Calculus Tutorial 3 Differential Equations

Calculus Tutorial 3 --Differential Equations Srinivasan Nenmeli-K

Introduction Differential equations form a large subject by itself.It can be considered as an extension of "Integral Calculus" too.Here we study elementary differential equations---the reason is that you learn integration better and also you find immeditate real-world applications which are fascinating.Further, many mathematical models of real world are constructed with these simple DE's/ [I use DE for differential equations.] First Order Ordinary Differential Equations We shall not use partial differentials---so they are called 'Ordinary' DE's. Let us take a simple example of a DE: the rate of change of a variable or derivative depends on the value of the variable at that instant. This is the key to DE's.We are finding the rate of change instantaneously through the derivative and its value, we assume or we find from data, depends on the variable. dy/dx= k y ------------------(1) Y is the variable we want to know, k is a constant.[K can be positive or negative --does not matter.K can be a fraction.] Let us take a practical example.Consider the population of a species, including humans.Let us take up the population of salmon fish in a lake, as a specific example here.The growth rate of this population can be proportional to salmon population at a given time. Hence we write: dP/dt= kP where t is the time and k ,a constant. If the population grows,then k is positive. Initally in the lake we have 100 salmons or y=100 when t=0. This is called the 'initial value.' Can we solve this DE? Find a relation between y and t or find y as a function of time. Let us rewrite this equation and then integrate:

Since we know the initial value,let us find the value of C. At t=0, ln(100) = 0 +C C= ln 100 Substitute for C now:

We have solved this DE and we have y as a function of t.Note that the inital value of 100 appears as a multiplier for exp(kt).Sine we have given a particular value for y at t=0, we call this "particular solution ' of this DE. In general ,we can write: y(t) = y(0) exp(kt)

------------------(2)

We will use this form of solution for all practical problems in this tutorial.[Instead of 't', you can write 'x' too.We are interested in growth/decay with time;hence this form!] Exponential growth and decay problems Example 1 The population of salmon increases in a lake and follows the equation: dp/dt= k t .The initial population is 100 at time t=0.Time is measured in years. After one year , a biologist estimated that there were 550 salmons in the lake. 1Find the general equation for the Salmon population as a function of time. 2 Estimate the population at the end of three years, if the same growth rate continues. 1.The first part we have already found: 2. For this particular case, we have to find P(0) and k. P(0) =100 [given]. To find k: After one year, at t=1, P(1)=550 Therefore:

To find population after three years P(3):

Note the exponential increase of the Salmon population.![The value of 'k' is important in these equations.] Example 2 The popualtion of India was found to be 200 million in the year 1920 by Census. The next census in 1930 showed a population figure of 280 million. Find the equation for this population growth and predict the likely population for the same trend in 1960. Let us take the figure for 1920 as the initial value t=0 [t in years.] P(t)= 200exp(kt) For the year 1930, t=10 ---->

280=200exp(10k) exp(10k)=1.4 Taking logarithm both sides: (10k)=ln(1.4) k= 0.336/10= 0.034 Therefore the equation for population growth: P(t)= 200 exp(0.034t) The predicted value for population for 1960: t=40 P(t)= 200 exp(0.034x40)=200exp(1.35) = 200x3.84=768 million. [The actual population of India was much less in 1960,due to decreasing growth rate or lesser k value.] {Note: Demographers,scientists who study popualtion of different countries, use more complicated models than this simple exponential growth.See the book by M Braun listed at the end of this tutorial.} Example 3 The tiger population in a wild-life preserve in South Asia was 12000 in the year 1950. It dwindled to 5000 in the year 1970 due to less forest cover. From this information, find the equation for tiger population,which is an exponential decay.Predict the tiger population in 2000 if no conservation efforts are taken. let t=0 correspond to the year 1950,called 'the base year'. P =12000exp(-kt) so that k is always positive. In the year 1970, taking t=20, P=5000 =12000exp(-20k) Find k first: exp(-20k)= 5/12=0.42 -20k= ln(0.42)=-0.87 k= 0.043 The tiger population equation: P(t)=12000exp(-0.043 t) To predict the population in the year 2000: Let t=50

P= 12000exp(-0.043x50)=12000exp(-2.16)= =12000 x 0.114= 1372 tigers. Doubling time and Half life Consider an exponential growth problem ,like the salmon population growth in Example 1. We can express the growth in terms of 'doubling time'----- the time it takes for the population to double itself.

Taking logarithms,

Note that doubling time is another way of expressing the growth constant,'k'. You can convert dougling time into k and then proceed. For exponential decay problems,we use the term "Half Life" --the time it takes for the variable to dacay to half its original value. Let

y(t)=y(0) exp(-kt)

At t', y(t)=(1/2)Y(0)

Then:

Again the same formula and we can realte half-life to 'k' Example 4 The half life of decay of certain species of beetles was found to be 21 days.If a laboratory had 2400 beetles, find out when the popualtion of beetles will reduce to 600. At t=0, p(0)=2400 P(t)=2400exp(-kt) To find k, we use the half-life : half-life= 21 days=0.693/k Therefore :

k= 0.693/21 = 0.033 p(t)=2400exp(-0.033t) To find t ,when P(t)=600 600=2400exp(-0.033t) 1/4=exp(-0.033t) ln(1/4)=-0.033t ln4=0.033t 1.386=0.033t t=1.386/0.033=42 days Check: We can check the result using the half-life data.After 21 days, the population will decrease to 1200 beetles and after another 21 days,it will decrease to 600beetles---so, a total of 42 days! Example 5 Drosophila or fruit flies are used often in Genetics research laboratories.[Why,because you can follow the changes over several generations of fruit flies in a short time.] Find the popualtion growth equation and doubling time if at t=2 days,we have 100 fruit flies and at t=4 days , 300 fruit flies are found. 100=P(0)exp(2k)

and 300=P(0)exp(4k)

Therefore:

300/100 = exp(4k)/exp(2k)=exp(2k) ln 3 = 2k = 1.1 k=0.55 To find P(0): 100 =P(0) exp(2x0.55) P(0)= 33 Therefore the population equation: P(t)=33 exp(0.55t) Doubling time T= 0.693/0.55=1.25 days. The number of beetles double every 1.25 days. In a month,you can study the changes over

24 generations !. Example 6 Madame Marie Curie extracted one gram of Radium , a radioactive element which decays into other elements,from a ton of Pitch Blende, a mineral of Lead and Uranium,[from Joachimstal mines] by a tedious chemical crystallisation process. [She received Nobel prize twice.!] If the half-life of Radium is 1620 years, how much of the original 1 gm of Radium will be found after 100 years. Half life= 1620 =0.693/k k=0.693/1620=0.00042 Amount of Radium= y(t)=y(0)exp(-0.00042t) Amount of Radium after 100 years : y(100)=1exp(-0.00042x100) y(100)= exp(-0.042)=0.958 grams. It had decayed about 0.05 grams in hundred years.! Example 7 John uses calculus to track his sales record of new cell phone models.He finds that the sales have doubled in 42 days.If the original sales quantity is 50000 cell phones per day, find the likely sales at the end of the quarter, after 90 days. doubling time T=42 =0.693/k k= 0.0165 Sales S(t)= S(0) exp(0.0165t) S(90)=S(0)exp(0.0165x90) = 50000 [ exp(1.485)] = 50000 [4.415]=220000 Check: In 42 days ,the sales will double to 100,000 and in 84 days, to 200,000.For 90 days, it is more than 200,000. Example 8 Carbon Dating in Archaeology:[This technique is due to Willard Libby, a Nobel Laureate in chemistry for 1960.] Carbon -14 is an isotope of carbon , which decays slowly over the years to other isotopes. The half-life of this unstable isotope is 5730 years. When a plant or a tree is living ,the C14 content is constant.When it dies, the C14 content slowly decays and decreases.By comparing the C14 of a fossil sample with that of a living plant today,we can find the date of the fossil. The C14 content of a fossil was found to be 15% of expected C14 of living plant of same species today.Find the age of the sample. Let the content be 100 % at the time of fossilisation. The C14 content: y(t)=y(0)exp(-kt) where t is in years. k= 0.693/5730=0.00012 0.15=1 exp(-0.00012t) ln (0.15)=-1.897 =- 0.00012 t The age of the fossil:

t= 15809 years.

Practice Problems 1 The population figures [in millions] for two large cities are given for 1990 and 2000.Find an exponential model for both the cities and predict the population figures for the year 2010. 1990 2000 Detroit ,Michigan 3.00 2.74 Dhaka,Bangladesh 4.22 6.49 [Note that for Detroit ,there is an exponential decay while for Dhaka ,an exponential growth] ------ Ans: Detroit: y=3exp(-0.009t) 2010: 2.5 million Dhaka; y= 4.2exp(0.043t) 2010: 10 million 2 Cobalt 60 isotope [used in radiation treatment for cancer patients] has a half life of 5.27 years. If a medical clinic buys 10 grams of this isotope, how much of Cobalt 60 will be present after 2 years. ----Ans: 7.85 grams 3 The demand for an electronic device decreased from the initial value of 10000 units in 1990 to 8500 in 2000.Using a exponential decay model for the demand ,predict the demand for 2010. ----Ans: 7225 units. -----------------------------------------------------------------------------------------------------------Modified exponential equations In many cases, the growth variable reaches a steady value after a long time,not increasing exponentially for ever. For instance ,the fish population will grow,but the rate of growth

keeps decreasing ,reaching a nearly zero growth rate.Then the fish popualtion reaches a nearly maximum value ,called 'saturation value'.[This often happens because the resources for growth,say food, becomes limited in a lake for fish population.]Such problems are modeled with differential equations as before,but the DE is modified. Example 9 The rate of growth of certain crabs in a sea shore is found to follow the equation: dP/dt= k(650 - P).When t=0, the population was 300;it increased to 500 at t=2,where t is in years. Find the population equation and estimate the crab population for the third year. First let us integrate the expression to solve for the differential equation: dP/(650 - P) = kdt Integrating both sides: - ln|650-P)|= kt +C When t=0, C= - ln(650-300)= - ln 350 -ln(650-P) =kt -ln350 ln(650 - P) =-kt +ln350 When t = 2, P=650 -350exp(-kt) 500 =650 - 350exp(-2k) Solving: k= 0.424 The Population Equation: ---> P= 650-350exp(-0.424t) For t=3, P= 551 crabs. Note: As t becomes large, the second term goes to zero and the population P tends towards P=650 crabs. The population increases from 300 and slowly reaches nearly 650 after a long time [asymptotically].Draw the graph of P versus t in a graph paper on in a graphing calculator and check your answer for P(3). Example 10 Newton's Law of Cooling Suppose you have a hot mug of coffee in your hand at temp of 150 deg F.The surrounding temp of air is 60 deg F.Then the coffee mug cools slowly, the temperature decreases exponentially and reaches the surrounding temp of 60 deg F.then there is no further decrease in temperature. Newton found that ,as a first apporximation, the rate of cooling is proportional to the temperature difference between the mug and the surrounding.(To). This is called "Newton's Law of Cooling." In the form of a DE ,this becomes: dT/dt= -k (T-To) where T is the temp of the mug and t ,the time. [A simple equation ,indeed.] Integrating:

dT/(T-To) =- k dt

ln(T-To)=- kt+C At t=0, T=150 C= ln(150 -To) Let us take To=60 deg F C= ln 90 = 4.5 ln (T-60)= -kt + 4.5 T-60 = exp(4.5)exp(-kt)= 90exp(-kt) To find k, we need another data:----> T =90 when t= 10 minutes. 90= 60 +90exp(-10k) exp(-10k)=30/90=1/3=0.33 - 10k=ln(.33)=-1.11 k= 0.011 The equation for the cooling: T=60+90exp(-0.011t) Note: You can plot T versus t to find the temperature at any time.Try this with graph paper or graphing calculator. Application Problems 1.Parachute Fall: A paratrooper drops from an aircraft and while falling,experiences air resistance [due to his parachute and his body] --- a drag force which slows his speed.{That is why he uses the parachute!] The basic DE for this problem is as follows: mdv/dt+ kv = mg { From Physics, you know that mdv/dt=ma is the force on the parachute due to its fall.. The drag or air-frictional term is 'kv', where k is a constant.We assume that this drag force is proportional to velocity v--a good assumption for low speeds. 'mg' is the weight of man + parachute, acting downwards.} Rewriting this equation : dv/dt + kv/m = g Solve this DE for Velocity v (t):

Ans:

The initial condition is that v=0 at t=0 [when the guy jumps out of aircraft.]

2 Weight -Control Problem This is a very simplified formulation of weight gain/loss problem.The answers will be very approximate.This method serves to illustrate the DE approach to such problems. The average energy burned by a person in the body is about 17.5 calories per pound of body weight.[This could vary a lot for an active person or athlete , compared to a couch potato.]If C is the calorie intake per day in calories, where w is the weight in lbs. a) Find the solution : w(t) b) Find the rate of weight loss per day if a person weighing 180 lbs,consumes C= 2500 cals per day. c) Suppose Jane weighing 180 lbs wishes to reduce her weight by 35 lbs.how many days will be required? d) Find the weight Jane after 40 days if she diets and consumes only 2000 calories. [Note: You can also do the calculations for weight gain if she consumes C=3000 cals.] a)

We can rewrite this equation: Let C=2500

This equation is exactly similar to the equation we had for Newton's law of cooling---Example 10. Therefore we write the solution: w-143=37exp(-kt) [C= ln(180 - 143)=ln (37)] where k=1/200=0.005 w= 143+37exp(-0.005t) b)

Setting w= 170 lbs, solve for t: 170=143+37exp(-0.005t) ln(27/37) = -0.005t -0.0317 = -0.005t t= 63 days It takes 63 days to reduce from 180 lbs to 170 lbs with daily intake of 2500 calories. c) Repeat the steps for w=145 lbs as in part b. Ans: t= 583 days or nearly 19 months. d) dw/dt= 4/7 -(1/200)W= -1/200(w -800/7) w= 114+66exp(-0.005t) If t=40, w=114 +66exp(-0.2)=114+66x0.82=114+54=168 Jane will lose 12 lbs in 42 days. 3 Rocket Flight: A rocket with an initial mass of 2000 kg is vertically moving with a constant force of 2000 Newton.It loses its mass in burning at the rate of 1 kg per second.Assuming that its velocity is 0 at t=0, find its velocity in 50 seconds. Acceleration

For t= 50 secs.

a= Force/mass

V(t) = 2000. ln [2/1.5] = 574 m/sec

-------------------------------------------------------------------------------------------------------------Logistic Growth Curve Consider this differential equation: dy/dt = ky(M-y)

The solution is : The solution is of the general form : ---------logistic-growth function You will come across this type of 'growth function' in many situations;medical research, business growth,chemical processes and so on. The notable feature of this function is that initially the growth is slow, and then rapid,the growth rate (dy/dt) is positiveand reaches a maximum;then the growth rate starts decreasing; dy/dt then approachess zero..there is saturation and no further growth. The point ,or value of t at which the growth rate,dy/dt is maximum is called the inflection point;This is the point at which the second derivative

.

Note that dy/dt will reach maximum when y=M/2 by inspection.! or 50% of total y value. You can see from that solution that as t goes to infinity, the denominator becomes one and y=a. ---the saturation value. This curve is also called Example 1: On a college campus, one student was infected with a deadly virus,V*.The spread of the virus among the students was given by the sigmoidal curve: where y is the number of infected students at time t. Find the number of students infected after 5 days.Find when 40% of the students would be infected. Find the time when dy/dt is a maximum. Plot the graph of y versus t. After 5 days :

y(t)=2000 = 5000/(1+4999exp(-0.8t) Solving t=10 days[approx]. dy/dt is max when the infected reach 2500. Solving for 2500, t=ln4999/0.8=10.6 days. Example 2 The sale of mobile phones in a city increase slowly for the first six months and then rapidly rises for thenext two years,tapering off to a total of 0.22 million.Fit the growth function to a sigmoidal curve with a =0.22, b=0.2 and c=0.1 whne t is in months. Find the growth rate after 1 year and the max growth rate.Draw the curve of y vs t .

Separation of Variables Method We have used this method in previous examples without giving it a name because it was 'obvious' to do.Now we shall see more difficult examples. In this method, we try to separate the y terms and bring them to the left side and all the x terms to the right.Then proceed to integrate. Example 11

Separating the variables:

Example 12

Solve:

Integrating:

y(dy/dx )=exp(x) ydy=exp(x)dx

Practice Problems 1 Solve:

y(x+1) + dy/dx= 0

2 Solve:

dy/dx= xysin x^2

3 Solve: dT/dt= - k(T-70) T(0)=140 -------------------------------------------------------------------------------------------------Standard Form and its solution The standard from for first order (linear) DE is given as follows: dy/dx + P(x) y = Q (x) To solve this , proceed as follows: Step 1 : Find the integrating factor:

--------------------(3)

Integrating factor= exp(I)= F Step 2: Multiply both sides by the integrating factor: F Then the solution: ---------------(4)

The steps involve two integrations. Example 13

Solve: dy/dx + (1/x)y= 3x+1 Here P(x)=1/x Q(x)= 3x+1

Solution:

Example 14 Here

Solve: dy/dx -y=cos x

P(x)=-1

Q(x) =cosx

Application Problem 1 An electric circuit has a resistance (R) and inductance(L) in series and attached toa EMF or voltage source.The current flowing in the circuit is given by the DE:

[Note: LdI/dt is the volatge drop across inductance.IR is the voltage drop across the resistance.The sum of these two is equal to the value of E] Solve for I as a function of time if E= constant =E0

Integrating factor: F = exp(Rt/L) Solution:

2. Solve for I for the same electrical circuit if E= sin 2t Ans: I= 1/(4L^2+R^2)(R sin 2t -2L cos 2t) + C exp(-R/L)t Practice Problems 1 dy/dx= -(1/x)y 2 dy/dx + 3x^2y= x^2.y^3 3 dy/dx + ytanx=secx+cosx

y(2)=4

ans xy=4 Ans: 1/y^2 = Cexp(2x^3) +1/3 Ans: y=sinx + (x+1) cos x

4 dy/dx +(1/x)y = xy^2

Ans: y= 1/(cx-x^2)

-----------------------------------------------------------------------------------------

System of First order equations Mathematical modellers are fond of writing a system of first order equations----two or more interconnected differential equations.We shall come across this 'system of ODE" in several

places: prey-predator models,econometric models,epidemic models and so on. These are called "Dynamic systems." Here ,for illustration,let us consider Epidemic models , a subject of ' Epidemiology.' A simple epidemic model, called 'SIR model' is described here. The population is divided into three groups: Susceptible {S}: those not infected,but likely to be infected by contact with infected and who do not have immunity for the disease; Infectives {I}: those already infected and will transmit the disease to the susceptible by contact;some of the infectives will be removed for isoaltion or quarantined and some would die also. Removed [R} removed from infectives into isoaltion and also died ones. We can picture this as state transitions: S ---> I ---> R We can write three first order diff equations for this situation: Since susceptibles decrease by becoming infected,we have a negative factor.The rate of change of S depends on the contact between S and the infectives and is taken as the product of population of S and I;this is called an 'interactive term' in these models: dS/dt = - a.S.I ----------------(1) The change in infectives is due to people getting infected from S and the loss of infectives due to death or isolation: dI/dt = aS I - bI ----------------(2) 'a' and 'b' are constants.'b' is a measure of those infected getting removed. The rate of change of R is simple: dR/dt = bI --------------------------(3) Note that the sum of these equations is zero: dS/dt + dI/dt + dR/dt =0 Why this happens? Total population N = S + I + R Since total population remains constant, we have dN/dt =0 therefore: ds/dt + dI/dt + dR/dt =0 How to solve this set of diff equation? Firstly, we will consider only equations 1 and 2.The third equation can always be worked out from the other two: dR/dt = - (dS/dt + dI/dt) Take a time interval of say ,one day.We can use the difference equation or Euler's method to solve them. Take the initial value of S and I.In this example , we take S = 1million,I =100 persons = 10^-4 million. Take a= 1 and b= 0.1 [Which means 10% of infected are isolated or removed.] So= 1 I = 0.0001 a= 1 b= 0.1 We now solve the equations as difference equations:

If we take

Now for the next day:

We increment the previous valus of S and I by the changes: It is easy to do the caluclations with a tabular format: Day

S

I

Day 1

1

10^-4

1.1.10^-4 =10^-4

10^-4-10-5=0.9x10-4

Day 2

0.99

1.9x10_4

0.99.1.1.9x10-4= 1.88x10-4

1.88x10-4 -1.9x10-5=1.69x10-4

Day 3

0.99-0.0188=0.97

3.59x10-4

0.97x1x3.6x10-4= 2.84x10-4

2.84x10-4-3.6x10-5= 2.54x10-4

Day 4

0.97-0.02=0.95

6.13x10-4

0.95x1x6.13x10-4= 5.8x10-4

5.8x10-4-6.13x10-5= 5.2x10-4

Day 5

0.95-0.05=0.90

11.33x10-4

0.9x1x11x10-4=9.9x10-4

9.9x10-4-11.1x10-5=8.8x10-4

You can continue with the table till I starts decreasing and reaches I=10-5 What do you observe for this table?:The infectives keep increasing ,while the susceptibles keep decreasing.The effect of isolation of 10% of infectives will be seen later...This is typical of many epidemics...In almost all cases,I will keep increasing,reaches a amximum and then start decreasing slowly.When I is very low,we can say that the epidemic has died down. The rapid increase in the first few days or few weeks ,will cause alarm. Day 6

0.9-0.09-0.81

20.1x10-4

0.81x1x20x10-4= 16x10-4

16x10-4-20x10-5= 14x10-4

Day 7

0.81-0016=0.794

34x10-4

0.794x1x34x10-4= 26.86x10-4

26.9x10-4-3.4x10-4= 23.5x10-4

day8

0.794-0.27 = 0.524

57x10-4

0.524x1x57x10-4=29.8x10-4

29.8x10-4-5.7x10-4= 24.1x10-4

day9

0.524-0.3=0.224

81x10-4

0.224x1x81x10-4=18x10-4

18x10-4-8.1x10-4= 10x10-4

day10

0.224-0.18=0.044

91.10-4

0.044x1x91x10-4=4x10-4

4x10-4-9.1x10-4=-5.1x10-4

day 11 0.044 -0.04=0 85.10-4 ------------------------------------------------In this second table, we find that the rapid spread of epidemic results in steady decrease of the population to almost zero population! The solution is very sensitive to the choice of a and b values we made.Since we took a=1,the full virulence of the epidemic blast is shown in the two tables above... Remember ,in the plague epidemic in the late renaissance period[1347 and later], nearly 33 % (1/3rd)of the population of Europe was wiped out,because the process of isolating the infected did not obtain at all. In recent plaque epidemic in Surat in India(1994) only 50 persons died,and a few hundreds were infected. Again, in the SARS global epidemic of 2003, effective isoaltion of diseased people took place and the epidemic was checked in time.The same thing did not happen during the flu epidemic in 1917 when nearly 1 million people died. One can also study the AIDS epidemic with similar models..see internet sources. Dynamics of S and I: Often we study the close relationship between the susceptible and the infective,at any given time. Taking the ratio of The crucial parameter is the ratio of b/a.Note that this ratio is independent of time.If b= 0.1,a= 0.001,then this ratio becomes: We can integrate this expression: dI = [1-100/s]ds I = s - 100lnS + C Such expressions are useful in relating I and S. When S= 1, I =10^-4 So, 10^-4 = 1+c c= 10^-4-1=-1 I = S -1 -100lnS For S= 0.8, that is 20% of population is already infected,we get I = -0.2 -100ln(.8) = -0.2+100(.233)= 23. The dynamics is very senstive to the values of a and b. This approach is called "state-space modelling', useful in all dynamical systems. A similar system of equations ,we find for Prey-Predator models and also Lanchester models for battles.Interesingly, many of these models have been modified and applied in busines situations. Nonlinear equations: This sytem of equation is non-linear and cannot be solved in many cases except by numerical methods.[Further,the methods of Laplace transforms provide easy way of solving a system of linear differential equations.]

Practice Problems

1 Solve the epidemic SIR models with s = 1 million.I=100 and a=0.1 and b=0.1 Draw a plot of I versus day .Find the max value of I before decreasing. The subject of epidemic models,controlled harvesting or culling of animals,prey-predator models form part of the growing filed of Mathematical Biology.

Prey-Predator Models

2 The Prey-predator problem is given by the set of equations: Prey:

dx/dt = ax-by

Predator: dy/dt= cy + dxy These are called Volterra and Lotka models ,given by Volterra and Lotka in 1926 and 1927. How would you solve this sytem of equations?. Find an expression for dx/dy and analyse the expression 3 In Snake lake, a biologist finds an intital popualtion of 100 snakes and 1000 frogs.A snake can eat up two frogs every day for a nice lunch.Snake popuation grow at the rate of 2% per month and also thrives by finding more forgs which is given by 0.1(#of frogs)(#of snakes).Frogs grow at the rate of 1% every month.Set up the system of diff equations and solve them. The equations are: Frog (x)---> dx/dt= 0.01x - 0.1xy Snake(y) ----> dy/dt= 0.02y + 0.1xy Using difference equations for each month:

Solve with xo=1000, yo=100 x

y

Month 1

1000

100

0.01.1000-0.002.100(1000)= 10-200=-190

0.02(100)+0.001(100) (1000)=102

Month 2

810

202

0.01(810)-0.002(810x202)=8-320=--312

0.02(202)+0.001(810) (202)=4+160=164

Month 3

498

366

0.01(498)-0.002(498)(366)=5-362=357

0.02(366)+0.001(498) (366)=7+181=188

Month 4

141

554

0.01(141)-0.002(141)(554)= 1-156=-155

0.02(554)+0.001(141) (554)=10+78=88

Month 5

0

642

0

0.02(642)= 12

The story of snake and frog is simple....The frog population has come to zero.The snakes have expanded to a population figure of 642 from 100 snakes...But now the snakes have no easy source of food..Thye must find another prey! Practice problem: Change the equation with :

You may find oscillations of prey and predator populations..when the prey is plentiful,the pop of predator is low and vice versa.Make a table to solve this problem Ref: see the website of Brandeis Univ : www.bio.brandeis.edu/biomath for further background --------------------------------------------------------------------------------------------------------

Second Order homogeneous equation

Here we use the second derivative in the equation: Consider the simple example of a vibrationg string with a constant frequency of vibration. is the angular frequency of vibration. [ where f is frequency-no of cycles per second] Then the amplitude of vibration:

This is a typical second order homogeneous equation. It is homogeneous because f(t)=0, with zero on the right side. We take up only the DE's with constant coefficients. Such equations are easy to solve with the solution in the form: y= exp(mx)

or y=exp(mt)

Example 15: solve Take the solution in the form:

y=exp(mx)

Then we get Substituting for the derivatives in the DE:

This algebraic equation is called the "Characteristic Equation" Solve this:

We have two solutions: This is the general solution. Here c1 and c2 are two constants,which can be determined for particular situations,as we will see later. I am introducing a notation here: I will write: This saves your time and mine as well. Example16: What if the two roots are equal? Solve:

The characteristic equation is:

The roots are : The solution is written as follows:

[See your text book for a proof of this solution.You can differentiate y and check with the given DE.] Example: What if the roots are complex? Solve: The characteristic equation:

m^2+2m+10 =0 The roots are -1+3i and -1-3i

If the roots are complex, a+bi and a-bi, the solution is as follows:

Therefore: Note: Even though the roots are complex, the solution is REAL. Applied Problem : Damped Oscillations A spring whose vibrations are damped follows a sinusoidal wave of particular frequency,but its amplitude keeps decreasing. The differential equation given above ,with the term having y' represents dampled oscillations. See the solution:the term exp(-x) gives decreasing amplitude or decaying vibrations. The most common situation is a ball or plate ,kept immersed in an oil pot and attached to a spring or your automobile suspension system. Practice Problems: Solve : 1 y'' - y' -30y =0

2 y''+6y'+9y=0 3 y'' -y' -6y=0 4 y'' +6y' +9y=0 5 y'' +y =0 6 y'' - 9y =0 7

y'' - 2y' +4y=0

Finding Particular Solutions We can find the particular solutions for these DE's--that is find the values of C1 and C2,if given further information.We may be given the values of y for two values of x or y(0) at x=0 and y'(0) at x=0.Then we can solve for C1 and C2. Example: Solve y'' - y' - 30y=0 The characteristic equation becomes:

Given: y(0)=1

y'(0)=-4

To find the particular solution,we apply the given initial condition: y(0)=1 ------------> 1=C1 + C2 y'(0)= -4 ---------> y'= 6C1 exp(6x) - 5 C2 exp(-5x) y'(0) = 6C1 - 5C2= -4 Solve for C1 and C2 ----> 6C1 + 6C2 = 6 C1= 1/11 C2= 10/11 The particular solution:

y= (1/11) [ exp(6x) +10 exp(-5x)]

Practice Problems Find the particular solution for the DEs: 1 y'' +16y=0 y(0)=0 y'(0)=2

Ans: y=1/2 sin 4x

2

Ans: y=exp(-2x)[2+5x]

y''+4y'+4y=0

y(0)=2 y'(0)=1

3 y'' +2y' +3y =0 y(0)=2 y'(0)=1

Ans: y= exp(-x)[2cos3x-(1/3)sin 3x]

-------------------------------------------------------------------------------------------------------------Summary and comments 1 We have included elementary differential equations. There are many DE's which cannot be solved by analytical expression or functions.We resort to numerical methods.Study them with DE's or in a separate course in 'Numerical Methods'. 2 There are simpler methods using power series and Laplace transforms.You can learn them.The' Laplace transforms' convert a differential equation into an algebraic equation, as we did substituting y=exp(mx) earlier.Then you solve the algbraic equations. 3 You may come across a set of DE's which can be solved.,called a 'system" of equations....such problems arise in modelling real world situations. 4 Differential equations come alive when you model real-world situations and solve for particualr conditions--very useful indeed. 5 There are excellent texts on Ordinary Diff Equations, but I strongly recommend the following two books: George Simmons: Differential Equations [with historical perspective][pub: McGraw Hill] M Braun : Differential Equations and its applications [Pub:Springer] Give your feedback through my email: [email protected] ----------------------------------------XXXXXXXXXXXXX--------------------------------------