ES10006: Core skills for economists: Mathematics 2

Transcription

1 ES10006: Core skills for economists: Mathematics 2 Seminar 6 Problem set no. 4: Questions 1c, 1d, 2, 4, 7, 8b

2 First-order linear differential equations From Chiang and Wainwright (2005, Section 15.1), the homogeneous differential equation dy + ay = 0 1 dy = a (1) y has the general solution y = Ae at (2) the non-homogeneous differential equation dy + ay = b has either the general solution (3) y = y h + y p = Ae at + b a if a 0 (4) or the general solution y = y h + y p = A + bt if a = 0 (5) 2/19

3 First-order linear differential equations Stability requirements Plots of equation (2) for A = 1 and α positive, zero, and negative (e ). α = 0.25 α = 0 α = 0.25 y y y t t t y = Ae at represents deviations from equilibrium. Stability, that is convergence to b/a in the general solution (4), requires α > 0. 3/19

4 Exercise 1 Part (c) Solve the first order differential equation 2 dy y = 12 given the initial condition y(0) = 10. From the reduced equation, 2 dy y = 0 dy y = 0 1 dy y = 1 4 and using (2), y h = Ae 1 4 t. For the particular solution, let y = k. Then, from the complete equation, 0 + k/2 = 12 k (= y p ) = 24 So the general solution is y = Ae 1 4 t + 24 (in terms of formula (4), a = 1/4 and b = 12/2 = 6). From the initial condition y(0) = 10, 10 = Ae A = 14. It follows that the unique solution is y = 14e 1 4 t /19

5 Exercise 1 Part (c) y(0) = 10 a = 1/4 > 0 y 24 as t 5/19

6 Exercise 1 Part (d) Solve the first order differential equation dy = 5 given the initial condition y(0) = 1. From the solution (5) dy = 5 y = A + 5t (this is the case of differential equation (3) with a = 0 and b = 5). Using the provided initial condition 1 = A A = 1 Consequently the unique solution is y = 1 + 5t 6/19

7 Exercise 2 Suppose that the change in the quantity sold of some commodity depends on the gap between the demand price P D and the supply price P S, so that dq = α (PD P S ), where α > 0. (6) The buyers inverse demand function is P D = a + bq (7) and the sellers inverse supply function is P S = g + hq. (8) If the value of Q at t = 0 is Q(0), find the expression showing quantity sold as a function of time; the equilibrium quantity sold in this market; the necessary conditions on the P D and P S parameters so that equilibrium is stable. 7/19

8 Exercise 2 Substituting the inverse functions (7) and (8) into equation (6) dq = α (a + bq g hq) dq From the reduced equation dq α(b h)q = 0 and using (2) we get Q h = Ae α(b h)t. α(b h)q = α(a g) (9) To find the particular solution let Q = k. Then dq/ = 0 and from (9) 0 α(b h)k = α(a g) k (= Q p ) = g a b h Thus the general solution is Q = Ae α(b h)t + g a b h where Q p = g a represents the equilibrium quantity. b h 8/19

9 Exercise 2 From the initial condition, Q(0) = Ae 0 + g a b h A = Q(0) g a b h and the unique solution can be written as Q = [Q(0) g a b h ] eα(b h)t + g a b h For the equilibrium to be stable we require α(b h) < 0. With α > 0 (see model specification) we need b h < 0 b < h that is the slope of the inverse demand line should be less than the slope of the inverse supply line. 9/19

10 Exercise 4 Consider the following IS-LM model: C = a + by gr I = I G = G L = ky hr M = M (consumption demand) (investment demand) (government demand) (money demand) (money supply) (Y denotes income and R denotes the interest rate.) In equilibrium, aggregate demand equals aggregate supply in the goods market, Y = C + I + G, and in the money market money demand equals money supply, L = M. Assume that the goods market clears instantaneously, but in the money market interest rates adjust gradually at a speed α > 0 in response to the gap between the demand for money and the supply of money. Derive the differential equation for R. Solve for R and determine the conditions on the parameters that must be satisfied for the equilibrium to be stable. 10/19

11 Exercise 4 We may write our assumption on the adjustment process of R as dr = α(l M) dr In addition, in the goods market = α (ky hr M) (10) Y = C + I + G Y = a + by gr + I + G Y = a + I + G 1 b Substituting for Y into equation (10) we get gr 1 b (11) dr = α [k ( a + I + G gr ) hr M] 1 b 1 b dr + α ( gk 1 b a + I + G + h) R = α [k ( ) M] (12) 1 b To simplify the notation let E = α ( gk 1 b We can re-write equation (12) as dr + E R = F a + I + G + h) and F = α [k ( ) M]. 1 b (13) 11/19

12 Exercise 4 We use equation (2) to solve the homogeneous version of (13), that is dr + E R = 0 Rh = Ae Et To find the particular solution let R = k. Then, from equation (13), 0 + E k = F k (= R p ) = F E Therefore, the general solution is R = R h + R p = Ae Et + F E We use the initial condition to find a value for the arbitrary constant, R(0) = Ae 0 + F E A = R(0) F E Therefore the unique solution is R = [R(0) F E ] e Et + F E 12/19

13 Exercise 4 For equilibrium stability we need E > 0, which (with α > 0) is equivalent to gk 1 b + h > 0 1 b > k g h From the goods market, that is equation (11), (14) R = (b 1)Y + I + G + a g and dr dy = b 1 = 1 b g g From the money market, that is equation (10), with dr α(ky hr M) = 0 R = ky M h and dr dy = k h = 0 at equilibrium, Thus stability, i.e. condition (14), requires that the slope of the IS curve is larger than the slope of the LM curve. 13/19

14 Exercise 7 The system q d = 100 3p q s = 20 + p dp = 0.2(q d q s) describes a market in which price converges only gradually to its steady state level. Price is initially at 40. Write down the definite (unique) solution for p. 14/19

15 Exercise 7 Substituting for q d and q s in the differential equation dp dp = 0.2(100 3p 20 p) + 0.8p = 16 From equation (2), dp + 0.8p = 0 ph = Ae 0.8t For the particular solution let p = k. Then, from the complete equation, 0 = k k (= p p ) = 20 Hence, the general solution to the differential equation is p = Ae 0.8t Finally, using the provided initial condition p(0) = 40 A = 20 we get the unique solution p = 20e 0.8t /19

16 Exercise 7 p(0) = 40 a = 0.8 > 0 y 20 as t 16/19

17 Exercise 8 Part (b) Solve the following non-autonomous differential equation: dy + t2 y = 5t 2 Hint: Use the integrating factor. The above equation is in the general form dy + a(t) y = b(t) where a(t) = t 2 and b(t) = 5t 2. The integrating factor is e a(t) = e t 2. Note that d [e a(t) d y] = [e a(t) ] y + e a(t) dy = a(t) e a(t) y + e a(t) dy = e a(t) dy [ + a(t) y] 17/19

18 Exercise 8 Part (b) Multiplying both sides of the given equation by the integrating factor: e t 2 [ dy + t2 y] = 5t 2 e t 2 d [e t 2 y] = 5t 2 e t 2 Integrating both sides of this equation: e t 2 y = A + 5t 2 e t 2 e t 3 3 y = A + 5t 2 e t3 3 (15) where A is the constant of integration. Consider the integral 5t 2 e t3 3 on the right-hand side of (15). Set z = t3 3. Then dz = t 2 and the integral becomes 5t 2 e t 3 3 = 5 e t3 3 t 2 = 5 e z dz = 5e z + k = 5e t k dz Substituting back into equation (15) we find the general solution: e t3 3 y = A + 5e t3 3 y = Ae t /19

19 References Chiang, A. C. and Wainwright, K. (2005), Fundamental Methods of Mathematical Economics, 4th edn, McGraw-Hill. Hoy, M., Livernois, J., McKenna, C., Rees, R. and Stengos, R. (2011), Mathematics for Economics, 3rd edn, MIT Press. 19/19