often with some initial conditions yt0 yewt0 ya The key theorem for second order differential equations homogeneous or not is the following an analog of Theorem 241 in the text Theorem 1 We proved the following theorem last time7 but let us restate it here Theorem 2 Principle of Superposition Using our initial conditions to determine 0102 we have the two equations Solving for 0102 we nd We can write this using determinant notation The denominator determinant is given a special name the Wronksian W of two functions 341342 is de ned to be If the Wronksian is non zero we know two facts 1 This second fact motivates our use of the term general solution77 for the solution y clyl ogyg We say 341342 are a fundamental set of solutions to our differential equation LINEAR INDEPENDENCE We say two functions g are linearly independent on in interval I if whenever k1ft k2gt 0 then k1 and k2 must both be zero for all t E I Otherwise we say the functions are linearly dependent The following theorem is an important one to keep in mind since it reduces the work needed to determine if two functions are linearly independent or not Theorem 3 Observe that this theorem does not say in any way that if for two functions g we have W gt 0 then g are linearly dependent For example the functions ft tzltl and gt t3 satisfy W gt 0 for all 71 lt t lt 1 but their linear independence and dependence is quite odd see problem 28 on p 159 Note that in our de nition of the Wronksian above we need to know what yl and yg are The following theorem though provides a way to calculate W without actually knowing y1y2 rst Theorem 4 This allows us to compute the 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oq s an awhchm Em we mag riglore am am MQBM XS a 905 Mon lb Mb Asswvximg oLLL eigewqus are reed QM gm abmvmeg W age0506 039 61M boundary VaLue 8 gt3gtO 3V0 r O U UT 0 WXLO 2gt0 739 1 i f mosh x Egmn3 x f c aquot quotm 6quot 1 339 a W WWW x gecehw 3 0O gt O 4 gt8 c 20 3 o s BAs anFK 2 A 20 mg isoLuiR csA is m mu Mmoerewg 30 szo 610 j 3A 3 Pam B 3WD 90 2 A 0 3 an eigemNMe WM am a A gmgmda on 2 3 Ear WW number E gt70 939gt O 2 90 C22 a gt ci K M t5 Aco l X X 3 N3 1 3 3 AH 3M5 4 Exgg bb v 9090 gt O 39 C Smygoc Nks w o Saw y 0 n 39gt 50 ll3 ML wow Wzn x201 A30 Z39Sr N 7 KC Q4 6 are QJQUNCQMeS 8 7 A olt X correseonc g 0 gt Lad Yeav s M G1 E nok alk benvcdxx S Mamp QCSQRC MCECOAS par HA 85AM JamAla roHQm a H w O a 313 gt 3 43 gts4tgt rQJvXCO V2quot r t i 3 AcosM av EgiNRGSx 0 6 FXAsmkbZS 1 PXBCOSH FP 9 8km xz m 0 MIKE 2A 3 O Rg 92A Pr SEE g 3 2amp 039 gfwsink gtXBampH 2 3 Dsh hgwg 7 0 ggsmm ngwm a 0 quot 2 7 gswn F O NM Azfq 333 GA QJESQAVWQ lt3 2 AC05YZ gt S nh 27 E gzxj K43 5quot O azAX E a ipv a3903 285 rfvzg Raga 96 1 O 39 gtxprwprzo 2Acj RSQ pD gab o 6805 GAME S x Q B AQQSCJ K ggk x3 by m lgt nm 6 EcosCRKB a39wyzbwwo AREva 0 g8 39339 9C 1 PE75 Esfa 1quot f SQ esm M ian gt n739lTrZ Azoru39z gr 1 M427K Handout Linear Systems Preliminaries amp Systems of Linear First Order ODEs Salman Butt April 6 2006 NOTES 1 Check to make sure you have your homework and not someone else s 2 Of ce hours this Friday will be adjusted to 3 330 and 430 5 EXERCISE Is the set of vectors 1 2 71 2 1 3 74 1 711 linearly independent or dependent EIGENVALUES AND EIGENVECTORS We say i not the zero vector is an eigenvector of a matrix A if there exists a number A such that Ai Af That is A takes the vector i to a multiple of itself In this case we say A is an 390 of A 1 t s and are important in many applications and nding them is an important technique to know in this class and beyond We rewrite our de ning equation Ai Af as This has a nonzero solution if and only if A is chosen so that Note that for a given eigenvalue there is an in nite number of associated eigenvectors 7 namely scalar multiples of the vector Lets do some examples Consider the matrices ii iii SYSTEMS OF FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS Suppose we have a system of rst order linear differential equations For illustrative purposes we will consider systems of two equations 901 P109901 q1t2 9175 902 P209901 q2t2 9275 We write this system as We begin by studying the constant coe icient homogeneous case ie where t 0 So we have the matrix equation i Af where A is a matrix of constants We begin by guessing that our solution is of the form f 175quot where 17 is a constant vector Differentiating and plugging into our linear system we nd Dividing by equot and rewriting our equation we nd So in order solve the differential equation we need to solve this algebraic equation Observe that this is precisely the eigenvalue equation so our T s are merely the eigenvalues of A and 17 is the eigenvector of A associated to 1 Thus our proposed solution x 175quot is a solution precisely when r and 17 are eigenvalues and associated eigenvectors of A In the 2 x 2 case we nd two different values of 1 As in the past the nature of our eigenvalues r ie if they are real and distinct repeated or complex will be critical in determining our nal solution We begin with the simple case real distinct roots In this case the two eigenvectors we nd will be linearly independent hence our general solution simply becomes i 011715 021725 M427K C Handout Problem 11224 Salman Butt November 137 2005 This handout describes how to do problem 24 in 112 of Polking et al s Di erentz39al Equations with Boundary Value Problems7 2nd edition We are given the differential equation 24 sin 901 y 0 1 We begin by checking that z 0 is an ordinary point this is clear since sinz and the function 1 are both smooth at 0 We thus can set our solution y to be a power series around 0 and compute its derivatives 00 y 2W n0 00 00 y E annxnil E annznil n0 n1 0 0 y Zannm 71mn 2 Zannm 71mn 2 0 n2 Now we want to substitute this into our differential equation In order to this7 we need to determine what to do with the sinz function We recall that the sinz function has the Taylor expansion 7 00 2n1 Sln72n1lm Now since we will only be computing the rst few terms in the expansion of our series solution7 we only need the rst few terms of this series expansion of sin m More precisely7 we approximate sinz by z 7 mil3 Plugging all this work into our differential equation 17 we get Zannm 71m 2 z 7 Zannmnil 2 0mm 0 n2 n1 n0 co 00 m3 0 co Z2P 2P1xp Zannz 1 7 Egannznil Jrganm 0 170 n1 00 n 00 n 7 00 M n2 00 n 7 Zan2n 2n 1m Zannz Z 3 m Z anz 7 0 n0 n1 n1 39 n0 00 n 00 n 7 00 w p 00 n 7 Z an2n 2n 1z Z annx Z 3 m Z anz 7 0 n0 n1 123 39 n0 i0 an2n 2n 1mn 0 annzn 7 i Wm 0 0mm 0 3 n0 n1 n3 n0 M427K C Handout Problem 5436 Salman Butt October 167 2005 This handout describes how to do problem 36 in 54 of Polking et al s Di erentz39al Equations with Boundary Value Problems7 2nd edition We are given the differential equation y y2sint7 ND 71 20 1 1 Method of Undetermined Coef cients We rst do this problem with the method of undetermined coef cients This entails determining the homogeneous solution yh and the particular solution yp In order to determine the homogeneous solution7 we consider the characteristic equation r210 2 which has roots r ii Thus yh 01 cost Cg sint Observe that the method of undetermined coef cients rst suggests to us to guess that yp A cos tB sin 25 But cost and sint are fundamental solutions to the homogeneous equation So using this guess would yield no results Thus we consider yp At cost Bt sint y Acost 7 tsin t Bsint tcos t y 72Asint7Atcost2BcostiBtsint Plugging into Equation 17 we have 72sintygypi2Asint2Bcost 3 which tells us that A 1 and B 0 So yp tcost Thus our full general solution is yt tcost 01 cost 02 sint 4 Taking the rst derivative and using the intial conditions7 we see that 01 71 and Cg 0 Thus the full solution is ytcost7cost 5 Method of Laplace Transform We begin by applying the Laplace transform E to both sides of Equation 1 E yn y s E 72 sint s gt 32148 7 8240 20 t Y 912139 Using the initial conditions and rewriting our resulting equation we get 72 s 1 Y 7 7 7 7 6 S 8212 321321 We know the inverse Laplace transform of the last two terms on the right hand side of the above equation but the rst term is more elusive We want to do is exploit the following property of the Laplace transform E If 05 8 F 8 7 where Fs E f We can apply the inverse Laplace transform to this property to get 751 71 Jr8 05 8 Thus if we can rewrite our rst term 7232 12 as 7F s for some function F whose inverse Laplace transform we do know we could determine E l 7232 12 We could begin by trying to integrate our fraction but the integral looks thorny lnstead let us consider a basic building block 7 ie a function whose inverse Laplace transform we know We take Fs s321 7 32 17 323 lt82 1 1 232 7327171 327171 Thus 7mg 7 Wily we can still use Observe that 7F s is the rst and last term of the right hand side of Equation 6 So using Equation 8 we see that which though it is not the fraction we were originally looking for 71321 t 371141875 t t 9 where ft E 1t 1s82 1 25 cost Thus we have that 1 2 E 1 777 252 25 10 821 3212 COS Finally we have as before 7 1 2 3 Mt E 1YSt E 1 i W i t tCOSt7 cost 11 M427K Handout PDEls The Heat and Wave Equations Salman Butt May 4 2006 1 Next week7 I Will continue to hold office hours Regular discussion will also be office hours THE HEAT EQUATION Recall the setup and solution for the heat equation Let s consider an example L 1 ms w e z THE WAVE EQUATION Our next PDE of interest is the wave equation Here is the setup We have an elastic string of length L tightly stretched between two supports along a horizontal line which we place on the z axis We pluck the string so that it vibrates in the vertical plane We let um t be the vertical displacement of the string at some point 0 g x g L and at some time t We ignore damping effects and assume the amplitude of motion is not too large this is to avoid pathologies in our solution Then um t satis es the differential equation This is the 1 dimensional wave equation see Appendix B for the derivation The constant a is the velocity of propagation of waves along the string We need some initial and boundary conditions We know the ends of the string are to remain xed so we have the boundary conditions The initial condition will be in terms oft and we have the second derivative of u with respect to t in our equation so we want two initial conditions Here fg will be given functions These functions must satisfy f0 fL 90 gL 0 in order to be consistent with our initial condition We need to now solve this problem As before it is an IV with respect to t and a BVP with respect to z But viewing the problem in the zt plane we see we have a BVP Type I g E 0 We will be looking at problems where g E 0 So we are looking at the EV As in the heat equation we assume we can write u as Plugging in for our PDE we nd as before Thus we have two ODEs The rst equation is the same as in the heat equation and we know the solution the boundary conditions will be homogeneous as in the heat equation case The second equation is new though We know how to solve this ODE yielding where k1 k2 are arbitrary constants Looking at the second condition which tell us that T 0 0 we nd that k2 must be 0 Thus letting k1 be normalized to 1 we nd our fundamental solution Sticking all these solutions into a linear combination we nd Finally we check our rst initial condition um 0 x to nd that But this just means 1 has a Fourier sine series expansion so our coef cients on are given by Thus once we determine these en s we are done Type U f E 0 We are now looking at the EV We again use separation of variables assuming that umt XzTt The problem for X will be as before so we have the solution there For T our initial conditions require that T0 0 As before our solution for T is But when we enforce that TO O we nd that k1 is forced to be 0 Thus Tt sinmratL after normalizing kg to be 1 Thus our fundamental solution is Putting all these fundamental solutions together we nd To determine the Gas we use the second initial condition which requires that we differentiate our solution and set if 0 Thus mraLcn is the n th coef cient in the Fourier sine series so we have Determining these en s7 we are done Type III 1 3E 0 9 3E 0 We now have the EV where f g are the given initial position and velocity functions respectively This problem can also be done using separation of variables7 but it can also be done by just adding our two solutions from above together To see this7 let 12z t be a solution to Type 17 wz7 t be a solution to Type H7 and let Mm t vzt wzt We want Mm t to be a solution to azum 7 utt 0 Let s plug in Thus u satis es the equation Does it satisfy the initial and boundary conditions Let s check Thus u v w is indeed the solution M427K Handout Second Order Homogeneous Equations with Constant Coe icients 1 Repeated Root Salman Butt February 207 2006 NOTES 1 Chris Freeman7 Samuel Kutscha7 and John Zhang come see me after class to The average score on the test was 5070 Right now7 the curve is shaping up to be 60 or greater A 40 or greater B Below 40 TBD 03 Let s discuss Matlab SECOND ORDER HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS 1 REPEATED ROOT We are looking at the differential equation W by Cy 07 where 11 c are constants Our characteristic equation is arzbrc0 which assume has discriminant 2 7 4m O7 and thus we have one repeated root r 719201 In such a case7 we know one Of our solutions is y1t 5quot but what is y2t7 A little bit Of work reveals that we can take yg to be A quick computation with the Wronksian proves that two functions are linearly independent and hence for a fundamental set Of solutions Let s see an example y 10y 25y 07240 2 20 1 The method by which we determine our second solution is called reduction of order We merely guess that our second solution is a function vt times our rst solution That is7 we guess y2t vty1t Plugging this into our differential equation7 we get a rst order differential equation for wt 1t We solve for w and integrate it to get vt Let s see an example of this technique in a more general setting Given one solution y1t t 1 to the differential equation 2t2y 3ty 7 y 07 nd a second linearly independent solution M427K Handout Applications of the Laplace Transform Step Functions and Discontinuous Forcing Functions Salman Butt March 31 2006 NOTES 1 I gave this puzzle before break Two men are standing in front of a house One man says to the other I want you to guess the ages of my 3 daughters Here are two clues Their ages add up to the sum of the number of this house The product of their ages is 3677 After a few minutes the second man turns to the rst and says I don t have enough information and need another clue77 The rst man says My eldest daughter has red hair77 What are the ages of the rst man s daughters The rst thing you want to do is write out all possible factorizations of 36 into 3 numbers 113612181312 149 166 334 343 362 419 433 491 914 922 941 123118213611 APPLICATIONS OF THE LAPLACE TRANSFORM STEP FUNCTIONS We will assume all of our functions below are piecewise continuous and of exponential order ie their Laplace transforms exist We de ne the unit step functions or Heaviside functions as 71005 where c 2 0 Here are the graphs of 74005 and 1 7 14025 The functions are helpful in dealing with functions that have jump discontinuities Their Laplace transform is easily computed E MD 8 2118 229 236 3112 32 616 623 632 66 a We will need to translate functions by some positive amount 0 which is done using these step functions So given a function ft de ned for t 2 O we can translate this function by c to get a new function gt uctft 7 c It should not surprise you that the Laplace transforms of ft and uctft 7 c are related Theorem 1 If the Laplace transform of 1 exists for s greater than some constant a Z 0 and if c a positive constant then for s gt a we have uctft08 Observe that if we use the above relation with ft 1 we nd As an example let s compute the Laplace transform of the following function D7 sint 0 tlt7r4 7 sintcost77r4 tin4 We also have the following result which proves very useful Theorem 2 If the Laplace transform of 1 exists for s greater than some constant a Z 0 and if c is a positive constant then for s gt a c we have E 60705 8 As an example of this theorem consider the inverse Laplace transform of G 152 745 5 M427K Handout Applications of the Laplace Transform Examples Salman Butt March 237 2006 NOTES 1 If you have any friends in engineering or geology that are graduating and looking for a full time position7 URS Corporation is hoping to hire a signi cant number of new employees I have a friend who works there7 and putting her name down as a reference on their online form would be helpful to your friends7 applications The name is Erica Allis EXAMPLES 1 Sketch the graphs of the following functions on the interval 25 2 0 u1t 2U3lttgt 7 6U4lttgt u3t sint 7 3 2 Find the Laplace transform of the following function 0 tlt1 flt t272t2 251 M427K Handout Non homogeneous Second Order Differential Equations Method of Undetermined Coef cients and Variation of Parameters Salman Butt February 21 2006 This handout discusses two techniques to solve second order differential equations that are non homogeneous First some notes NOTES 1 If you have not gotten your midterm back yet see me after class or in office hours to For the love of all that is holy come pick up your old homeworks from the box attached to my office RLM 11114 03 Remember that your are required to come to office hours at least one and ask a question or explain a solution to me NONHOMOGENEOUS EQUATIONS We are looking at equations of the form Llyl y WW 6175 W 1 where we now drop the assumption that gt 0 for all t We have the following theorem Theorem 1 We will often refer to 01y1t 02y2t as the associated homogeneous solution yht the book refers to it as the complementary solution yct and Yt will be called the particular solution sometimes denoted by ypt There are a few methods to solving second order non homogeneous equations For the moment we will be interested in two techniques the method of undetermined coefficients and variation of parameters Both of these techniques are aimed at determining ypt thus they assume that one knows or can calculate the homogeneous solutions y1y2 Bear in mind that there are other techniques eg the Laplace transform which we may cover later in the course which compute the entire solution at once ie compute y and yp together METHOD OF UNDETERMINED COEFFICIENTS The method of undetermined coe icients U0 is predicated on the idea of guessing the form of yp based on the form of gt In our guess we have unknown constants 7 the undetermined coef cients 7 which we solve for by plugging our guess for yp into our differential equation and deducing the values of our unknowns Here are some key ideas to bear in mind when using U0 1 The rst step in using U0 is to rst determine y1y2 With that in hand we then think of gt as the sum gt 91t gnt where gilt are either exponential cosine sine or polynomial functions For each 91 we make a guess and set the guess for our particular solution to be ypt Y1t Ynt Keep in mind that for Y should not be yl or yg for any i If our table leads to that guess for Yi we are in an exceptional case which is discussed below The following is table describing the guesses one should make For the exceptional cases we merely use the table above to guess and if this is either yl or yg we multiply Y by 25 If tYi is not y1y2 we are done otherwise multiply once more to get tZYi which will most assuredly not be yl or yg To nish nding y we merely set y clyl ngg yp and solve for 01 and 02 using the initial conditions Let s see some examples 2 32 2y 36 4t7y017y0 0 y 22 2y 2008 27290 ZMO 0 METHOD OF VARIATION OF PARAMETERS Variation of parameters VP is a more general method than UC7 but requires that certain integrals exist and are computable We begin as before with the differential equation My y pty qty gt As in UC7 we need to be able to determine the homogenous solutions yhyg so 197 q will either be constants or 341342 will be given The idea behind VP is to guess that the particular solution is ypt u1ty1t u2ty2t where U17u2 are unknown functions of t Plugging this into the differential equation7 we are ultimately led to conclude by details made explicit in the text that the particular solution is Let s see some examples y2yy t5et y y seczt


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