Differential Equations

Transcription

1 Universit of Differential Equations DEO PAT- ET RIE Definition: A differential equation is an equation containing a possibl unknown) function and one or more of its derivatives. Eamples: sin + + ) + e The order of a differential equation is the order of the highest derivative appearing in it. We shall usuall be looking for functions which satisf a given differential equation and other natural conditions. These are called solutions of the differential equation d.e. for short). In this course we will onl stud ver simple first order d.e. s. Separable Equations Definition: A first order d.e. is said to be separable if it can be written in the form g)h) Such equations can be manipulated so as to have the variables and separated: d g)h) is equivalent to d d h) g)d, and a general solution is given b d h) g)d and will look like H) G) + C

2 Universit of Eample: is separable, because it can be re-written as DEO PAT- ET RIE and then can be put into the form d d or d d. Integration gives d d or + C This is equivalent to saing that all solutions of must satisf C for some constant C. Graphicall, we see that the famil of hperbolas C, C, plus the pair of lines and, constitutes the set of solutions of the given differential equation. Specifing a point, ) through which such a hperbola must pass is equivalent to what is called an initial condition. For eample, if the solution is to pass through the point,), we must take C

3 Universit of Constant Relative Rate of Change The most important differential equation we have so far encountered is k or k, that is, situations where the relative rate of change is a constant k. This is easil solved b separating variables: d dt k becomes d kdt, which we integrate: d kdt results in ln kt + C. Taking eponentials of both sides, we get DEO ET PAT- RIE e kt+c e C e kt or ±e C e kt. We usuall write this in the form t) )e kt. The Basic Technique: Finding Parameters from Data Points If we are given that a variable has constant relative rate of change and two values of the variable at two different times, we can find the parameters ) and k. Suppose we are given that t) at time t and t) at a later time t. This means that the graph of passes through the two points t, ) and t, ) Since we know that t) )e kt, we get two equations in the two unknown parameters ) and k: )e kt and )e kt.

4 Taking ratios, and letting t t > ), we get: Universit of DEO PAT- ET RIE ekt e kt ekt kt e kt t ) e k, an equation involving onl one unknown parameter, k. Taking natural logarithms, we get: ln ) ln e k) or ln ln k or k ln ln ln ). We can now find ) b substituting this value of k into the equation )e kt : )e ln ln t. Taking logarithms, we get: ln ln ) + ln ln t,so ln ) ln ln ln t ln t ln + t ln t ln t ln ln t ln t Therefore ) e ln ) e and therefore ln t ) t ln t t e ln t t t t ) ln t ln + t ln ) t t, i.e., ) ) t t t) )e kt ) t t ) ln e t t t ) )) t e ln t t ) ) t t t ) t t ) t t t t ) ) t t t t Thus we have a ver useful formula: t) ) t t t t 4

5 The above calculation ma appear to be formidable to those who are not et adepts in the manipulation of log and eponential functions. Quite often in practical situations one will just work with specific numbers as in the following eamples. However, it also often happens that such calculations have to be done man times in a da, and ma have to be delegated to persons who, with a calculator or computer, can use a formula derived for them, but can t do the basic math. In such cases the professional will derive the applicable general equation and ver possibl program it into the calculators and/or computers. Universit of DEO ET PAT- RIE Eample: A certain function Nt) satisfies the eponential growth law. If N) and N6) 6, what is N4)? Solution Direct Computation): We have Nt) N)e kt, and: N6) 6 N)e 6k N6) N)e k is easil solved b dividing the second equation into the first) for e k, and N) ) ) 5, so Nt) 5 t, and thus N4) 5 4 Solution Using Formula): We have t, N, t 6, and N 6, so we have ) ) 6) t 6 6) t ) 6 t Nt) ) t 6 ) t ) ) and thus N4) ln ) 5 t

6 Universit of Applications: DEO PAT- ET RIE ) Population Growth Eample: 8 bacteria. A bacteria culture starts with 5 bacteria and after hours there are a) Find an epression for the number of bacteria after t hours. b) Find the number of bacteria after 4 hours. c) When will the population reach,? Solution Direct Computation): t) )e kt 5e kt. Since ) 8 5e k), we have e k 8 5 k ln 6, and k ln 6. 6, so t ln 6 a) t) 5e 56) t b) 4) 56) 4, 59 c) t) 56) t so we must have t if 6) t 5 6, ln 6 ln 6, or t ln 6 ln 6 4.4hours) Solution Using Formula): We have t, 5 ), t, and 8, so we have t) 8) t ) t ) 4 5 8) t 5) t 5) ln ) ) t 56) t

7 Universit of Eample: The population of Saskatoon is projected to be DEO PAT- ET RIE Pt) e.t 998), where t is measured in ears. The population in will be P) e. 998) e.) e.,.74968) 5, 5, and in it will be P) e. 998) e.) e.,.46767) 49, ln 5 5 ) Radioactive Deca For the radioactive isotope Radium-6 the value of k is.459, assuming that the unit of time is a ear. If we had grams of it which could be ver dangerous) right now, then in t ears we would have t) e.459t grams left. Thus in one ear we would have ) e.459).99956) grams, so that.44 grams would have disappeared. In ears there will be ) e.459).9574) grams, so that 4 grams would have disappeared. In ears there will be ) e.459) ) grams, so that 5 grams would have disappeared. In 59 ears there will be 59) e.45959).5) 5 grams, so that 5 grams would have disappeared. The number 59 is called the half-life of the isotope. 7

8 Universit of / / DEO PAT- ET RIE Problem: A radioactive isotope is weighed in a lab. At time t there are grams present, and at time t there are grams. Find a formula for its half-life. Solution Using Formula): t) ) t t t t t t,sot ). We find the value of t for which t) : We must have ln + ln ) t t t t t t. We take logarithms: t t [ t t ) ln t t ) ln ],so t t ) ln + t t ) ln t t ) ln t t ) ln,or t t ) ln + t ln t ln tln ln ) t ln + t ln,or t t ) ln + t ln ln ) tln ln ),or t t t ) ln + t ln ln ) ln ln t + t t ) ln ln ln. Thus the half-life is t t t ln ln ln 8

9 Universit of ) Continuousl Compounded Interest Eponential functions are often used to model the growth of invested funds b approimating the compounding of interest. DEO ET PAT- RIE If P is invested initiall at an annual rate of i% compounded n times per ear, it will grow to the value P + i ) n in one tear, and to the value P + i n t in t ears. n n) Taking the limit as n,weget lim P + i ) n t P lim + i ) n ) t ) P lim n n n n n eln + i n ) t )) t n ln + P lim i en n n P lim ln + i ) n e n n t P lim n e i n + i n n t ) i t P lim e + n i P e lim n n i + i n ) t P e it so we have Pt) P e it 4) Miing of Chemicals A tank with volume V is full of water in which a chemical is dissolved at a concentration of c grams per litre. Keeping the amount of solution in the tank constant, fresh water is added to the tank and mied thoroughl with the solution which is drained out of the tank at the same rate, r litres per minute. What is the concentration of chemical in the tank as a function of time t? Solution: Let be the amount of chemical in the tank at time t, and let ct) be the concentration of chemical in the tank at time t. Then ct). The rate of change V of is given b rct) r V r V,so which becomes, on separation of variables: d dt r V d r V dt which can be integrated as above: 9

10 d r V dt and thus ln r V t + C Since, we can remove the absolute value signs: ln r V t + C Taking eponentials, we get e C e r V t )e r V t if we let f V, the flush time, or the length of time it takes for one volume V of the r tank to pass through it, we get t) )e t f Epressed in terms of the concentration, we have: ct) ) V t e f Eample: Blue Stewart.5-) A tank contains 5 litres of brine with a concentration of. kg per litre. In order to dilute the solution, pure water is run into the tank at the rate of litres per minute and the resulting solution, which is stirred continuousl, runs out at the same rate. a) How man kilograms of salt remains after minutes? b) When will the concentration be reduced to. kilograms per litre? Solution: Let t) kg be the amount of salt in the tank at time t, and let ct) be the concentration t) kg of salt in the tank at time t. We have ct) 5l, and t) ct) l [ ] min t) kg l 5l min t) 75 kg min

11 t) ce kt,so t) cke kt t) 75 cekt 75 c 75 ekt or cke kt c 75 ekt. Thus k t, and therefore t) ce Since ) 45, we have c 45. Thus t) 45e t 75. a) ) 45e 75 45e 5 kg) b) t) 45e t 75 if e t 75 or t ln ln. Therefore t 75ln ln ).4min) Eample: Blue Stewart.5-) As before, but instead of pure water, brine with a concentration of. kg of salt per liter is used. Solution: Let t) kg be the amount of salt in the tank at time t, and let ct) be t) kg the concentration of salt in the tank at time t. We have ct) 5l, and t) [. kg ] l ct) l [ min. kg l ] t) kg l [ 5l min t) 75 t) 5 + ce kt,so t) cke kt t) 75 5+cekt 75 c 75 ekt ] kg min or cke kt c 75 ekt. Thus k t, and therefore t) 5 + ce Since ) 45, we have ce 75 so c. Thus t) 5 + e t 75.

12 Universit of DEO PAT- ET RIE a) ) 5 + e e 5 5kg) b) t) 5+e t 75 if e t 75 or t ln. Therefore t 75 ln 5min) 75 5) Newton s Law of Temperature Change When a small object with initial temperature T is introduced into a temperature controlled environment whose temperature, called the ambient temperature, is kept at A, Newton s Law sas that the rate of change of of the temperature T of the small object dt is proportional to the difference between T and A: kt A) where k is called dt the thermal coefficient of the small object. dt dt Separating variables, we get T A kdt, T A kdt, ln T A kt + C, T A e C e kt If we assume that T >A, then we have T>A, and T A T A, and k<, so that T A e C e kt or T A + e C e kt. Since T T when t, we have T A + e C e k A + e C,soe C T A and thus T A + T A)e kt. Since k<, lim e kt, and therefore lim T A. Because of this, t t we often write T instead of A: T T + T T )e kt In practice, this equation is assumed to be known to hold and the problem is to use phsical observations of temperature to determine the constants that appear in it and then to determine when a certain target temperature will be attained.

13 Eample: Green Stewart 9.4-4) A thermometer is taken from a room where the temperature is to the outdoors, where the temperature is 5. After one minute the thermometer reads. Use Newton s Law of Cooling to answer the following questions: a) What will the reading on the thermometer be after one more minute? b) When will the thermometer read 6? Solution: We have Tt) T + T T )e kt 5 + 5e kt Since T), we have 5 + 5e k,soe k 7 5. Thus Tt) ) t. ) a) T) ) t ) t ) b) Tt) if or t ln 7 ln 5 or 5 ln 5 t ) ln 7 5 ln 5 ln 5 ln 7.55min) T t

14 Universit of 6) Oil Well Depletion Oil is pumped continuousl from a well at a rate proportional to the amount of oil left in the well. Initiall there were,, barrels of oil in the well; 6 ears later, 5, barrels remain. DEO ET PAT- RIE a) Give a formula for the amount of oil, Bt), in the well in barrels) as a function of time t in ears). Solution: Bt),, e kt must satisf B6) 5,, so B6),, e 6k 5,, so e 6k. Thus Bt),, e kt,, e 6k t 6,, e 6k ) t 6,, ) t 6,, ) t 6 B/, t b) At what rate was the amount of oil in the well decreasing when there were 6, barrels of oil remaining? Solution: B t) ln ) t Bt) ln )6,, ln 69, 5barrels/ear) 6 6 c) It will no longer be profitable to pump oil from the well when there are fewer than 5, barrels remaining. The plan is to pump oil for 4 ears, will the well be profitable for this length of time? Solution: ) B4),,,,,, ) 6, 5 > 5,. 6 the answer is YES!. 4

15 Universit of Logistic Growth Equation Another important and related differential equation is the logistic growth equation used to model biological populations: DEO PAT- ET RIE km ) where k> is a growth rate factor and M is the of the ecosstem. carring capacit Separation of variables gives: d M ) kdt kt + C, and we have d M ) M + ) d d M M M M ln M ln M M ln M Thus we have ln Mkt + C) M + M d M Taking eponentials, we get: M emkt+c) e Mkt e MC Denoting ) b, we get: M emk) e MC e MC, so we have M M emkt Since represents a population, we must have >. It is not necessar to assume that <Min order for this model to work well, but we shall do so for the purposes of this eample. We then have M M e Mkt, which ma be solved for : M + M )e kmt 5

16 4) The table below gives the number of east cells in a new lab culture. Timehours) Yeast cells a) Plot the data and use the plot to estimate the carring capacit for the east population. b) Use the data to estimate the initial ralative growth rate. c) Find both an eponential model and a logistic model for this data. d) Compare the predicted values with the observed values, both in a table and with graphs. Comment on how well our models fit the data. 7 Solution: The carring capacit K 675. We estimate that initiall the population doubles ever two hours, so we take k ln. Our eponential growth model is Pt) 8 t, and our logistic model is Pt) e t ln t t t 6

17 Universit of 8) Biologists stocked a lake with 4 fish and and estimated the carring capacit to be,. The number of fish tripled in the first ear. a) Assuming that the size of the fish population satisfies the logistic equation, find an epression for the size of the population after t ears. b) How long will it take for the population to increase to 5,? DEO ET PAT- RIE Solution: Pt) Pt) Using the form P K P + K P )e kt with P P) 4, and K,, we have 4, 4 +, 4)e 4,,, kt 4 + 9, 6e kt + 4e kt Using P), we can solve for k: 5 Thus e k 4 6., a) Pt) + 4 6, b) Solve for t: 5 ) t, ) t 6 ) t 6 4 ) t 6 tln ln 6) ln 4 so t, + 4e,so+, k) 4e k) ln 4.68 ears. ln 6 ln 7