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Modeling with 1St Order ODE Modeling with 1st order ODE Now, we move into one of the main applications of differential equations. Mathematical modeling is the process of creating a mathematical representation of some phenomenon in order to gain a better understanding of that phenomenon. It is a process that attempts to match observation with symbolic statement. During the process of building a mathematical model, the modeler will decide what factors are relevant to the problem and what factors can be de-emphasized. Once a model has been developed and used to answer questions, it should be critically examined and often modified to obtain a more accurate reflection of the observed reality of that phenomenon. Although problems may require very different methods of solution, the following steps outline a general approach to the mathematical modeling process: 1. Identify the problem, define the terms in your problem, and draw diagrams where appropriate. 2. Begin with a simple model, stating the assumptions that you make as you focus on particular aspects of the phenomenon. 3. Identify important variables and constants and determine how they relate to each other. 4. Develop the equation(s) that express the relationships between the variables and constants. 5. Verifying and refining a model (test model). Mixing and Population Models Mixing Model In these problems we will start with a substance that is dissolved in a liquid. Water will be entering and leaving a holding tank for the liquid. The water entering the tank may or may not contain more of the substance dissolved in it. Water leaving the tank will of course contain the substance dissolved in it. Let xt() be the amount of the substance dissolved in the liquid in the tank at any time t. What’s the relationship we can find out to solve for ? For this type of question, we have dx() t Rate of change of = Input rate-Output rate since Rate of change of x(t)= . dt Thus we have differential equation, dx() t x' = Input rate-Output rate dt where, Input rate=flow rate of liquid entering x concentration of substance in liquid entering Output rate=flow rate of liquid exiting x concentration of substance in liquid exiting Notice: concentration at time t=(amount of substance at time t)/(volume of liquid at time t). Example 1: A brine solution of salt flows at a constant rate of 6 L/min into a large tank that initially held 500 of brine solution in which was dissolved 0.5 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the rate of 3 L/min. If the concentration of salt in the brine entering the tank is 0.05 kg/L, determine the mass of salt in the tank after t min. Also determine the concentration of salt in the tank at time t. Let yt()represent the mass of salt in the tank after t min. Input rate=6 0.05 yt() Output rate=3 500 (6 3)t yt() y ' 6 0.05 3 (this is a first order linear linear differential equation.) 500 (6 3)t Solving this linear differential equation, we have c150 t 0.45 t 2 y 5003 t Since y(0) 0.5, we have c 150 0 0.45 0 0.5 500 3 0 c 250 250 150tt 0.45 2 yt() 500 3t y( t ) y ( t ) 250 150 t 0.45 t 2 The concentration at time t is Volume at tt500 3 (500 3t )2 Population Model These problems are similar to mixing problems. Let Pt()be the population in a given region at time t. Rate of change of = Rate at which enters the region- Rate at which exits the region Thus, we have the following differential equation, dP() t = Rate at which ) enters the region- Rate at which exits the region dt There are several special cases of this model. Logistic model: dP 2 k13 P k P dt PP(0) 0 or dP k12 P(1 k P ) dt PP(0) 0 Example 2: For the following population model (in billions) dP PP144 2 dt P(0) 7 Describe the behavior of P(t) as t ? dP Solve the separable differential equation: PP144 2 dt dP dP 1 144 dt dt () dP t c PP144 2 PPPP(1 144 ) 1 144 P P cet ln t c etc 1 cet P 1 144PP 1 144 1 144cet P(0) 7 ce0 7 7 c 1 144ce0 1007 cet 11 Pt() tt1 1007 144ce 1 144 c e 144 et 7 1 t , et 0, P ( t ) 144 Heating and Cooling of Building Let Tt()) represent the temperature inside a building at time t. Here we just consider the effect of the outside temperature M(t) on the temperature inside the building. Experimental evidence has shown that this factor can be modeled using Newton’s law of cooling: the rate of change in the temperature Tt() is proportional to the difference between the outside temperature M(t) and the inside temperatureTt(), i.e., dT(t) k[M (t) T(t)] dt Notice: Here we do not consider other factors. Just consider the effect due to the outside temperature. Also, we call 1 the time constant for the building. k Example 3: A bottle of orange juice being taken out of refrigerator at 20 C warms up 50 C in 5 min while sitting in a room of temperature 230 C . How warm will the orange juice be if left out for 15 min? Let Tt()be the temperature of orange juice. t 0 is the time when orange juice taken out of refrigerator. Room temperature Mt( ) 230 . By Newton’s law of cooling dT kT(23 ) dt T(0) 2 Solving it, we have T23 cekt T (0) 2 2 23 ce0 c 21 Te 23 21 kt T (5) 5 5 23 21e5k 18 ln( ) k 21 0.0308 5 Te 23 21 0.0308t At time t 15, T23 21 e0.0308 15 9.8( 0 C ) .
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