1.4 Application Separable Equations and the Logistic Equation

Transcription

1 1.4 Applicaion Separable Equaions and he Logisic Equaion If a separable differenial equaion is wrien in he form f ( y) dy= g( x) dx, hen is general soluion can be wrien in he form f ( y ) dy = g ( x ) dx + C. Thus he soluion of a separable differenial equaion reduces o he evaluaion of wo indefinie inegrals. Hence i is emping o use a compuer algebra sysem such as Maple or Mahemaica ha can compue such inegrals symbolically. We illusrae his approach using he logisic differenial equaion dx d 2 = ax bx (1) ha models a populaion x() wih birhs (per uni of ime) proporional o x and deahs proporional o x 2. If a = 0.01 and b = , for insance, Eq. (1) is dx d x = 0.01x x = (100 x). (2) Separaion of variables leads o dx d = C x(100 x) = +. (3) Any compuer algebra sysem gives a resul of he form ln( x) ( x 100) = + C. (4) You can now apply he iniial condiion x(0) = x0, combine logarihms, and finally exponeniae in order o solve (4) for he paricular soluion x () = 100 xe 100 x x0e (5) of (2). The direcion field and soluion curves shown in Fig in he ex sugges ha, whaever is he iniial value x 0, he soluion x ( ) 100 as. Can you use (5) o verify his conjecure? 12 Chaper 1

2 The secions ha follow illusrae he use of Maple, Mahemaica, and MATLAB o carry ou he procedure oulined above. You migh warm up for he invesigaion below by applying a compuer algebra sysem o solve Problems 1 28 in Secion 1.4 of he ex. Invesigaion For your own personal logisic equaion, ake a = m/n and b = 1/n in (1), wih m and n being he larges wo disinc digis (in eiher order) in you suden ID number. (i) Firs generae a slope field for your differenial equaion and include a sufficien number of soluion curves ha you can see wha happens o he populaion as. Sae your inference plainly. (ii) Nex, use a compuer algebra sysem o solve he differenial equaion symbolically, and use he symbolic soluion o find he limi of x() as. Was your graphically-based inference correc? (iii) Finally, sae and solve a numerical problem using he symbolic soluion. For insance, how long does i ake x o grow from a seleced iniial value x 0 o a given arge value x 1? Using Maple Firs we inegrae boh sides of our separaed differenial equaion as in Eq. (3). soln := in(1/(x*(100-x)),x) = in(1/10000,)+c; soln : = ln( x) ln( x) = in(1/10000, ) + C Then we apply he iniial condiion x(0) = x0 o find he consan C. C := solve(subs(x=x0,=0,soln),c); We subsiue his value of C and simplify. C : = ln( x0) ln( x0) soln := simplify(100*soln); 1 soln : = ln( x) ln( x) = + ln( x0) ln( x0) 100 Nex we exponeniae boh sides of his equaion. 1.4 Applicaion 13

3 soln := simplify(exp(lhs(%)) = exp(rhs(%))); soln x e x0 : = = x x0 x() = solve(soln, x); Using Mahemaica e x0 x ( ) = x0 + e x0 Firs we inegrae boh sides of our separaed differenial equaion as in Eq. (3). Inegrae[1/(x(100-x)),x] == Inegrae[1/10000,] + c log( x) 1 log( x 100) == c Then we apply he iniial condiion x(0) = x0 o find he consan c. c = Firs[ soln /. {->0, x->x0} ] log( x0) 1 log( x 0 100) We subsiue his value of c and simplify. Expand[100*Firs[soln]] == Expand[100*Las[soln]] 1 log( x) log( x 100) == log( x0 100) + log( x0) Nex we exponeniae boh sides of his equaion. 14 Chaper 1

4 Exp[Firs[soln]] == Exp[Las[soln]] // Simplify x e /100 x0 == x 100 x0 100 Solve[ soln, x ]; x e x0 e x x x = Firs[x /. soln] 100e x0 e x0 x Using MATLAB Here we solve he logisic equaion in (2) using he MATLAB "symbolic oolbox" inerface o he Maple kernel. We begin by separaing variables and inegraing each side of he resuling equaion. However, i is more convenien now o work wih "everyhing on one side of he equaion", as in dx d C x(100 x) = So we sar by "declaring" our symbolic variables and evaluaing he wo inegrals in his equaion. syms x C in(1/(x*(100-x)),x) - in(1/10000, ) - C 1/100*log(x)-1/100*log(-100+x)-1/10000*-C We are acually hinking here of he equaion 0, bu he righ-hand side zero is suppressed hroughou. I simplifies he equaion a bi by muliplying hrough by *soln log(x)-log(-100+x)-1/100*-100*c 1.4 Applicaion 15

5 Then we apply he iniial condiion x(0) = x0 o find he consan C. soln0 = subs(soln, {,x}, {0,'x0'}) soln0 = log(x0)-log(-100+x0)-100*c C = solve(soln0, C) C = 1/100*log(x0)-1/100*log(-100+x0) We subsiue his value of C simply by evaluaing he presen implici soluion. eval(soln) log(x)-log(-100+x)-1/100*-log(x0)+log(-100+x0) x = solve(soln, x); prey(x) x0 exp(1/100 ) x0 + x0 exp(1/100 ) 16 Chaper 1