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098008A - 8008A SEMESTER 2 2009 THE UNIVERSITY OF SYDNEY...

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Unformatted text preview: 8008A SEMESTER 2 2009 THE UNIVERSITY OF SYDNEY FACULTIES OF ARTS, ECONOMICS, EDUCATION AND SCIENCE MATH 1013 LIFE SCIENCES DIFFERENTIAL AND DIFFERENCE EQUATIONS November 2009 LECTURERS: L Poladian TIME ALLOWED: One and a Half Hours MARKER’S USE ONLY This examination has two sections: Multiple Choice and Extended Answer. The Multiple Choice Section is worth 50% of the total examination; there are 25 questions; the questions are of equal value; all questions may be attempted. Answers to the Multiple Choice questions must be entered on the Multiple Choice Answer Sheet. The Extended Answer Section is worth 50% of the total examination; there are 3 questions; all questions may be attempted; questions are of equal value; all working must be shown. Calculators will be supplied; no other calculators are permitted. THE QUESTION PAPER MUST NOT BE REMOVED FROM THE EXAMINATION ROOM. * PAGE 1 OF 20 8008A SEMESTER 2 2009 PAGE 10 0F 20 Extended Answer Section Answer these questions in the space provided. 1. (a) Find the general solution to Xn+2 — 5Xn+1 + 6Xn = 6. [3 marks] (b) Use separation of variables to find the solution to (cl—z: = e‘y(3t2 — 1) With y(0) = O. [4 marks] ((3) A refrigerator maintains a constant temperature of 2°C. A hot cup of liquid at 98°C is placed in the refrigerator and after 10 minutes the temperature of the liquid has fallen to 75°C. Assume Newton’s Law of Cooling for the temperature (in degrees Celsius) after t minutes: T(t) 2 Tag + fie—kt. Find the values of Teq, A and k. [3 marks] 8008A SEMESTER 2 2009 PAGE 11 OF 20 WW-»W-WM-W_MWW ..._.,._..... m“... 8008A SEMESTER 2 2009 PAGE 12 OF 20 8008A SEMESTER 2 2009 PAGE 13 OF 20 2. The interaction of two species is modelled by the following differential equations: 23': 31:—-4y y’ = ~2x+y where 3:05) and y(t) denote the population sizes as a function of time t in years. (a) Show that IE” — 423' — 51; = 0. [3 marks] (b) Find the general solution for $05). [2 marks] (c) Give the corresponding general solution for y(t), [1 mark] (d) Give the particular solution corresponding to z(0) = 900 and y(0) = 600. [2 marks] (e) For this particular solution, one of the species becomes extinct. Find the time t when this occurs. [2 marks] 8008A SEMESTER 2 2009 PAGE 14 OF 20 8008A SEMESTER 2 2009 PAGE 15 OF 20 8008A SEMESTER 2 2009 PAGE 16 OF 20 3. The spread of an epidemic is modelled by the logistic equation dp —~ 2 0.01 — dt 23(1 p) where p represents the fraction of‘ the population with the disease, and t is the time in days. 1 p(1 — p) ' (b) Hence, show that the general solution to the logistic equation has the form ln|p] —ln|1—p] =0.0lt+C where C' is an arbitrary constant. [2 marks] (c) Determine the value of C’ if initially 10% of the population is infected. [1 mark] (d) Thus, determine how long it takes for half the population to become infected. [1 mark] (e) What fraction of the population is infected after 365 days? [2 marks] (f) At large times, p ——> 1 and thus ln ]p] —> 0. Use this to show that the long term behaviour of the epidemic is given by p(t) Rd 1 —— Ae—kt. What are the values of the constants A and k? [2 marks] (a) Use partial fractions to decompose [2 marks] 8008A SEMESTER 2 2009 PAGE 19 OF 20 Notes Separable Differential Equations: d”; — —F<t>a<z> 52:) =/ F(t) dt Linear Difference Equations with constant coefficients: Consider the second order linear difference equation aXn+2 + bXn+1 + an = 0. Let 7‘1 and T2 be the roots of the auxiliary (or characteristic) equation: 0,7“2 + 67" + c = O. o The general solution has the form Xn = Av";I + B 7‘3, Where A and B are arbitrary constants, provided 7‘1 75 7‘2. Linear Differential Equations with constant coefficients: Consider the second order linear difierential equation adZX dX 0:de +bE+ +CX — 0. Let k1 and 192 be the roots of the auxiliary (or characteristic) equation: (1k2 + bk: + c = 0. o If the discriminant is positive, I)2 — 4ac > O: the general solution has the form X (t) = Ask1t -I— Bel“2t Where A and B are arbitrary constants. c If the discriminant is negative, b2 — 4ac < 0: the general solution has the form X (t) = e"t [A sin(wt) + B cos(wt)] Where A and B are arbitrary constants and n_ b _ |b2—4ac| Iti—'—2a .allld w———————2—a-——— 8008A SEMESTER 2 2009 PAGE 20 OF 20 Exponentials and logarithms: Derivatives: Integrals: O a = 1 ea+b ___ eaeb 8111(2) = 11:, :I: > 0 ealn(z) z“, 1: > 0 111(1) = 0 ln(ab) = 111(a) + 1n(b), a > O, b > O 111(33“) = a1n(a:), m>O 111(ez) II (E n tTL—l k ekt l t 19 cos (kt) —k sin(kt) f0?) / NW 77. tn+l t n + 1 + 0, n 55 —1 1 1 6'“ is“ + C, 1: ¢ 0 sin(kt) 2% (308091?) + 0, 19 ¢ 0 cos(kt) lflsinUct) + C, Is 74 O THIS IS THE LAST PAGE OF THE QUESTION PAPER. ...
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