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# Class Note for MATH 254 at UA

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Date Created: 02/06/15

MATH 254 Lectures 34 0 Seporoble rst order ODEs Applications 0 Direction elds 0 First order autonomous ODEs This week we will be studying rst order ODEsi jg my 21gt We examine three approaches First as already discussed last week we study separable ODEs for which ay 9Ihy 2 2 and which are solved by getting all 17s on one side y s on the other and integrating For y values such that My 0 dy 7 I I 7 ltgtd 23gt gzdzc 24 Second we will look at the method of direction elds In this method we simply observe that the differential equation 21 tells us that the slope of the solution curve at zy is given by This allows us to draw at each point zy a unit tangent vector to the solution curve at that point We will see that from the collection of such unit tangent vectors it is easy to see quantitatively what the solution does Third we look at autonomous odels where the RHS fz y is a function only of y the dependent variable Example 1 Quiz 1 Q1 in ow 2m Valve 1 tank is kept well stirred out ow Given At time t 0 mini7 valves A amp B are opened If 1t lbs amti of salt in tank after t mini E 7 E 25 For z 107 dI 772d 1 1 5 7m1hiflm C n m 01 Cl 1h7777710 n m Taking the exponential of both sides7 we obtain 6721671C 1 il 10 t l7 l 7 whereas if 1t gt z z z 7 i i 7 7 if l 10 10 ll Which Sign is l 10 i We certainly know its Sign at t 0 because we were given 10 5 But does the sign of l 7 always remain the 10 E W 10 i The answer is YES because in order that l7 W change How do we solve for 1t If 0 lt 1t lt 107 m same as that of l 7 2 sign7 zt would have to cross the line I 10 But it cannot cross I 10 because I at z 10 i 0 and so the solution curve zt is horizontal therei 7dt When we study autonomous rst order equations later this week7 you Will see this even more clearly So7 What you do When you nd is to remove the absolute value signs by Writing IO 7 2 l 7 Ae 39 10 Where A can be positive or negative depending on Whether 0 lt 10 lt 107 or 10 gt10 Note A 6710 gt 0 ifO lt 10 lt 10A 76quot10 lt 0 if 10 gt 10 In our case 10 5 so that and therefore A The solutions of our initial value problem is 1 m 10 lt17 if i Graph of solution zt 10 t 5 mini zt How long does zt take to get to 7 lbs 1 l l 1t1 7E 10 lt17567392 1gt 567211 2E or i2t1 ln2 or t1 Ean rninutesi Example 2 Time of death Q2 MT 7 Tan 26 For T Tan T fTTm ikdt MT 7 Tairi ikt C iT Tairi e k ec T 7 Ta AM At t 0Tt T0 A T0 7 Ta Tt 7 Ta T0 7 Tage k We know at t tag Ttd 88i6 Fi Therefore since T0 986 FTair 68i6oF 20 30e k di An hour later It td 1Ttd 1 78 60Fi Hence 10 30eklt d1gti Divide and nd 2 ek k 1n2i The unit of k is inverse hoursi From the rst equation 3 kid in E in 3 7 1n 2 1n 3 id 71112 m 1 tag 2 06hr 2 36min Since ta corresponds to 2 am the time of death is 124 arni Example 3 Radioactive Decay We build our mathematical model from the observed fact that the rate of decay of a radioactive isotope eigl thorium 234 within a substance is proportional to the amount of isotope present at the given time t Let the amount of isotope present at time t Note the amount can be measured as the number of radioactive atoms or the mass of the radioactive component eigl OlOOlkg of C050 Then dQ 7 ile dt We have 77 because Q decays7 k7 is constant of proportionality measured in units of inverse time eg years l7 and Q7 is the amount present at time t Let us suppose that at time t to we know there is the amount Q0 presentl Solve the ilvlpi dQ 7 E 7 imam Q0 27 This equation is both separable and linear see Lectures 56 to come Linear you will meet this method next week dQ EMQ 7 0 LE e d a Qekt 0 Qek c aconstant Q Geek At time t to Q Q0 Qoe ka wh Separable dQ 7 E kQ dQ i ikdt Q logelQl ktdr constanti Since Q gt 0 Q constant e because em cant is constant or Q Qoeikaito The idea for using this simple idea and for which a Nobel Prize was awarded to Willard Libby in 1960 is as follows Suppose we know the decay rate k and the amount of a particular isotope Q0 that would be present in a piece of tissue at the time of death to then by measuring at the present time we can determine t 7 to This process is called Radioactive Dating Many times the isotope used is an isotope of carbon C14 then the process is often called Radiocarbon Dating but many other radioactive elements are used as well Example Suppose Q0 100 mg and we observe that after 7 days is 8204 mg From this information we know 16 because Q7 100e 7k 8204 where time t is measured in units of days 1 100 1 k 7 loge N 0 03days After what time T will there be 50 mg left 1 l 50 100 exp7kT T E loge 2 or 24 days The constant k is often written as 1 T where T in units of time is the half life of the isotope Namely after T units of time the amount of radioactive isotope left QT 7 Qoexp bgegT k loge 2 T QoeXPIOgeTl 2 As an exercise look up the halflives for several radioactive substances eigi Carbon 14 and learn how the carbon dating method works in practice 6 Example 4 Population dynamics Left on its own and with access to unlimited resources the population ofa species will grow at a rate proportional to the population77 Malthus7 law Let yt be the population at time t Then dyTit km 28 If nothing changes then yt yo expkt 7 to 29 grows exponentially In class we will discuss the validity of such a model and possible variations but it is not hard to see why it might be valid lndeed from 1800 to 1845 the population in Ireland grew in just such a fashion mainly because of the introduction of a new staple diet 7 the potato But we all know from experience that unlimited growth is a rare phenomenon and that more often then not there are other in uences which slow the growth phenomenon down For example one might very well argue that the growth rate k should be a function of y itself because the bigger y is the greater the competition for the nite amount of resources Also there may be other species about which act as predators We will study this effect later For now let us look at a model where k the growth rate decreases with The simplest model would be d g k 7 am 210 which we will rewrite as dy 7 y a klt17 f y 211 Equation 211 is known as the logistic equation For y 0K dy ya 7 M W To proceed you will have to do a partial fraction 1 A B ylt1eykgty1eyK A review of the method to nd A amp B will be given in your recitation section For 1 now I Will tell you A 71B 7 Then Z lt1gt 1HWFJHH7yN ln 4 17 yK Integrating both sides we nd 9 7 ln 7 1675 C or y kt C l1 7 yKl e e or y m 17 yK Ae Where A can be positive or negative y0 If 039 39 A i y1sg1ven 17y0K Aekt t 7i 1ekt For 0 lt y0 lt K A is positive and the graph of yt looks like 2t K t The population grows exponentially at the beginning but then saturates When yt approaches Ki If y0 gt K then is negative but its absolute value is greater than unity The denominator 1 ek can never be zero for 0 S t lt 00 The graph of yt looks like W Again7 the population saturatesi What happens in the nonrealistic in this context case Where y0 lt 0 Then AK is negative and lies between 71 and 0 Then there is a nite time to such that K emu 77 Ar K remember 71 1s pos1tlve and greater than one W t Direction elds and SLUPES Consider the rst order ODE dy i 21 d1 f I y Given fzy we know the slope of the solution through the point z Example Consider equation 13 with Q1 Q2 3 galmin C1 l lbgal and G 100 galsi Namely the in ow has a rate of 3 gallons per minute with a concentration of 1 lb per gallon while the well mixed out ow has the same rate of 3 gallons per minute The total volume of brine in the tank stays the same Here I is time y is amount of salt dy 3y Eigiml 213 y 200 m 100 quotquotquotquot w fff f ff I At each point z y draw a directed arrow of unit length whose angle 9 with the z axis satis es tant9 fz y 37 Wgoy We will take the angle 9 which lies between 7 and gig lt 9 S For example if tant9 ioo t9 lf tant l 9 El lf tant9 71 t9 7 The convention is to draw the arrows in the direction of increasing 1 time Since in this case is independent of z the directed arrows have the same direction for all I at the same yl Check that 1001 7 e is a solution of 213 It has the property y0 0 Note that the solution has the property that everywhere along it the direction arrows are its unit tangentsl This is a very convenient way to visualize what the solution does We will study how to take full advantage of this using the UA toolkit program SLUPESl This program takes the rst order equation dy 7 Gzy 7 numerator dz 7 Fzy 7 denominator 2 14 and will draw the direction eld the solution and a graph of an expression for the solution that you put in 10 As an example we will solve7 equation 13 with Q1 Q2 3 C1 l G 100 and C0 0 ie dy 3 7 7 7 0 0 d1 1 7 y UA SoftwareToolkitsSLOPES To access log in password Main Menu Teaching NZ UA Software a Toolkits ZlSLUPES Function Operation Disk Operation Other Operations 7 use aI I OWS 7 gt press enter Create function Edit I use arrowsl Plot slopes Quit gt numerator 3 7 3 ylOO gt denominator l gtDisplay graph on square screen YNEE arrows gt x coordinate on left of screen 0 lt7 arrowsdeletenurnbers bx coordinate on right of screen 100 gt y coordinate on bottom of screen 0 by coordinate on top of screen 150 gt initial x value 0 gt initial y value 0 gt 100 7100 exp73 1100 Exact solution with y0 0 gt Plot slopes move arrow down You will obtain a screen displaying the z and y axes ll gtDisplay graph on square screen YNEE arrows gt x coordinate on left of screen 0 lt gt arrowsdeletenumbers gt x coordinate on right of screen 100 gt y coordinate on bottom of screen 0 gt y coordinate on top of screen 150 gt initial x value 0 gt initial y value 0 gt 100 7100 exp73 1100 Exact solution gt Plot slopes move arrow down You will obtain a screen displaying the z and y axes Homework Use SLUPES to draw the direction eld for ailiy bi21yzl Also for the initial condition a y0 2 b y0 1 ll draw the graph e of the exact solution 2 write an expression on your printout for the exact solution 3 use the integrator to numerically calculate the solution should coincide with e Remember the unit arrow is simply the tangent to the exact solution which passes through the base of the arrowi You may also use other software to plot direction elds if you wish 12 dy 3y D t l f 1rec 10n ed or dz 3 100 3 ffxfiIxffixfrw39zxfflf f 31271 fife ftfla fffi a nr 8quot draws dimction eld quotd pm V r at i npat nlutimx and i numerically inmgt esf ljl mr I f u I v if FIGURE 1 13 Another example dy 7 1 dz 7 y z y tan19 197glt19lt 0 y 0 0 0 I f 0 y 0 ioo 7r2 a a 71 77r4 a is any real number 7a a 1 7r4 1 J3 7 46 71 J3 7r6 J3 1 7 43 7 1 7r3 y 1 1 1 I 14 At each point 17y we have simply placed a unit line whose directions makes the angle 9 with the z axis These little lines are the unit tangent vectors to the solutions curve of the point I It is easy to see that the solutions are circles in this case In general7 it takes a long time drawing gures by hand The program SLOPES and the software available in your book does it automatically As HWY7 I ask you to draw the direction elds for dyi sinz 80 y dyiizzg bl T Do them rst by hand and then by SLUPESi Examples a ky iyK Take kKli dy b 3 y17 M 7 y 15 Example Slopes for dt y1 y2 y I i aquot at w w h 7 Jquot W a 1 rquot r w M I 1 939 m3 w 130 H r w w o quot 1 1w H Q 8 I 9 aquot 052 Wm 11quot I 1 v38 NaV I r 0 t2va 39 rms at 1 V x w a 9 mi 4 x M I M M i quotV 8 f I W 88 I u m Iquot 4quot W I soquot W 51 x x x I 39E 394 I V39 gt A t I 4quot W 11 I p A in FIGURE 2 16 First order autonomous equations my my 2 15 215 is autonomous when fz y does not depend on the independent variable 1 We can solve 215 by the separable method ie for My 0 dy ix 0 216 We can also solve it using direction elds But we can also get a very good picture of what the solution does by simply graphing My vsi yl Suppose My A BCVDEy the graph of My looks like this If one were to begin with a y0 between B and C say what would happen d At all points between B and C My gt 0 and therefore dig My is positive there 1 Thus as 1 increases will increase and so at an instant of time 11 later y0 will become yzl which is now closer to C than y0 but which still lies between B and Cl Repeat the argument Since yzl lies in B CMyzl gt 0 and therefore is again positive continues to increase until eventually it reaches C When y d reaches C dig My at C 0 and so does not change anymore Likewise if I we start with y0 in C D My lt 0 and will decrease as 1 increases Again it will end up at C We can now draw arrows in each of the intervals indicating which direction the solution moves as 1 increases 17 lfyou begin with y0 in A B A fool lfyou begin with y0 in B D A C If you begin with y0 in DEyx A 00 Observe remains in the same interval it began for all 1 Let us return to example a given that k is 5710 loge 2 years ll This means that if K 00 so that 211 reduces to 28 then the time for the doubling of the logg 2 50 population is 50 years Why Because yt 2y0 yo exp50 This is roughly what happened in lreland between 1800 and 1845 Then the famine struckl By 1860 the population of lreland had fallen back to its 1800 gure of about four million where it remains to this day Consider 0 lt y0 lt Kl Then gt 0 for all yt until yt approaches K from belowl Likewise if we begin at y0 gt K lt 0 and stays negative until yt approaches K from above The general solution of 211 was given on page 8 The value y 0 is an exact solution of 211 as is y Kl If we begin at y0 0 then y 0 is the unique solution for all t 2 0 But from looking at the direction eld picture it is clear that the exact solution y 0 is qualitatively different from the exact solution y Kl Why Well suppose you begin very close to y 0 say y0 5 gt 0 Then it is clear that gt 0 and yt continues to grow away from y 0 We say that the exact or equilibrium solution y 0 is unstable because if we start near it we diverge away from it as time increases On the other hand the exact or equilibrium solution y K is stable because if we start near it then as t increases the solution moves back towards it Let us generalize this ideal Consider the autonomous rst order ODE y 2 17 Equilibrium We say that ye is an equilibn39um solution of 217 if 0 Stability 1 We say ye is stable in the Lyapunov sense if for any given 5 in some range 0 lt 5 lt 50 we can nd a 65 such that if 7 yel lt 65 then 7 yel lt 5 for all t gt 0 18 ldea You start near ye you stay near ye Remark 50 measures some distance beyond which you can t nd a 6r 2 An equilibrium point which is not stable in the sense of l is said to be unstable 3 We say ye the equilibrium point7 is asymptotically stable if a it is stable in the sense of l and b there exists a constant C gt 0 such that if 7 yel lt C7 then lirntH00 7 yel 0r Remark You start near ye you tend towards ye and end up there as t 7 Go The basin of attraction of the asymptotically stable equilibrium point ye is the set of initial conditions y0 such that the solution of 217 beginning at y0 ends at ye as t 7 col Example 37 fy kyl 7 kK gt 0 ye 0 is an unstable equilibriurnr ye K is an asymptotically stable equilibrium with basin of attraction 07 fkmliy gt 0y increases for 0 lt y lt K d 0171 lt 0 y 0 y K unstable stable equilibrium equilibrium y decreases y decreases with time for y gt K for y lt 0 We often test for stability by letting yM and treating y t as very srnalllr Then we ask if y t 7gt 0 as t 7 col 1Note here y t is a new function of t de ned by yt 7 115 Do not confuse it with which we also sometimes write 1 19 Example Substitute y ye y K y in 211 to get W 7 K y E 7 Mk y l 7 K k K i lt lt Kgt 2 7k 7 k y y K Now if y is very small7 then we can disregard the term My compared to the term Icyi We can then solve 17 iky to get y t yO exp7kt so that any initial perturbation yO away from the equilibrium solution y ye K decays With time as t A 00 We say that y ye is linearly stablei 20 Practice questions for quiz given in Week 3 and based upon rst and second week lecturesi Qili Radio carbon dating is based on the fact that living material wood plants bones accumulates a small amount of Carbon 14 a radioactive isotope of carbon At the plants death the isotope begins to decay at a rate proportional to the amount presenti Let be the amount of the isotope in grams at time t measured in years Write an equation for relating the rate of change ie 13 with the amount present iiei Show that the time 739 for to halve in value ie is independent of Q0 the starting amounti Given the half life 739 of C14 is 5568 years how long has some previously living matter been dead if the remains have only 20 of the original amount contained when the plant was alive Qi2i Draw by hand the direction elds for the ODE dy z Eid gi a Find 97 lt 9 S where tant9 at the points a 000a 0 a ia ia u u ia for a gt 0 b At each of these points with a l 2 3 draw a unit tangent vector which makes an angle 9 with the z axis c Draw curves whose tangents at the designated points match the unit tangent vectors you have put down d Use the separable method to solve exactly and plot the curves that go through the points 1 0 01 11 22 71 71 l 71 72 2 How do these match up with the curves you drew for part Q3 The behavior of the amplitude of a buckled beam satis es the ODE dA E 7A17A27A a Using the graphical method nd limtxar00 At if A0 lt 0 ii 0 lt A0 lt 1 iii 1 lt A0 lt 2 iv A0 gt 2 b Solve exactly given A0 Qi4i A cylindrical tank of radius R1 is drained through a circular hole of radius R2 centered on its base Write down the ODE for ht the height of the water in cmsi at time t measured in seconds 21 Given h0 ho CInsi7 R1 4 CIns7 R2 1 Ini7 solve for ht and nd an expression t17 the time at Which the tank is drained take 9 103 CInseCQ If he 5 In What is t1 in seconds 22

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