15-20 - CHAPTER 15 Exercise 15.1 1(a With a = 4 and b = 12...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 15 Exercise 15.1 1. (a) With a =4 and b =12 ,w ehav e y c = Ae 4 t , y p = 12 4 =3 . The general solution is y ( t )= Ae 4 t +3 .S e t t i n g t =0 ,weget y (0) = A ,thus A = Y (0) 3=2 3= 1 . The de f nite solution is y ( t e 4 t . (b) y c = Ae ( 2) t , y p = 0 2 . The general solution is y ( t Ae 2 t e t t i n g t ,we have y (0) = A ; i.e., A =9 . Thus the de f nite solution is y ( t )=9 e 2 t . (c) y c = Ae 10 t , y p = 15 10 = 3 2 .T h u s y ( t Ae 10 t + 3 2 e t t i n g t y (0) = A + 3 2 , i.e., A = y (0) 3 2 3 2 = 3 2 . The de f nite solution is y ( t 3 2 ¡ 1 e 10 t ¢ . (d) Upon dividing through by 2, we get the equation dy dt +2 y . Hence y c = Ae 2 t , y p = 3 2 ,and y ( t Ae 2 t + 3 2 e t t i n g t eg e t y (0) = A + 3 2 , implying that A = y (0) 3 2 h ed e f nite solution is y ( t 3 2 . 3. (a) y ( t )=(0 4) e t +4=4(1 e t ) (b) y ( t )=1+23 t (c) y ( t )=(6 0) e 5 t +0=6 e 5 t (d) y ( t ¡ 4 2 3 ¢ e 3 t + 2 3 1 3 e 3 t + 2 3 (e) y ( t )=[7 ( 1)] e 7 t +( 1) = 8 e 7 t 1 (f) After dividing by 3 throughout, we f nd the solution to be y ( t ¡ 0 5 6 ¢ e 2 t + 5 6 = 5 6 ¡ 1 e 2 t ¢ Exercise 15.2 1. The D curve should be steeper then the S curve. This means that | β | > | δ | ,or β < δ , which is precisely the criterion for dynamic stability. 99
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2. From (15.9), we may write α + γ =( β + δ ) P . Hence (15.10 0 )canberew r i t tena s dP dt + j ( β + δ ) P = j ( β + δ ) P ,or dP dt + kP = dP dt + k ( P P )=0 . By (15.3 0 ), the time path corresponding to this homogeneous di f erential equation is ( t )= (0) e kt .I f (0) = 0 ,then ( t ;i .e . , P ( t P . If (0) 6 =0 ( t ) P ( t ) P will converge to zero if and only if k> 0 . This conclusion is no di f erent from the one stated in the text. 3. The price adjustment equation (15.10) is what introduces a derivative (pattern of change) into the model, thereby generating a di f erential equation. 4. (a) By substitution, we have dP dt = j ( Q d Q s j ( α + γ ) j ( β + δ ) P + j σ dP dt .T h i s c a n be simpli f ed to dP dt + j ( β + δ ) 1 j σ P = j ( α + γ ) 1 j σ (1 j σ 6 =0) The general solution is, by (15.5), P ( t A exp h j ( β + δ ) 1 j σ t i + α + γ β + δ (b) Since dP dt i f Q d = Q s , then the intertemporal equilibrium price is the same as the market-clearing equilibrium price ³ = α + γ β + δ ´ . (c) Condition for dynamic stability: 1 j σ > 0 σ < 1 j . 5. (a) Setting Q d = Q s , and simplifying, we have dP dt + β + δ η P = α η The general solution is, by (15.5), P ( t A exp ³ β + δ η t ´ + α β + δ (b) Since - β + δ η is negative, the exponential term tends to zero, as t tends to in f nity. The intertemporal equilibrium is dynamically stable. (c) Although there lacks a dynamic adjustment mechanism for price, the demand function contains a dP dt term. This gives rise to a di f erential equation and makes the model dynamic. 100
Background image of page 2
Exercise 15.3 We shall omit all constants of integration in this Exercise.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/25/2009 for the course FIN FIN534 taught by Professor Stevenjordan during the Spring '09 term at Korea Advanced Institute of Science and Technology.

Page1 / 47

15-20 - CHAPTER 15 Exercise 15.1 1(a With a = 4 and b = 12...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online