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# Diff Equ & Lin Algebra MATH 2250

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Math 2250 Numerical Computations In this handout we will study numerical methods for approximating solutions to first order differential equations The fact is most differential equations do NOT have simple formulas for their solutions despite all the examples you ve seen in which they do In the case that no nice formula exists for the solution one must approximate it numerically A month from now we will see how higher order differential equations can be converted into first order systems of differential equations and that there is a natural way to generalize what we are doing now in the context of a single first order differential equation to this more complicated setting So understanding today s material will be an important step in understanding numerical solutions to higher order differential equations and to systems of differential equations We will be working through selected material from sections 2426 of the text The most basic method of approximating solutions to differential equations is called Euler s method after the 1700 s mathematician who first formulated it If you want to approximate the solution to the initial value problem dydx fxy yx0y0 first pick a step size h Then for x between x0 and x0h use the constant slope fx0y0 At xvalue xl x0h your yvalue will therefore be yl y0 fx0y0h Then for x between x1 and xlh you use the constant slope fxlyl so that at x2xlh your yvalue is y2y1fxlylh You continue in this manner It is easy to visualize if you understand the slope field concept we ve been talking about you just use the slope field where you are at the end of each step to get a slope for the next step It is straightforward to have a programmable calculator or computer software do this sort of tedious computation for you In Euler s time such computations would have been done by hand A good first example to illustrate Euler s method is our favorite DE from the time of Calculus namely dydx y say with initial value y0l so that yexpx is the solution Let s take h02 and try to approximate the solution of the xinterval 01 Since the approximate solution will be piecewise affine we only need to know the approximations at the discrete x values x002040608l Here s a simple do loop to make these computations gt restart clear any memory from earlier work gt XO00 xnl0 yOlO n5 hxn xOn specify initial values number of steps and size x00 mu10 y0210 n5 h12000000000 gt fxy gty this is the slope function fxy in dydx fxy in our example dydx y fxy gty gt xx0 yy0 initialize xy for the do loop x0 y10 39gt print x y expx gt for i from 1 to n do gt k2 fxy current slope use to suppress output gt yz y hk new y value via Euler gt x x h updated x value gt printxyexpx display current values gt and compare to exact solution gt od odll ends a do loop nygg 2200000000 7430083706 9025013499 2400000000 8916100447 1102317638 2600000000 1069932054 1346373804 2800000000 1283918465 1644464677 3000000000 1540702158 2008553692 Notice your approximations are all a little too small in particular your nal approximation 2488 is short ofthe exact value of exple271828 The reason for this is that because of the form of our fxy our approximate slope is always less than the actual slope We can see this graphically using p1TsiVithplotswithlinalg Warning the protected names norm and trace have been redefined and unprotected gt xvalvectornl yvalvectornl to collect all our points xval arrayl 6 yval arrayl 6 gt xval l xO yval l yO initial values xval1 0 7 yval1 10 gt paste in the previous work and modify for plotting 39gtforifromltondo gt xxval i current x gt yzyval i current y gt k2 fxy current slope gt yvalil y hk new y value via Euler gt xvalil x h updated x value gt od odll ends a do loop gt approxsolpointplotseq xval i yval i i1 nl gt exactsolplotexpt tO l color black gt used t because x was already used above 39 gt displayapproxsolexactsol o 032 04 06 08 1 t If you connect the Euler dots in your mind the picture above is like the one in gure 243 on page 109 of the text The upper graph is of the exponential function the lower graph is of your Euler approximation The reason that the dots lie below the true graph is that as y increases the slope fxyy should also be increasing but in the Euler approximation you use the slope at each lower point to get to the next higher point It should be that as your step size h gets smaller your approximations get better to the actual solution This is true if your computer can do exact math which it can t but in practice you don t want to make the computer do too many computations because of problems with roundoff error and computation time so for example choosing h00000001 would not be practical But trying h001 should be instructive Since the width of our xinterval is l we stepsize h001 by taking an nValue of subintervals to belOO We keep the other data the same as in our rst example The following code only prints out approximations when h is a multiple of 01 gt XO00 xnl0 yOlO nlOO hxn xOn x0 0 xn 10 y0 10 n 100 h 001000000000 gt y0 1 0 i y0 10 gt x xO yy0 x 0 gt gt gt gt gt gt gt gt gt gt gt for i from 1 to n do k2 fxy current slope yz y hk new y value via Euler x x h updated x value if fracilOO then printxyexpx fi use the if test to decide when to print the command frac computes the remainder of a quotient it will be zero for us if i is a multiple of 10 0d 0100000000011046221261105170918 02000000000L220190040L221402758 03000000000L347848915L349858808 0400000000014888637341491824698 05000000000L644631822L64872127l 06000000000L816696698L822118800 0700000000020067633692013752707 0800000000022167152192225540928 0900000000024486326772459603111 L00000000027048138332718281828 e can also see this graphically gt VVVVVVVVVVVVV xvalvectornlyvalvectornl to collect all our points xvalarrayl101 yvalarrayl101 xvallx0 yvally0 initial values xval1 0 yvaqL0 paste in the previous work and modify for plotting for i from 1 to n do xxvali current x yyvali current y k2 fxy current slope yvalil y hk new y value via Euler xvalil x h updated x value od odll ends a do loop approxsolpointplotseqxvaliyvali i1nl exactsolplotexpttOl color black used t because x was already used above displayapproxsolexactsol 26 24 22 Actually considering how many computations you did with n100 you are not so close to the exact solution In more complicated problems it is a very serious issue to nd relatively ef cient ways of approximating solutions An entire field of mathematics numerical analysis deals with such issues for a variety of mathematical problems The book talks about some improvements to Euler in sections 25 and 26 If you are interested in this important eld of mathematics you should read 25 and 26 carefully Let s summarize some highlights below For any time step h the fundamental theorem of calculus asserts that since dydx fxyx gt restart gt yxh yx Int dydt tx xh xh h d yd yx yx dt t gt yxhyx Intftyttxx xh W h yxl Kt yt dt The problem with Euler is that we always approximated this integral by hfxyx ie we used the lefthand endpoint as our approximation of the average height The improvements to Euler depend on better approximations to that integral Improved Euler uses an approximation to the Trapezoid Rule to approximate the integral The trapezoid rule for the integral approximation would be l2hfxyxfxhyxh Since we don t know yxh we approximate it using unimproved Euler and then feed that into the trapezoid rule This leads to the improved Euler do loop below Of course before you use it you must make sure you initialize everything correctly We ll compare it when n5 to our rst unimproved attempt 7 gt XOO y0lxnl0n5 h xn XO n XXO yy0 f XIY gtY x0 0 y0 l xn 10 n 5 h 02000000000 x 0 y l f I x y gty gt for i from 1 to n do gt klfxy left hand slope gt k2fxhyhkl approximation to right hand slope gt k klk2 2 approximation to average slope gt y yhk improved Euler update gt x Xhz update x gt printxyexpx gt 0d 02000000000 1220000000 1221402758 04000000000 1488400000 1491824698 06000000000 1815848000 1822118800 08000000000 2215334560 2225540928 1000000000 2702708163 2718281828 Notice we almost did as well with improved Euler when n5 as we did with n100 in unimproved Euler One can also use Taylor approximation methods to improve upon Euler by differentiating the equation dydx fxy one can solve for higher order derivatives of y in terms of the lower order ones and then use the Taylor approximation for yxh in terms of yx See the book for more details of this method we won t do it here In the same vein as improved Euler we can use the Simpson approximation for the integral instead of the Trapezoid one and this leads to the RungeKutta method which is widely used in real software You may or may not have talked about Simpson s Rule in Calculus it is based on a quadratic approximation to the function f whereas the Trapezoid rule is based on a first order approximation Here is the code for the RungeKutta method The text explains it in section 26 Simpson s rule approximates an integral in terms of the integrand values at each endpoint and at the interval midpoint RungeKutta uses two different approximations for the midpoint value 39gt xx0 yy0 n5 hxn xOn x0 yl n5 h12000000000 39gt for i from 1 to n do gt klfxy left hand slope gt k2fxh2yhkl2 lst guess at midpoint slope gt k3fxh2yhk22 second guess at midpoint slope gt k4fxhyhk3 guess at right hand slope gt kkl2k22k3k46 Simpson s approximation for the integral gt xxh x update gt yyhk y update gt printxyexpx display current values gt 0d 02000000000L221400000L221402758 04000000000L491817960L491824698 06000000000L822106456L822118800 0800000000022255208252225540928 7 100000000027182511362718281828 gt Notice how close RungeKutta gets you to the correct value of e with n5 Even with code like RungeKutta there are subtleties and problems which particular problems will cause We will not go into those here there are good examples in sections 2426 and the homework problems Math 2250 2 Summer 2007 Linear Algebra Computations with Maple gt restart gt withlinalg load the linear algebra library Warning the protected names norm and trace have been redefined and unprotected De ning matrices and vectors Use the Matrix command De ne it row by row with each row inside and note the number of brackets you need to put Also use upper case calling the Matrix function gt A Matrix 1234l5l6789 1 A1 4 5 6 gt B Matrix 1200l5l2280 1 1 2 0 Bi 0 5 2 2 8 0 gt v Matrix 114131 1 1 v 4 3 You can also setiup entries of a matrix automatically For example de ne the ij entry of a matrix by the following function gt f ij gt ij fiij iJ Then you can set up an 8x8 matrix with entries as above by typing gt G Matrix8f 39 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 G 5 6 7 8 9 10 11 6 7 8 9 10 11 12 7 8 9 10 11 12 13 8 9 10 11 12 13 14 9 10 11 12 13 14 15 So that the 11 entry is 11 2 and the 12 entry is 123 and so on Matrix operations gt AB adding two matrices 2 4 3 4 10 8 9 16 9 gt AB matrix multiply using not 39 7 36 4 16 81 10 25 126 16 gt AB Error in rtableProduct invalid arguments gt Av matrix and vector multiplication 18 42 66 gt A22 you can take the square of a matrix so that A 2 AA 30 36 42 66 81 96 102 126 150 gt AA Error in rtableProduct invalid arguments gt A34 you can raise the matrix to any power you want 7560 9288 11016 17118 21033 24948 26676 32778 38880 Identity Matrix and Zero Matrix gt I3 Matrix33shapeidentity 3x3 identity matrix 1 0 0 31 0 1 0 0 0 1 gt Z3 Matrix33 3x3 zero matrix 0 0 0 Z3 0 0 0 0 0 0 Row Operations You need to know three function calls mulrow swaprow addrow Note all of these functions start with a lower case gt M Matrix 22 l 63 3536542l3 J 2 2 71 6 M 3 73 5 36 5 4 2 13 The lnction mulrow will multiply a row by a constant mulrowA24 will multiply row 2 of matrix A by 4 Another example below gt M1 mulrowM 1 12 multiply row 1 of matrix M by 12 1 1 1 73 2 M1 3 e3 5 36 5 4 2 13 The function addrow will multiply a row and then add it to another row addrowA135 will multiply row 1 of matrix A by 5 and then add it to row 3 So addrowmatrixiname rowia rowib coe means coe krowia rowib gt M2 addrowMll2 3 3rowl row2 1 1 1 73 2 M2 0 6 E 45 2 5 4 2 13 gt M3 addrowM213 5 5rowl row3 1 1 1 73 M3 0 6 E 45 2 0 71 3 28 2 Finally swaprow will just swap two rows swaprowA 13 will swap row 1 and row 3 of matrix A gt M4 swaprowM323 swap row 2 and 3 1 1 i 73 2 M4 0 1 2 28 2 0 6 E 45 2 Augmenting matrices and reduced row echelon form There is an automatic way to reduce a matrix to the reduced row echelon form gt RREFM rrefM l 0 0 l RREFMi 0 1 0 1 0 0 l 6 To augment two matrices together gt Av MatrixAv gt AB MatrixAB gt B inv gt inverseBv as a matrix gt MatrixinverseBv MatrixinverseB order to do operations after fijnv 5 8 Computing determinant and inverse this is what 2 3 1 5 6 4 8 9 3 3 1 2 0 take the inverse of matrix B you need to set this as a Matrix in 1 2 0 2 1 0 l 4 E happens if you don39t assign the inverse 0 2 1 0 l 4 4 3 1 75 2 8 l 2 1 H gt Ainv MatrixinverseA Maple will tell you if the inverse doesn t exist our matrix A above is singular Error in inverse singular matrix gt MatrixBA l you can also compute inverse by raising to the power of l 2 0 2 1 0 l 2 4 Z l i 4 2 V detB compute the determinant using det 8 gt detA this confirms that indeed A inverse does not exist as the determinant is zero 0 Eigenvalues and Eigenvectors gt e eigenvalsA compute the eigenvalues of matrix A 153 331573433 2 2 2 2 Note that one of the eigenvalue is zero which again shows that matrix A is singular not invertible gt evalfeigenvalsA numerical values 0 1611684397 71116843970 gt v eigenvectorsA eigenvectors 7 3g 1713 13x 1 V7 a 2 22 4 44 1573a331 71734331734331 01 1 1 2 2 2 224 44 72 1 gt Vll the first eigenvalue 3 2 2 gt vl2 the multiplicity of that eigenvalue L gt NH 31 gt v2l v23 1 the corresponding eigenvector 14345l4 3V51 2 22 4 44 the second eigenvalue and its corresponding eigenvector Modeling with rst and second order systems of DE s Math 22504 Friday November 16 2001 Example 1 Glucose insulin model adapted from a discussion on page 339 of the text quotLinear Algebra with Applicationsquot by Otto Bretscher Let Gt be the excess glucose concentration mg ofG per 100 ml ofblood say in someone s blood at time t hours Excess means we are keeping track of the difference between current and equilibrium quotfastingquot concentrations Similarly Let Ht be the excess insulin concentration at time t When blood levels of glucose rise say as food is digested the pancreas reacts by secreting insulin in order to utilize the glucose Researchers have developed mathematical models for the glucose regulatory system Here is a simpli ed linearized version of one such model with particular representative matrix coefficients It would be meant to apply between meals when no additional glucose is being added to the system dG E 1 4G dH 1 1 H dt Explain understand the signs of the matrix coef cients Now let s solve the initial value problem say right after a big meal when m E gt restartwithlinalg withplots gt Amatrix22 l 4 l l gt eigenvects1 l 2000000000 1 l 2000000000 0 I 0l1 l 2000000000 1 l 2000000000 0 I 0 l1 Extract a basis for the solution space to his homogeneous system of differential equations from the eigenvector information above Solve the initial value problem Here are some pictures to help understand what the model is predicting l Plots of glucose vs insulin at time t hours later gt G t gtlOOexp 1t cos 2t Ht gt50exp ltsin2t L 39gt plotGtHttO30colorblacktitle Excess Glucose and Insulin concentrations Excess Glucose and Insulin concentrations 0 5 1b 15 30 20 E 2 A phase portrait of the glucoseinsulin system 7 gt pictlfieldplot lG 4H lG lH G 4O lOOH 15 40 soltnplotGtHttO30colorblack displaypictlsoltntitle Glucose vs Insulin phase portrait G r Ss fnso rihi 39 H 0 39 I I Q r39 39 rtllll 1H 1 1 IIIIII 074 o 2 A IllI III or 2 2220 O 11 Example 2 Tanks This is example 2 from our text page 413 We have a cascade ofthree tanks see gure 732 page 413 The tank volumes are V120 V240 V350 gallons The water ow rate is r10 gmin The initial amounts of salt are Xl015 X200 X300 Show that the initial value problem for this system is dx1 dt 5 0 0 x1 dx2 5 25 0 x2 dt 0 25 2 x3 dx3 dt x10 15 x20 0 x30 0 olve this system using gt Amatrix 5 O O 5 25 O 0 25 2 eigenvects1 5 1 1 2000000000 1666666667 25 1 0 1 5000000000 2 1 0 0 1 You should get x10 3 0 0 75 725 72 x2t 5 e 0 6 30 e 0 1 125 e 0 0 x30 5 5 1 We can plot the three solute amount VS time to see what is going on gt xlzt gt15exp 5t x2t gt 30exp 5t30exp 25t x3t gt25exp 5t 150exp 25t125exp 2t 39 gt plotxl t x2 t x3 t tO 30colorblack title salt contents in each tank salt contents in each tank Example 3 Spring systems This is Example 1 page 427 We have a massspring system with two masses and two springs see gure 743 m12 m2l kl100 k250 Derive the second order system Mx Kx ie equation 13 on page 427 d2 x 2 0 dtz 150 50x1 0 1 d2 x2 50 50 x2 dtz E which is equivalent to the system d2 x dtz 75 25H x d2 x2 50 50 x2 dtz This is like a single massspring equation except now it s a system What is the dimension of the solution space Hint What size first order homogeneous system is it equivalent to If we look for solutions of the form coswtv how is the angular velocity w related to eigenvalues of the matrix A where gt Amatrix22 752550 50 75 25 14 50 50 25 1 1 21 100 1 1 11 gt eigenvects1 Find the general solution and explain geometrically in terms of the springs what s going on Math 22503 Friday August 27 2004 First order differential equations notes for 1113 HW for Wednesday September 1 bold means hand in others optional You already got this assignment on Wednesdsay and it s also posted at our home page 11 3 4 715 1619 20 27 30 34 36 46 12 5 7 10 14 20 25 26 36 44 13 3 6 811 12 14 21 25 32 33 Course home page httpwwwmathutaheduNkorevaar2250fa1104 Tuesday problem sessions 730820 am in EMCB 101 10451135 am in LS 101 1255145 pm in EMCB 105 5550 pm in LCB 219 Maple and Math lab introductions All are in LCB 115 Tuesday August 31 11501240 Wednesday September 1 2250 Friday September 3 10451135 11501240 What is a differential equation What is the order of a differential equation How do you transform a sentence about rates of change for a function into a differential equation Can you do 11 34 How do you check whether a function solves a differential equation Can you do 11 4 Or see example 2 below A First order differential equation is any equation which can be written as dy xy EJ0 W 1 y dx is a differential equation Often we can rewrite a general DB in the more convenient form For example d j x y Can you do so for the example above No matter how we write the differential equation the goal is to understand the functions yx which make it true Sometimes we want the particular solution yx which satisfies an initial condition 1 yxo y0 Then the corresponding problem is called an initial value problem The reason for the name quotinitial conditionquot is that our independent variable will often be time quottquot rather than x and the initial condition will be the value of the solution function at time t0 for example an initial population in a population growth problem In section 12 the book focuses on the easiest kind first order differential equations to solve namely ones of the form dy dx 4m Notice the unknown function yX does not appear on the right hand side of this DE so you are simply looking for antiderivatives of fX with respect to X yJ xdxC You got good at antidifferentiation in Calculus Basic graViational attraction problems in physics are of this form when quottquot is the variable Sometimes more complicated problems also reduce to antidifferentiation Example 1 Solve this initial value problem and then graph the solution function dy dx x 3 3712 Here s how Maple would solve the problem gt withDEtools load differential equations package gt dsolvediffyx xx 3yl2yx l 2 9 yx x 3 x gt Maple is a word and math processing package think Microsoft word that does math You can buy Maple from the bookstore for around 100 or use it on campus computer systems where it already eXists Example 2 Solve the initial value problem dy dx y x Y00 Hint You CANNOT solve this problem by pure antidifferentiation since you can t antidifferentiate the yX on the right side if you don t know it yet We WILL learn an algorithm for solving this sort of DE in section 14 For today you may use the fact that the general solution to this differential equation is given by yxx l Cex Step I verify that yX does solve the DE Step 2 nd a value of C in the general solution to solve the IVP Step3 graph the solution function Slope Fields And now for something which seems completely different but isn t The geometric interpretation of rst order differential equations is connected with slope elds The use of the word quot eldquot is analogous to how it s used with the pre xes quotforce electric magneticquot quotwhea quot Slope elds will be discussed in more detail on Monday and we introduce them now The differential equation dy dx x y is telling us that the slope of the solution graph at a point xyx is given by the function fxy We can use this fact to approximate solutions especially when exact formulas cannot be found By hand or computer we can draw a picture of the slope eld determined by f ie at each point xy assign a slope using the formula fxy The natural way to indicate these slopes is with small line segments see below Solution graphs to the differential equation will be tangent to these slope elds If there is an initial condition yx0 y0 then the solution graph will pass through the point x yO Example 1 The initial value problem 7 Ly dx Y1 2 x 3 lgicture of slope eld Slope field icture for example 1 w u U W t 7 s wyy ss3va4 6x lt 77 M ii l An A A xyyyyyygt7 v a l 3 black yx x X 0 0y y x 0AV i Maple solution 0 d yx xl e dx this is example 2 W dy dx Y0 abth 1 y blacklinecolor 2 Slope field and solut on for example 2 yxx 0yx color diffyxx line 0 title Slope field and solution for example 2 make a picture DEplotdeqtnyxx 2 arrows dsolvedeqtny0 y0 Geometric picture and the Maple commands that generated it gt deqtn 39 39 Example 2 The initial value problem 39gt dydx nv Some real examples Example 2 page 12 13 Example 4 page 1516 MATH 22501 MassSpring systems December 1 2008 This maple document covers section 74 and contains commands you will need for the Earthquake exploration you re doing at the end of that section Here is how Maple computations work out the running example in today s notes gt withlinalg gt with plots with DEtools gt Mmatrix 20 01 Kmatrix 150 50 50 50 A evalminverse M ampK 150 50 K 50 50 75 25 A 50 50 3910011125112 Therefore the natural frequencies of this system are the 10 and 5 and the two fundamental modes correspond to the masses moving in opposite directions with equal amplitudes and angular frequency 10 and in parallel directions with amplitude ratio of two and angular frequency 5 as discussed on page 2 of today s handwritten notes gt eigenvectors A Now let s consider the forced system with force vector equal to coswt050 ie the second mass is being forced periodically In other words the system Mx Kx F where Fcoswt050 this is Example 3 on page 327 and worked out in today s notes 39gt FOevalminverseMampvectorO50 The F0 in the normalized equation 30 page 436 needs serious modification for Maple project Idenarrayl2l2identity the 2 by 2 identity matrix Aleftomega gtevalmA omega 2Iden the matrix function multiplying c on the left side of 32 comega gtevalm inverseAleftomegaampFO the solution vector comega to 32 obtained by multiplying both sides of equation 32 on the left by the inverse to Aleft gt comega see equation 35 page 437 and our hand work on page 3 of today s notes 1250 50 7502 2500 125 0204 2500 125 02 104 The vector cw as above times the function of time coswt is a particular solution to the forced oscillation problem we are considering If we assume that our actual problem has a small amount of damping then we expect that this particular solution is very close to the steady periodic solution to the damped problem as we just discuseed and also on pages 437438 text Study resonance phenomena for these slightly damped problems by plotting the maximum amplitude for the individual masses in the steady state solutions to the undamped problems Maple has the command norm to measure this maximum amplitude ready for you to use in your Maple project 39 gt norm c omega 75 02 1250 ma 2 4 50 2 4 2500 125m0 2500 125a 0 I The following graph of the maximum amplitude norm for the undamped particular solution shows that in the slightly damped problem we expect practical resonance when omega is near 5 or 10 radians per second gt plot normc omega omega0 15maxamplitude0 15 numpoints200 color black 14 12 10 maxamplitude 8 6 4 2 0 i 21 10 1 2 1 4 6 8 omega This is qualitatively the picture on page 437 figure 7410 although they plotted the Euclidean magnitude of comega rather than the maximum of the individual amplitudes Notice how we get Maple to label the axes as desired We can get a plot of resonance as a function of period by recalling that 2PiTomega gt plot norm c 2Piperiod period0 l 3 maxamplitude0 15 numpoints2 OO color black 14 12 10 maxamplitude 8 6 4 2 0 015 1 1 5 2 215 3 penod COMMENTS FOR THE EARTHQUAKE PROJECT 1 Students are often confused by the forcing term in equation 2 of page 440 namely gt E omega A2cos omegat b E 02 cos0 t b where b is the transpose of lllllll They ask how can the earthquake be forcing all seven stories it seems like it s just shaking the bottom one Well the students are correct but so is EdwardsPenney The authors talk about an opposite inertial force being the reason for this forcing term and there s a detailed discussion in some notes I ve posted with this project on our project page Here s a brief summary Think of the ground as the zeroth story In the rest frame it is shaking with oscillation Ecoswt And so its acceleration is its second time derivative namely Ew 2coswt If you write down the inhomogeneous system of EIGHT second order DE s for the accelerations of stories zero thru seven the forcing well accelerating term is Ew 2coswtl0000000 as you would expect Call the solution 8vector to this system yt then see what the shaking looks like to someone on the ground by letting xt 7tEcoswt llllllll Then the zeroth story component of xt will be identically zero and the other seven components will satisfy equation 2 on 440 exactly as the authors claim This is worked out in detail in the notes at httpwww math ntah edn Nkuicvaal eaquot 39 ndf Very important note also repeated on your Maple project 2 For large matrices the eigenvect command won t work well unless you enter at least one decimal number if all entries are rational numbers expressed without decimal points Maple tries to find the eigenvalues and eigenvectors algebraically and exactly instead of numerically and often fails Make sure at least one of your matrix entries has a decimal point in it Math 22501 November 11 2008 Guess the resonance game using convolution formula section 104 gt with plots with inttrans i the library inttrans includes Laplace We are considering the undamped forced harmonic oscillator X t Xt ft with initial data X0V00 When we take the Laplace transform of this equation we deduce 1 F s X 2 s 1 so that the convolution theorem implies Xt sin Kt Since the unforced system has a natural angular frequency 00 l we expect resonance when the forcing function has the corresponding period 2 7c of 2 TE We will discover that there is a surprising possible error in our reasoning w 0 Example 1 A square wave forcing function with amplitude l and period 2 TE Let s talk about how we came up with the formula which works until I 11 TB 39 t gt l2sum l AnHeaViside t nPi nO 10 Heaviside was an early user of the unit step function and so Maple names it after him 10 f t gt l 2 2 1quot Heavisidet n TE n 39gt plotfttO30colorblacktitle square wave forcing function square wave forcing function 5 NW D We U U U UL gt xt gtintsint tauftautauOt convolution formula for the solution 0 o x t gtJsint 17t17d17 0 You actually could use the convolution formula to work out Xt by hand if you wished to do so 39 gt x t you actually could work this out by hand 1 l 2 HeaVisidet 3 TE 2 HeaVisidet 2 HeaVisidet 3 TE cost 2 HeaVisidet cost 2 HeaVisidet 4 TE 2 HeaVisidet TE 2 HeaVisidet 4 TE cost 2 HeaVisidet TE cost 2 HeaVisidet 5 TE 2 HeaVisidet 2 TE 2 HeaVisidet 5 TE cost 2 HeaVisidet 2 TE cost 2 HeaVisidet 6 TE 2 HeaVisidet 9 TE cost 2 HeaVisidet 6 TE cost 2 HeaVisidet 10 TB 2 HeaVisidet 7 TE 2 HeaVisidet 10 TB cost 2 HeaVisidet 7 TE cost 2 HeaVisidet 8 TE 7 2 HeaVisidet 8 TE cost 2 HeaVisidet 9 TE cost As expected we get resonance 39 gt plot x t t0 3O colorblack title resonance response resonance response Example 2 triangle wave forcing function same period 39gt gt gtintfuuOt l5 this should be a triangle wave I g t gtJiudu 15 0 gt plotgttO30colorblack title triangle wave forcing function triangle wave forcing function 15 1 05 0 10 15 2 2 5 3 0 o5 1 15 gt yt gtintsint taugtautau0t y raj sint l7 g l7 d1 0 gt plotyttO30colorblack title resonance response msonanceiesponse 15 10 mi O N 04 2 30 5 t 10 15 Example 3 Now let s force with a period which is not the natural one This square wave has period 2 39 gt h t gt l2sum l AnHeaViside t n nO 30 30 h t gt 1 2 2 1quot HeaVisidet n n70 39 gt plotht t0 30colorblack title forcing function forcing function 05 1 39 gt zt gtintsint tau htau tau0 t plot z t t0 30colorblack title resonance response z t gtJ sint 17h17d17 0 resonance response 22AM m 3 v WK 06 Example 4 A square wave which does not have the natural period so we don t expect resonance 39 gt kt gtsumHeaviside t 4Pin Heaviside t 4Pin Pi n0 5 5 k t gt 2 Heavisidet 4 7 TE HeaVisidet 4 n T TE quot0 39 gt plotkt t0 60colorblacktitle forcing function period gt forcing function period 2 0 1 0 2b 3P 4 0 50 6b gt wt gtintsint tau ktau tau0 t t wt gtJsint Ek cd c 0 gt plot wt t0 60 colorblack title resonance response resonance response 0 MMM 5 VVVVV 10 10 Hey what happened How do we need to modify our thinking if we force a system with something which is not sinusoidal in terms of worrying about resonance 7 Math 22504 Friday August 23 2001 First order differential equations notes for 12l3 HW for Wednesday August 30 11 3 4 7151619 20 27 28 33 35 36 12 Remember what a differential equation is What is the order of a differential equation Don t we get to see some REAL differential equations When did the body die A First order differential equation is any equation which can be written as For example is a differential equation Often we can rewrite the general DB in the more convenient form m m i a A F l l JN n H d j x y Can you do this for the example above No matter how we write the differential equation the goal is to understand the functions yX which make it true Sometimes we want the particular solution yX which satisfies an initial condition f yxoy0 Then the corresponding problem is called an initial value problem The reason for the name quotinitial conditionquot is that our independent variable will often be time quottquot rather than X and the initial condition will be the value of the solution function at time t0 for example an initial population in a population growth problem ones of the form dy dx x Notice the function yX does not appear on the right hand side of this DE so you are looking for i In section 12 the book focuses on the easiest kind first order differential equations to solve namely i antiderivatives of fX with respect to X y J x dx C You got good at antidifferentiation in Calculus Graviational attraction problems in physics are often of this form when quottquot is the variable Sometimes more complicated problems also reduce to antidifferentiation 7 Example 1 Solve this initial value problem and then graph the solution function dy dx yl 2 x 3 7 Example 2 Solve the initial value problem Ly dx y x Y00 Hint You CANNOT solve this problem by pure antidifferentiation since you can t antidifferentiate the yX on the right side if you don t know it yet We WILL learn an algorithm for solving this sort of DE in section 14 For today you may use the fact that the general solution to this differential equation is given by 7 yxx l Cex Step I verify that yX does solve the DE Step 2 nd a value of C in the general solution to solve the IVP Step3 graph the solution function 7 Slope Fields And now for something which seems completely different but isn t The geometric interpretation of first order differential equations is connected with slope fields The use of the word quotfieldquot is analogous to how it s used with the pre xes quotforce electricquot quotmagneticquot quotwhea quot Slope fields will be discussed in more detail on Monday and we introduce them now The differential equation dy dx x 7 is telling us that the slope of the solution graph at a point xyx is given by the function fxy We can use this fact to approximate solutions especially when exact formulas cannot be found By hand or computer we can draw a picture of the slope eld determined by f ie at each point xy assign a slope using the formula fxy The natural way to indicate these slopes is with small line segments see below Solutions graphs to the differential equation will be tangent to these slope fields If there is an initial condition yx0 y0 E then the solution graph will pass through the point x yo Example 1 The initial value problem 7 Ly dx Y1 2 x 3 T Picture of slope field 3 ield picture for example 1 gt withDEtools gt gt deqtndiffyx arrowsline dirgrid3030 make a picture Xyx dsolvedeqtny00 DEpiotdeqtnyxx 22y00y 31 colorblacklinecolorblack e lll N sV gtwv 2 N sV Y gty w m q 11 N sV gtwv iiiHHRRRQQEF Wi q 0 Ex I sp3ww lllll k V wy w m lli 2W 1J Example 2 The initial value problem 39gt dydx y x YO0 dy dxyx Y00 E Geometric picture Maple commands left in this time differential eqtns command library this is example 2 Maple solution X Yxi title Slope field and solution for example 2 B deqtn EC yx yx x xl he 2 b P X x m m e r z m n 0 m 2 r lo 4 1 ww S I d z M V pane A e P Io S JT 7 Some real examples Example 2 page 13 Example 4 page 1617 Example from Section 36 Math 225037 Fall 2007 Elementary row and column operations and determinants Let A be a square matrix 1 If B is the matrix obtained from A by multiplying any row or column of A by a constant 57 then detB cdetAl 2 If B is the matrix obtained from A by interchanging any two rows or two columns7 then detB 7 detAl 3 If B is the matrix obtained from A by adding any multiple of a row or column of A to a different row or column7 then det B det Example 1 5 4 A 2 72 0 0 1 3 We Will nd the determinant of A by using elementary row or column operations to turn this matrix into a triangular matrix7 and then simply multiply the diagonal elements Interchange 2nd and 3rd rows 1 5 4 0 1 3 2 72 0 Subtract 2times 1st row from 3rd row 1 5 4 0 1 3 0 712 78 Add 12times 2nd row to 3rd row 1 5 4 0 1 3 0 0 28 We used rule 2 once and rule 3 three times7 so 15 4 1 A2 72 070 0 1 3 0

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