SYSTEMS AND TENSES. English for Scien/sts Maria Cris/na Teodorani

Transcription

1 SYSTEMS AND TENSES English for Scien/sts Maria Cris/na Teodorani

2 SYSTEMS AND TENSES Scien.fic subjects essen.ally deal with both systems descrip.ons and working systems. In order to simply describe systems or state something about them, or tell about the way they usually work we use the simple present tense. But as soon as the system is at work, whatever system, we use the present con3nuous.

3 SYSTEMS AND TENSES Present simple: statements generally acknowledged, well- known behaviours, descrip.ons, shared results in Math models, conclusions and theory modeling. Present con.nuous: system s behaviour while interac.ng with something as being emphasized at the.me of speaking. Past simple and past con.nuous: same logic

4 GRAMMAR REVIEW: PRESENT PASSIVE DESCRIBING A SYSTEM (Present simple): ACTIVE: The solu.ons of a first order differen.al equa.on (Subject) produce (Verb) a slope field (Object). PASSIVE: A slope field (Subject) is produced (Verb) by the solu.ons of a first order differen.al equa.on (0bject).

5 GRAMMAR REVIEW: PRESENT PASSIVE SYSTEMS AT WORK (Present con.nuous): ACTIVE: The solu.ons of a first order differen.al equa.on (Subject) are producing (Verb) a slope field (Object). PASSIVE: A slope field (Subject) is being produced (Verb) by the solu.ons of a first order differen.al equa.on (0bject).

6 GRAMMAR REVIEW: PRESENT PASSIVE General rule for the present passive: Present simple: Subject + am/is/are + past par3ciple (3 rd column of the verbal paradigms). Present con.nuous: Subject + am/is/are + being + past par3ciple (3 rd column of the verbal paradigms).

7 TEXT BUILDING: REDOX REACTIONS(1) Redox is an acronym that stands for reduc.on/oxida.on. During a chemical reac.on, or equa.on, some reactants are being transformed into some products. We generally associate an oxida.on state to the charge an atom would have if all bonds to atoms of different elements were 100% ionic. Thus the oxida.on number is connected to the charge.

8 REDOX REACTIONS Let's consider, for instance, the molecule of sodium chloride. We know that sodium is an alkaline metal and that it has one valence electron in group I, while chlorine is an halogen of group VII that just needs 1 electron to have full 8 valence electrons in its shell. Consequently, in the forma.on of NaCl, Na is going to give electrons and Cl is going to get them.

9 REDOX REACTIONS As a result of this we can write Na + Cl -. Here Na + means +1 charge because the sodium is giving the electron, while Cl - means - 1 charge because the chlorine is ge>ng it. The bond is ionic. If the bond were covalent, we would focus on par.al posi.ve or nega.ve charges. In the forma.on of the molecule of water, the oxygen is gaining 2 electrons from the 2 hydrogens, which are losing them, the Hs being more electroposi.ve and the O being more electronega.ve.

10 REDOX REACTIONS Consequently the oxida.on number of hydrogen in H 2 0 is +1, while the oxygen's is - 1. As a result of this we can say that in a molecule of water the hydrogen are oxidized by the oxygen: the electrons are taken away from them, so that they have a posi.ve charge. Now, let's study the following combus.on:

11 REDOX REACTIONS CH 4 + 2O 2 CO 2 + 2H 2 O Here a molecule of methane is reac3ng with two molecules of oxygen in order to produce a molecule of carbon dioxide plus 2 molecules of water plus some heat (being esothermic, the reac.on produces more heat than you put into it). In CH 4 an atom of carbon is bounded with 4 hydrogens.

12 REDOX REACTIONS While reac3ng, being more electronega.ve, the carbon is taking 4 electrons from the hydrogens, so its charge is going down by four. As a result its oxida.on number is - 4, while the hydrogen's is +1. Thus, we can write C - 4 H +1. In CO 2, the carbon's oxida.on state is +4, which means that it is giving up 4 electrons, and really it only has 2 electrons to give up, for it has 4 electrons in its valence shell.

13 REDOX REACTIONS So, what is ge>ng oxidized and what is ge>ng reduced? Let's write down the first half reac.on: C - 4 C e - Here carbon is going from an oxida.on number of - 4 on the leh side of this equa.on, to an oxida.on number of +4 on the right side: 8 electrons are being taken away from carbon, so it is being oxidized.

14 REDOX REACTIONS As for the second half reac.on: 4O + 8 e - 4O - 2 we are shown 4 oxygens with a zero oxida.on state (being in the elemental form) turning into 4 oxygens with a - 2 oxida.on state, so each of these oxygens are taking 4 electrons, the two of them, thus there are 8electrons.

15 REDOX REACTIONS The oxida.on state, that is, the hypothe.cal charge, is going down, or it is being reduced by carbon, as well as the carbon above is being oxidized by oxygen. Finally, what is the oxidizing agent, what is the thing that is oxidizing? Oxygen, of course. is the oxidizing agent, while carbon is the reducing agent.

16 REDOX REACTIONS Redox can also be reviewed from a biological point- of- view. Biologists usually say oxida.on deals with losing hydrogen atoms, while reduc.on deals with gaining hydrogen atoms, though the essen.al meaning stays the same. The reac.ons within cells which result in the ATP (adenosine triphosphate) synthase using energy stored in glucose are referred to as cellular respira.on. It requires oxygen as the final electron acceptor.

17 REDOX REACTIONS The equa.on for aerobic respira.on is C 6 H 12 O 6 + 6O 2 6CO H 2 O + energy Here we are combining glucose with molecular oxygen so that cellular respira.on is being made for. We end up with 6 carbon dioxides and six molecules of water, while the energy produced is made up of some heat and about 38 ATPs.

18 REDOX REACTIONS Glucose is completely broken down to CO 2 + H 2 O though, during fermenta.on, it is only par3ally broken down. Let s take a look at the half reac.ons: H 12 6 H 2 (read: Hydrogen 12 yielding to 6 hydrogens 2) says the hydrogen preserves a +1 oxida.on number (o.n.) on both sides of the equa.on so that nothing is happening with respect to oxida.on and reduc.on,

19 REDOX REACTIONS while C 6 6C + 24e- shows the number of electrons lost by carbon in cellular respira.on, the carbon being oxidized by the oxygen. Finally, O 6 + 6O e- 6O 2 + 6O emphasizes the fact that these 24 electrons are the same electrons carbon is losing, so that oxygen, which is gaining electrons, is being reduced by carbon.

20 REDOX REACTIONS Where does the energy come from? It is produced because the electrons are going from a higher energy state, or level, to a lower one (lower orbitals are more stable): C 6 H 12 O 6 + 6O 2 6CO H 2 O oxidized reduced oxygens taking e - Therefore carbon is losing hydrogens, while oxygen is gaining hydrogens.

21 REDOX REACTIONS Use the following redox reac.ons to build a coherent text with appropriate verbs in the present tense (both ac.ve and passive forms). Zn + CuSO4 à ZnSO4 + Cu Fe + 2HCl à FeCl2 + H2 H3PO3 + MnO4 à HPO4 + Mn 2H2 + O2 à 2H20

22 TEXT BUILDING: ENTHALPY(2) We want to figure out and quan3fy the concep.on of enthalpy of forma.on of a substance. To this extent, let s consider a P(V) diagram, where P is the pressure and V is the volume of the system (3):

23 ENTHALPY The area under the curve in the clockwise direc.on is the net work done by the system. Here ΔU = 0 - that is, there is no change in the internal energy of the system. We know that ΔU = Q W = 0 where Q is the heat applied to the system and W is the work done by the system.

24 ENTHALPY We define enthalpy as H = PV. If we consider a finite change in H we can write ΔH = ΔU + Δ(PV) = Q W + Δ(PV) Here ΔH is a state variable because it is the sum of other state variables. If we consider a system with a piston we can write ΔH = Q PΔV + ΔPV The change in enthalpy will equal Q if the last two terms cancel out.

25 ENTHALPY Under what condi.on? If and only if the pressure is constant, then we can factor it out, so we get ΔH = Q P Δ V + PΔV = Qp that is, heat at constant pressure. In so doing we are kind of squeezing out the previous diagram, for we are making of the forth path and the return path the same exact path that is, a horizontal line from A to B.

26 ENTHALPY As a result, no net work is being added in going from A to B. Most chemical reac.ons are at constant pressure (1 atm). In this case we define enthalpy as the heat content when pressure is constant. Let s consider the following reac.on as an example:

27 ENTHALPY C(s) + 2H2 (g) à CH KJ of heat released How much heat is being added to the system? We realize that H(ini.al) = reactants > H(final) = products and that Q p = - 74 KJ (change in enthalpy). Since this heat is added to the system, what is the change in enthalpy of the reactants system rela.ve to the products system?

28 ENTHALPY H f H i = ΔH = - 74 KJ that is, H f is lower than H i by 74 KJ, so we conclude that H f is at a lower level of energy, or it is more stable and that the reac.on is esothermic, that is, the heat is released by the system. Most chemical reac.ons are at constant pressure (1 atm). We thus define enthalpy of forma.on of a substance as the heat content when pressure is constant.

29 ENTHALPY Calculate the varia.on of the standard enthalpy of forma.on for the process, at constant pressure, C(diamond) à C(graphite) knowing that C graph + O 2 (g) à CO 2 ΔH f = - 393,5 KJ/mole C diam + O 2 (g) à CO 2 (g) ΔH f = - 395,4 KJ/mole You must write a coherent text including numerals and chemical equa.ons.

30 TEXT BUILDING: MECHANICS APPLIED TO MACHINES (4) Dealing with mechanics applied to machines means dealing with power transmission. Actually, mechanical power is the rate at which work is being provided once the system is in movement. In symbols: P = dw/dt = Fds/dt = Fv W = Fds = F cosθ ds = F t ds

31 MECHANICS APPLIED TO MACHINES where P is power, W is the work done, F is the exerted force and F t its tangent component, while s denotes movement in space and v velocity. We define work, generally speaking, as energy transferred by force and, in the same sense, we define energy as the ability to do work. Physically speaking, power is nothing else than a sent unit of work per second.

32 MECHANICS APPLIED TO MACHINES We are going to see how power is being transmifed as long as a system moves. In systems such as cars and the like, power is transmifed via fric.on wheels, cogwheels, belts, joints, rod/crank systems, flywheels and similar systems in such a way that a torque is generated between driving and driven, moving and resis.ng structures.

33 MECHANICS APPLIED TO MACHINES A simple example of power transmission between two shahs not too distant from one another is that of fric.on wheels. The figure below shows the scheme of such a transmission.

34 MECHANICS APPLIED TO MACHINES

35 MECHANICS APPLIED TO MACHINES We have two wheels whose diameters are D1 and D2. The first one is placed along the driveshah and has an angular velocity ω m and a torque M m. The second one belongs to driven shah and has an angular velocity ω u and a torque M r.

36 MECHANICS APPLIED TO MACHINES If we call I the distance shown in the figure below we get the interaxis I = ½(D 1 + D 2 )

37 MECHANICS APPLIED TO MACHINES If the wheels do not slide, the velocity of the contact point on wheel 1 will equal the velocity of the contact point on wheel 2, so that V 1 = V 2. Consequently ½(ω 1 D 1 ) = ½(ω 2 D 2 ). That is to say: i = ω 1 / ω 2 = D 2 / D 1 The transmission ra.o i depends on the diameter of the two wheels. We size the diameters of the two wheels correctly using this last equa.on (together with that of the interaxis I).

38 MECHANICS APPLIED TO MACHINES On a prac.cal level, the uses of fric.on wheels are rather limited, though being silent and having a regular transmission, for we can use them only at low powers. For high powers the force T = f R must be elevated, but since the fric.on coefficient for commonly used materials such as steel or cast iron is rather low (f= ), there should be very high pushing forces R, so that shahs, pins, bearings, etc., would be strongly stressed.

39 MECHANICS APPLIED TO MACHINES Since fric.on wheels undergo a huge radial stresses in order to ensure their adherence they do not provide high power transmissions. Anyway, star.ng from two ideal fric.on wheels we can obtain a series of cogs on their external surfaces that is, a series of projec.ons on the edge of a wheel transferring mo.on by engaging with another series and alterna.ng with empty spaces.

40 MECHANICS APPLIED TO MACHINES Once in mo.on these cogs are being easily interpenetrated; in this case, power transmission is no longer due to fric.on but to the pushing force that each cog of the driving wheel is exer.ng on those of the driven wheel. In this way, provided that the built cogs are strong enough, we can transmit high powers.

41 MECHANICS APPLIED TO MACHINES It is possible to convert rota.onal mo.on into linear (transla.onal) mo.on using the pinion/ rack mechanism, where the pinion s rota.onal mo.on is being converted into a transla.onal mo.on by the rack. Given a gear, we define the pinion as the cogwheel with the smallest diameter and the wheel as that with the largest diameter. The interaxes is the distance between the axis of the two wheels.

42 MECHANICS APPLIED TO MACHINES If ω 1 is the angular velocity of the pinion and ω 2 the angular velocity of the wheel, then we define the transmission ra.o as i = ω1 / ω2

43 MECHANICS APPLIED TO MACHINES An example of this is the car s steering- wheel mechanism. While driving, the steering s rota.on is being converted into the transla.on of the elements ac.ng upon the wheels.

44 MECHANICS APPLIED TO MACHINES As shown in the figure below, the force transmifed from the driving wheel to the driven one is the tangent component F T.

45 MECHANICS APPLIED TO MACHINES It is Ft = Fcosθ = Cm / R1 = Cr / R2, where Cm is the machine torque and Cr is the resis.ng torque. Of course the radial component Fr is not responsible for mo.on and cons3tutes a solicita.on all over the shah on which the wheels are keyed. Its module is Fr = Fsinθ. This suggests we ought to render the pressure angle θ very small in order to increase the value of F T.

46 MECHANICS APPLIED TO MACHINES The value of the angle of pressure affects the minimum number of cogs that a wheel can have. In prac.ce, we assign the number of cogs as a func.on of the pressure angle and of the transmission ra.o using the following formula: Z min = 2 / [i 2 + (1+2i)sin 2 θ - i] ½ As for the minimum cogs number (Zmin) as a func.on of θ and i we have the following table:

47 MECHANICS APPLIED TO MACHINES i θ= θ= θ=

48 EXERCISES Write a text describing the previous table. Describe how the pinion/rank system works with respect to your own car in mo.on: what happens if you have to turn right, leh, or get straight on? Write a coherent text in your area using appropriate tenses.

49 DIFFERENTIAL EQUATIONS (DE) (5) Differen.al equa.ons are useful for modeling and simula.ng phenomena and understanding how they operate. They are available in different nota.ons: y + 2y = 3y (read: the second deriva.ve of y (or y prime prime ) plus twice the first deriva.ve of y is equal to three y)

50 DIFFERENTIAL EQUATIONS (DE) Func.on nota.on: f (x) + 2f (x) = 3f(x) (read: the second deriva.ve of the func.on with respect to x plus.) Leibnitz s nota.on: d 2 y/dx dy/dx = 3y (read: the second deriva.ve of y with respect to x twice plus.).

51 DIFFERENTIAL EQUATIONS (DE) The solu.on of a DE is a func.on or class of func.ons. One solu.on is y 1 (x) = e - 3x : it sa3sfies the equa.on y + 2y = 3y. In fact since the first deriva.ve is y 1 (x) = - 3e - 3x (read: y subset 1 prime of x equals nega.ve 3.mes e (raised) to the nega.ve 3x) and the second deriva.ve is y 1 (x) = 9e - 3x (read: y subset 1 prime prime of x ), then subs.tu.ng into the equa.on we realize we get an iden.ty, so that it works. No.ce that y 1 is a solu.on, but not the only one : for example y 2 (x) = e x is another solu.on.

52 PARTICULAR LINEAR SOLUTIONS TO DIFFERENTIAL EQUATIONS.(6) Let s consider dy/dx = - 2x + 3y - 5. A solu.on is a linear func.on in the form y=mx + b. Therefore we have to find out the m and b that make this linear func.on sa.sfy the DE. In order for y=mx + b to sa.sfy the DE this has to be true for all x in the linear equa.on. So dy/ dx = m = - 2x + 3y - 5, so that m = - 2x + 3(mx + b) - 5 and m = (3m- 2)x + 3b - 5.

53 PARTICULAR LINEAR SOLUTIONS TO DIFFERENTIAL EQUATIONS. This has to be true for all x, so that 3m- 2=0 and m=2/3=3b- 5. Consequently b=17/9 and the solu.on is y = (2/3)x + 17/9. EXERCISE. Find out any similar DE and write down a coherent text describing it.

54 SLOPE FIELDS (7) We use slope fields to visualize solu.ons of differen.al equa.ons. In other words, the solu.ons of a first order DE produce a slope field that is, their graphical representa.on. Given dy/dx = - x/y let s say we do not know the solu.on and want to give a general sense of what a solu.on might look like.

55 SLOPE FIELDS To this extent we create a table labeling its columns x, y and dy/dx. When x=0 and y=1, then dy/dx = 0. It means if the solu.on goes through the point (0,1), then the slope is going to be 0 (marked as a short horizontal line parallel to the x- axis). At the point (1,1) (read: 1 comma 1 or ordered pair 1,1 ) the slope is - 1 (a nega.ve downward slope);

56 SLOPE FIELDS at (1,0) the slope is undefined; at (- 1,- 1) the value of the slope is again nega.ve 1, while if the solu.on goes through (1,- 1), we will have a posi.ve upward slope of 1. When plo ng the table onto a graph a slope field is being produced. The solu.ons, corresponding to some condi.ons, look like circles.

57 SLOPE FIELDS (8) EXERCISE. The graph below shows the slope field of dy/dx=x 2 - x- 2, with the blue, red, and turquoise lines being (x 3 /3)- (x 2 /2)- 2x+4, (x 3 /3)- (x 2 /2)- 2x, and (x 3 /3)- (x 2 /2)- 2x- 4 respec.vely. Write a coherent introduc.on/data descrip.on/conclusion text including the graph and the Math equa.ons converted into words. Use the present passive in both the system s descrip.on and func.oning.

58 SLOPE FIELDS (8)

59 SEPARABLE DIFFERENTIAL EQUATIONS(9) Given the DE dy/dx = - x / ye x^2, we want to find the par.cular solu.on that is going through the point (0,1). This ordered pair is being given as our ini.al condi.on. Separa.ng the variables means get the dy and dx on separate sides. Therefore mul.plying by ydx both sides of the equa.on we are being leh with ydy = - xe - x^2 dx, so that, integra.ng both sides, we get

60 SEPARABLE DIFFERENTIAL EQUATIONS ydy = - ½ 2xe - x^2 (read: the integral of y (.mes) dy is equal to nega.ve one half.mes the integral of 2x (.mes) e to the nega.ve x squared). As a result we get y 2 /2 + C 1 = ½ e - x^2 + C 2, where C 1 and C 2 are two constants. (read: y squared over 2 plus C 1 is equal to one half.mes e to the nega.ve x squared, which is the an.- deriva.ve, plus C 2 ).

61 SEPARABLE DIFFERENTIAL EQUATIONS At the point (0,1) we get ½= ½ + C, where C = C 2 C 1, so that y = e - x^2/2, which is the par.cular solu.on that sa3sfies our ini.al condi.on.

62 SEPARABLE DIFFERENTIAL EQUATIONS EXERCISE. Find out any separable DE and build a coherent text. As usual, numerals and/or diagrams must be a part of the text. Equa.ons must be wriˆen in words (just once for each new equa.on and only if quan..es are different). The text must be structured according to an introduc.on/data descrip.on/conclusion layout. Support your text with samples of separable DE applica.ons to any phenomena, using appropriate connec.ves when paragraphing.

63 MODELING WITH DIFFERENTIAL EQUATIONS (10) Let s use DE for modeling popula.on. If P stands for popula.on and t stands for.me, say, in days, then dp/dt stands for the rate of change of popula.on with respect to.me. If we consider the rate of change as being propor.onal to the actual popula.on, then we can write dp/dt = KP, which is a reasonable statement since the larger the popula.on the larger the rate of growth at any given.me.

64 MODELING WITH DIFFERENTIAL EQUATIONS Integra.ng both sides of the equa.on we get dp/dt = kp, or ln P = Kt + C 1. (read: the logarithm of the module of P (or the absolute value of P) is equal to Kt plus C subset 1). Therefore P = e Kt+C1 = e Kt e C1 = C e Kt.

65 MODELING WITH DIFFERENTIAL EQUATIONS P = e Kt+C1 = e Kt e C1 = C e Kt (read: the module of P (or the absolute value of P) is equal to e to the kt plus C subset 1, which is equal to e to the kt power.mes e to C subset 1, which is in turn equal to a constant C.mes e to the kt). If P>0, then P= C e Kt. EXERCISE. Model anything with DE and write down a coherent text.

66 HOMOGENEOUS DIFFERENTIAL EQUATIONS (11) Given dy/dx = f(x,y), if we algebraically manipulate this statement so that dy/dx = F(y/x), then we can make a variable subs.tu.on that makes it separable. So for example if dy/dx = (x+y)/x then dy/dx = 1 + y/ x. By se>ng y/x = V, then we get y=xv, so that dy/dx = V + x dv/dx, or dv= (1/x)dx, from which V = ln x + C. Unsubs3tu3ng back we get y/x = ln x + C or y = xln x + xc. To figure out C we need some ini.al condi.ons.

67 SECOND ORDER DIFFERENTIAL EQUATIONS (12) A second order DE is in the form a(x)y + b(x)y + c(x)y = d(x) (read: a of x.mes the second deriva.ve of y with respect to x (or y prime prime ) plus ) where coefficients are func.ons of x. The associated linear homogenous equa.on is in the form Ay + By + Cy = 0, where A, B and C are constants.

68 SECOND ORDER DIFFERENTIAL EQUATIONS Let s consider the following example: y + 5y + 6y = 0 In order to find out a solu.on we have to ask ourselves if there is any func.ons that when taking its 1 st, 2 nd, 3 rd, 4 th, 5 th, n th deriva.ve it essen.ally becomes the same func.on. The func.on we are searching for is e x. Let s set y = e rx.

69 SECOND ORDER DIFFERENTIAL EQUATIONS Then (e rx ) + (e rx ) + 6 e rx = 0 or e rx (r 2 + 5r + 6) = 0 where r 2 + 5r + 6 = 0 is the characteris.c equa.on, from which we get r 1 = - 2 and r 2 = - 3, so that y 1 = e - 2x and y 2 = e - 3x. EXERCISE. Write a coherent text with an appropriate layout about any second order DE and their applica.ons.

70 LISTENING EXERCISES Listen to the following videos, take notes of the key- words while listening and write down a coherent text using appropriate phrasal/preposi.onal verbs and logical connec.ves. hˆps:// reduc.on/redox- oxida.on- reduc.on/v/balance- and- redox- reac.ons1 hˆps:// reduc.on/redox- oxida.on- reduc.on/v/prac.ce- determining- oxida.on- states hˆps:// of- forma.on hˆps:// prep/mcat/biomolecules/principles- of- bioenerge.cs/v/enthalpy- 1 hˆps:// angular- momentum/torque- tutorial/v/ rela.onship- between- angular- velocity- and- speed hˆps:// angular- momentum/torque- tutorial/v/ rela.onship- between- angular- velocity- and- speed hˆps:// angular- momentum/torque- tutorial/v/constant- angular- momentum- when- no- net- torque hˆps:// angular- momentum/torque- tutorial/v/cross- product- and- torque hˆps:// equa.ons/second- order- differen.al- equa.ons/ complex- roots- characteris.c- equa.on/v/complex- roots- of- the- characteris.c- equa.ons- 1 hˆps:// equa.ons/second- order- differen.al- equa.ons/ undetermined- coefficients/v/undetermined- coefficients- 1

71 REFERENCES 1) Listening exercise at hˆps:// reduc.on/ redox- oxida.on- reduc.on/v/introduc.on- to- oxida.on- and- reduc.on All Khan Academy content is available for free at 2) Listening exercise: hˆps:// 3) Graph at hˆp://en.wikipedia.org/wiki/pressure_volume_diagram 4) Adapted from 5) Listening exercise: hˆps:// equa.ons/first- order- differen.al- equa.ons/differen.al- equa.ons- intro/v/differen.al- equa.on- introduc.on 6) Listening exercise: hˆps:// equa.ons/first- order- differen.al- equa.ons/differen.al- equa.ons- intro/v/finding- par.cular- linear- solu.on- to- differen.al- equa.on 7) Listening exercise: hˆps:// equa.ons/first- order- differen.al- equa.ons/differen.al- equa.ons- intro/v/crea.ng- a- slope- field 8) From hˆp://en.wikipedia.org/wiki/slope_field 9) Listening exercise: separable- differen.al- equa.ons- introduc.on 10) Listening at hˆps:// equa.ons/first- order- differen.al- equa.ons/modeling- with- differen.al- equa.ons/v/modeling- popula.on- with- simple- differen.al- equa.on 11) Listening. hˆps:// equa.ons/first- order- differen.al- equa.ons/homogeneous- equa.ons/v/first- order- homegenous- equa.ons 12) Listening exercise: 2nd- order- linear- homogeneous- differen.al- equa.ons- 2