Modeling First Order O.D.E s

Transcription

1 Modeling Fist Ode O.D.E s 1- Rate of Gowth and Decay The I.V.P. (1)! dx = kx x(t 0 ) = x 0 whee k is a constant, occus in many physical sciences models that involve gowth o decay. Fo example, in biology, it is often obseved that the ate at which cetain bacteia gow is popotional to the numbe of bacteia pesent at any time. Ove shot intevals of time the population of small animals, such as odents, can be pedicted faily accuately by the solution of the I.P.V. (1). The constant k can be detemined fom the solution of the diffeential equation by using a subsequent measuement of the population at time t > t 0. In physics I.V.P.(1) povides a model fo appoximating the emaining amount of a substance which is disintegating though adioactivity. Also I.V.P. (1) detemines the tempeatue in a cooling body. In chemisty the amount of a substance emaining duing cetain eaction is also descibed by I.V.P. (1). Example: 1) A cultue initially has N 0 numbe of bacteia. At time t = 1 hou the numbe of bacteia is measued to be 3/2 N 0. If the ate of gowth is popotional to the numbe of bacteia pesent at any time, detemine the time necessay fo the numbe of bacteia to tiple. Let s wite the I.V.P. that models the poblem:! dn = kn N(0) = N 0 Fist solve the fist ode equation dn = kn it is sepaable, and can be expessed as dn N = kt integating, we get ln N = kt + c, whee N > 0 o N(t) = ce kt Using the initial condition: N(0) = N 0 = ce 0 = c then N(t) = N 0 e kt Since at time t = 1 hou the population is 3/2N 0,

2 then N(1) = 3/2 N 0 = N 0 e k o 3/2 = e k and solving fo k, k = ln(3/2) = Theefoe, N(t) = N 0 e t Now, we must find the time t when the numbe of bacteia tiple the oiginal numbe. 3N 0 = N 0 e t!!!!!3 = e t taking natual logaithm both side ln 3 = t t = ln = 2.71!hous Remak: The function e kt inceases as t inceases if k > 0 and deceases as t inceases if k < 0. Thus, poblems descibing gowth, such as population, bacteia, o even capital ae chaacteized by a positive value of k. Meanwhile, poblems involving decay, as in adioactive disintegation will yield a negative k value. 2) A beede eacto convets the elatively stable uanium 238 into isotope plutonium 239. Afte 15 yeas it is detemined that 0.043% of the initial amount A 0 has disintegated. Find the time t it takes fo one-half of the atoms in the initial amount A 0 to disintegate (half-life). A(t) is the amount of plutonium emaining at time t. So we can ceate the I.V.P.! da = ka A(0) = A 0 The solution of the sepaable equation is A(t) = A 0 e kt,

3 If 0.043% of the atoms of A 0 have disintegated, then % of A 0 emains. To find k, we solve A 0 = A 0 e 15k 15k = ln ( ) k = ln( ) =! Hence, A(t) = A 0 e! t Half-life 1 2 A 0 = A 0 e! t 1 2 = e! t! t = ln 1 % 2& ' =! ln 2!0.693 t =! = 24,180!yeas 2- Cooling: Newton s law of cooling states that the ate at which the tempeatue T(t) changes in a cooling body is popotional to the diffeence between the tempeatue in the body and the constant tempeatue T 0 of the suounding medium. dt = k ( T! T 0 ) whee k is the constant of popotionality. Example: When a cake is emoved fom a baking oven its tempeatue is measued at 300ºF. Thee minutes late its tempeatue is 200ºF. How long will it take to cool off to oom tempeatue of 70ºF? The poblem can be descibed by the I.V.P. dt = k( T! 70) % T(0) = 300 We solve the sepaable equation dt T! 70 = k ln T! 70 = kt + d T! 70 =!ce kt T(t) = 70 + ce kt

4 Using the initial condition when t = 0, T = 300, we get 300 = 70 + c o c = 230 then T(t) = e kt Since T(3) = 200, then e 3k 200! 70 = = !!o!!k = 1 13% ln 3 23& ' =! Theefoe, T(t) = e! t We want to detemine how long it will take to each oom tempeatue. Notice that lim T(t) = 70, so the equation 70 = T(t) does not have a solution. t! Intuitively we expect that the cake will each oom tempeatue afte a easonably peiod of time. How long? Let s ceate the following table t (min) T(t) º º º º º º The table shows it will each oom tempeatue in about 30 minutes. 3- A Mixtue Poblem At time t = 0 a tank contains Q 0 lb of salt dissolved in 100 gal of wate; see figue. Assume that wate containing ¼ lb of salt /gal is enteing the tank at a ate of gal/min and that the well-stied mixtue is daining fom the tank at the same ate. Set up an I.V.P. that descibes this flow pocess. Find the amount of salt Q(t) in the tank at any time. Find the limiting amount Q L that is pesent afte a long peiod of time.

5 We assume that salt is neithe ceated no destoyed in the tank. Theefoe vaiations in the amount of salt ae due solely to the flows in and out of the tank. Moe pecisely the ate of change of salt in the tank /, is equal to the ate at which salt is flowing in minus the ate at which it is flowing out. In symbols, = ate!in!!!ate!out The ate at which salt entes the tank is the concentation ¼ lb/gal times the flow ate gal/min o /4 lb/min. To find the ate at which salt leaves the tank, we need to multiply the concentation of salt in the tank by the ate of outflow gal/min. Since the ate of flow in and out ae equal, the volume of wate in the tank emains constant at 100 gal, and since the mixtue is well-stied, the concentation thoughout the tank is the same, namely, [Q(t)/ 100] lb/gal. Theefoe, the ate at which salt leaves the tank is [ Q(t)/100] lb/min. Thus the diffeential equation govening the pocess is = 4! Q 100 the initial condition is Q(0) = Q 0 The equation can be witten as Q = 4 it is a linea equation in the dependent vaiable Q and independent vaiable t, whee P(t) = /100. The integating facto is e So! 100 e! e d e % Q' &' = e int egating = e Q. 100 = e 4 e Q = ( e = 25e + c 4 Q(t) = 25 + ce! Using the initial condition, we get C = Q 0 25 Then Q(t) = 25 + ( Q 0! 25)e! Q(t) = 25 1! e! % ' + Q 0 e! & 4

6 Fom this equation, we see than when t inceases without bound, then Q(t) 25, so the limiting value Q L = 25. Assume Q 0 = 50, and = 3, find the time t afte which salt level is 60% of Q 0. Then, Q(t) = e!0.03t Since 60% of 50 is 30, we want to find t such that Q(t) = = e!0.03t 5 = 25e!0.03t 1 5 = e!0.03t ln 1 % 5& ' =!0.003t ln 1 % 5& ' t =!0.03 = 53.6!min Remak: If the mixed bine solution is pumped out of the tank at a ate faste o slowe than the ate at which is pumped in, then the equation is diffeent fom the one in the pevious example. Example: Initially 50 lb of salt is dissolved in a lage tank holding 300 gal of wate. A bine solution is pumped into the tank at a ate of 5 gal/min, and a well-stied solution is then pumped out at a slowe ate of 3 gal/min. If the concentation of the solution enteing the tank is 2 lb/gal, detemine the amount of salt in the tank at any time. How much salt is at t = 50 min? The ate at which the salt entes the tank is: IN = (5 gal/min)(2 lb/gal) = 10 lb/min The solution is accumulating at a ate of (5 3) gal/min = 2 gal/min. Afte t min, thee ae t gal of bine in the tank. The ate at which salt is leaving the tank is:! Q(t) OUT = (3 gal/min) lb / gal t % & = 3Q(t) lb / min t The diffeential equation govening the pocess is: = IN! OUT = 10! 3Q(t) t t Q = 10 It is a linea equation with P(t) = 3/( t), so the integating facto is! e t = e 3 2 ln ( 300+2t ) = ( t) 3 2

7 ( t) ( t) Q = ( t) t 10 d!( t) 3 2 Q(t) = ( t) int egating ( t) Q(t) = % 10( t) = 5 ( t) 5 2 Q(t) = 2( t) + c( t) & 3 2 Using the initial condition Q(0) = 50, 50 = 2(300) + c (300) -3/2, c = -2.8x10 6. Q(t) = 2( t) + (! )( t)! 3 2 When t = 50, Q(50) = 2( ) ( )-3/2 = 450 lb c