Chap15 - Solucionario alpha chiang fundamental methods of mathematical economics PDF

Preview

Summary

Solucionario alpha chiang fundamental methods of mathematical economics...

Description

CHAPTER 15

Exercise 15.1

1.

3.

a = 4 and b = 12, we have yc = Ae−4t, yp = 124 = 3. The general solution is y(t) = Ae−4t + 3. Setting t = 0, we get y(0) = A + 3, thus A = Y (0) − 3 = 2 − 3 = −1. The definite solution is y (t) = −e−4t + 3. 2t 0 (b) yc = Ae−(−2)t , yp = −2 = 0. The general solution is y (t) = Ae . Setting t = 0, we have y (0) = A; i.e., A = 9. Thus the definite solution is y (t) = 9e2t . 15 (c) yc = Ae−10t , yp = 10 = 32 . Thus y (t) = Ae−10t + 23 . Setting t = 0, we get y (0) = A + 32 , 3 i.e., A = y (0) − = 0 − 3 = − 3 . 2 2 2 ¢ ¡ 3 The definite solution is y (t) = 1 − e−10t . 2 dy (d) Upon dividing through by 2, we get the equation dt + 2y = 3. Hence yc = Ae−2t , yp = 32 , and y(t) = Ae−2t + 32 . Setting t = 0, we get y(0) = A + 23, implying that A = y(0) − 23 = 0. The definite solution is y(t) = 32 . (a) With

(a) (b) (c) (d) (e)

y(t) = (0 − 4)e−t + 4 = 4 (1 − e−t) y(t) = 1 + 23t y(t) = (6 − 0) e5t + 0 = 6e5t ¢ ¡ y(t) = 4 − 32 e−3t + 32 = 3 13 e−3t + 23 y(t) = [7 − (−1)] e7t + (−1) = 8e7t − 1

(f) After dividing by 3 throughout, we

y(t) =

¡

¢ 5

0− 6

¡ 5

e−2t + 65 = 6

find the solution to be

e

1 − −2t

¢

Exercise 15.2

1. The D curve should be steeper then the S curve.

This means that

which is precisely the criterion for dynamic stability.

99

|

− β|

>

| |,

δ

or

−β

< δ,

α + γ = (β + δ ) P ∗ . Hence (15.100 ) can be rewritten as dP dt + dP + kP = kP ∗ , or dP + k (P − P ∗ ) = 0. By (15.30 ), the time ∗ j (β + δ ) P = j (β + δ ) P , or dt dt path corresponding to this homogeneous differential equation is ∆(t) = ∆(0)e−kt. If ∆(0) = 0

2. From (15.9), we may write

, then

∆(t)

and only if

= 0; i.e.,

P (t)

=

∗ . If

P

∆(0)

= 0, then 6

∆(t) ≡

P (t)

− P∗

will converge to zero if

0. This conclusion is no different from the one stated in the text.

k >

3. The price adjustment equation (15.10) is what introduces a derivative (pattern of change) into the model, thereby generating a differential equation. 4.

dP dt

(a) By substitution, we have

=

j

( Qd

− Qs )

=

(α + γ )

j

− j (β + δ ) P

+ j σ dP .

dt

This can

be simplified to

dP dt

j(β+δ ) P

+ 1−j σ

j(α+γ )

= 1−j σ

(1 − j σ = 0) 6

The general solution is, by (15.5),

h

P (t)

(b) Since

=

dP dt

A exp

= 0 iff

i

− j1(β−+jσδ ) t

Q

d

=

Q

s,

α+ γ

+ β +δ then the intertemporal equilibrium price is the same as the

³

market-clearing equilibrium price

α+ γ

´

= β +δ

(c) Condition for dynamic stability: 1 − j σ

>

. 0, or

σ

<

1

j.

5. (a) Setting

dP dt

Q

d

=

Q

s,

and simplifying, we have

β +δ α + η P = η

The general solution is, by (15.5),

³

P (t)

(b) Since -

=

A exp

´

− βη+δ t

+ βα+δ

β +δ η is negative, the exponential term tends to zero, as t tends to infinity.

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 differential equation and makes the model

dynamic.

100

Exercise 15.3

We shall omit all constants of integration in this Exercise.

R

u = 5, w = 15, and u dt = 5t, solution formula (15.15) gives ¢ ¡ ¡ R ¢ y(t) = e−5t A + 15e5tdt = e−5t A + 3e5t = Ae−5t + 3

1. Since

The same result can be obtained also by using formula (15.5). 2. Since

R

u = 2t, w = 0, and u dt = t2 , solution formula (15.14) gives us y(t) = Ae−t2 . R

u = 2t, w = t, and u dt = t2 , formula (15.15) yields ³ ³ ´ ´ R y(t) = e−t2 A + tet2 dt = e−t2 A + 21 et2 = Ae−t2 + 12 1 Setting t = 0, we find y (0) = A + 2 ; i.e., A = y (0) − 12 = 1. y(t) = e−t2 + 12 .

3. Since

Thus the definite solution is

R

u = t2 , w = 5t2 , and u dt = t33 , formula (15.15) gives us ´ 3 ³ 3 ³ 3 3 ´ R 3 y(t) = e− 3 A + 5t2 e 3 dt = e− 3 A + 5e 3 = Ae− 3 + 5 Setting t = 0, we find y (0) = A + 5; thus A = y (0) − 5 = 1. 3 y(t) = e− 3 + 5.

4. Since

t

t

t

t

t

The definite solution is

t

dy

y = −et .

5. Dividing through by 2, we get dt + 6

Now with

R

u = 6, w = −et, and u dt = 6t,

formula (15.15) gives us

¡

R

¢

¢

¡

y(t) = e−6t A + −ete6tdt = e−6t A − 17 e7t = Ae−6t − 71 et 1 Setting t = 0, we find y (0) = A − 7 ; i.e., A = y (0) + 17 = 1. y(t) = e−6t − 17 et R

u = 1, w = t, and u dt = t, the general solution is ¡ R ¢ y(t) = e−t A + tet dt = e−t [A + et (t − 1)] [by Example 17, Section 14.2] = Ae−t + t − 1

6. Since

101

The definite solution is

Exercise 15.4

1.

(a) With

M

= 2yt3 and

R

F (y, t)

Step ii:

∂F 2 2 ∂ t = 3y t +

Step iii:

ψ (t) =

Step iv:

F (y, t)

2 3 =

(b) With

M

Step ii:

∂F ∂t =

Step iii:

ψ(t) =

Step iv:

F (y, t)

y

3 + t2 =

y t

M

R

y (t)

N

=

y

R

Step ii:

∂F ∂t =

Step iii:

ψ(t) =

Step iv:

F (y, t)

y

+ y2t =

N

=

y

3 + ψ (t)

y t

3 + 2t; thus

ψ0 (t) = 2t

[constant omitted]

y t

³

or

R

y (t)

t(1

N

= =

0

0 dt =

=

yt

c−t2 t

y (1

´ 31

+ y ), we have

+ 2y )dy + ψ (t) =

+ y 2 + ψ (t) =

R

t3

3 + t2 , so the general solution is

=

=

¡ c ¢1 2

3 + 2t, we have ∂ M ∂N 2 ∂ t = 3y = ∂ y .

2t dt = t2

= t (1 + 2y ) and

F (y, t )

=

3y 2 t dy + ψ (t) =

3 + ψ 0 (t) =

c

Step i:

yt

k

2 3 + k , so the general solution is

= 3y 2 t and =

ψ0 (t) = 0

y t

or

F (y, t )

= 3y 2 t2 ; thus

ψ0 (t) = N

0 dt =

=

c

Step i:

(c) With

R

∂M = 3y 2 t2 , we have ∂ t = 6yt2 = ∂∂N y.

2yt3 dy + ψ (t) = y 2 t3 + ψ (t)

Step i:

y t

=

N

N

=

y (1

yt

∂M ∂N . ∂ t = 1 + 2y = ∂ y

+ y 2 t + ψ (t)

ψ0 (t) = 0

+ y ); thus

k

+ y 2 t + k , so the general solution is

c

¡

(d) The equation can be rewritten as 4y 3 t2 dy + 2y 4 t + 3t2 N

¢

dt

∂N 3 ∂M = 2y 4 t + 3t2 , so that ∂ t = 8y t = ∂ y .

Step i:

F (y, t )

=

R

4y 3 t2 dy + ψ (t) =

Step ii:

∂F 4 ∂ t = 2y t +

Step iii:

ψ(t) =

Step iv:

F (y, t)

4 2 + t3 =

y t

R

ψ0 (t) = N

3t2 dt = t3

4 2 + ψ (t)

y t

= 2y 4 t + 3t2 ; thus

ψ0 (t) = 3t2

[constant omitted]

4 2 + t3 , so the general solution is

=

y t

c

or

³

y (t)

=

c−t3 t2

´41

2.

102

= 0, with

M

= 4y 3 t2 and

(a) Inexact; y is an integrating factor. (b) Inexact; t is an integrating factor. 3. Step i:

F (y, t )

∂F = ∂∂t ∂t

Step ii: Step iii: Step iv: Setting

=

ψ (t) = F (y, t) F (y, t)

R R

M dy

+ ψ (t)

M dy

+ ψ (t) =

R¡

N

=

R

0

− ∂∂t

M dy

R +

M dy

R

¢

N; dt

N dt

−

thus

=

R

ψ0 (t) = N − ∂∂t

N dt

R ¡∂ R ∂t

−

R ¡∂ R ¢

M dy

∂t

R

M dy

¢

M dy

dt

dt

= c, we obtain the desired result.

Exercise 15.5

1.

(a)

2

i. Separable; we can write the equation as ii. Rewritten as

dy dt

2

y dy + t dt = 0.

+ 1t y = 0, the equation is linear.

(b) i. Separable; multiplying by (y + t), we get ii. Rewritten as and

=

m

−1.

dy dt

=

−2ty−1 ,

Define

z

y dy

+ 2t dt = 0.

the equation is a Bernoulli equation with

R

= 0,

T

=

−2t

= y 1−m = y 2 . Then we can obtain from (15.240) a linearized

equation dz

− 2(−2)t dt = 0

dz dt

or

+ 4t = 0

(c) i. Separable; we can write the equation as

y dy

ii. Reducible; it is a Bernoulli equation with

R

+ t dt = 0. = 0,

T

=

−t,

and

m

=

−1.

(d)

1 i. Separable; we can write the equation as 3y2 dy ii. Yes; it is a Bernoulli equation with

R

= 0,

T

− t dt = 0

= 3t,

m

= 2.

2.

ln

yt

2

2

y dy + t dt = 0 after

(a) Integrating

cancelling the common factor 2, we get ln y + ln t = c, or

= c. The solution is

yt

Check:

=

e

dy dt

c

=

=

k

−kt−2

or =

y (t)

−ty

=

k t

(consistent with the given equation).

103

1 1 2 2 y+t , and integrating we get 2 y + t = c.

(b) Cancelling the common factor

Thus the solution

is

¡

¢1

¢1

¡

y (t) = 2c − 2t2 2 = k − 2t2 2 ¢− 1 dy 1 ¡ 2t 2 Check: dt = 2 k − 2t2 (−4t) = − y

(which is equivalent to the given differential

equation).

y dy + tdt = 0, we get 21y 2 + 12 t2 = c, or y 2 + t2 = 2c = A. Thus the solution ¢ 12 . Treating it as a Bernoulli equation with R = 0, T = −t, m = −1, and is y (t) = A − t2 1 − m 2 z = y = y , we can use formula (15.24 ) to obtain the linearized equation dz + 2tdt = 0, dz or dt = −2t, which has the solution z = A − t2 . Reverse substitution then yields the identical

3. Integrating

¡

0

answer

y2 = A − t2

¢1

¡

y (t) = A − t2

or

1 −2 3 y dy − tdt = 0, 1 The solution is y (t) = A− 23 t2 .

4. Integrating

2

we obtain

− 31 y −1 −

12 2t

c

= , or

y −1 = −3c − 23 t2 = A − 23 t2 .

R = 0, T = 3t, and m = 2. 32 = −3t, which has the solution z = A − 2 t . Since

Treating it as a Bernoulli equation, on the other hand, we have

dz dt

dz + 3t dt = 0, or z = y1−m = y−1 , we have y(t) = 1z , which represents an identical solution. Thus we can write

5. The derivative of the solution is the other hand that

dz dt

At + 2) − 2 = 2At + 2.

2(

=

dz dt

2z − 2. t

At + 2. But since z

= 2

The linearized equation itself implies on =

At2 + 2t,

the latter result amounts to

Thus the two results are identical.

Exercise 15.6

1. (a) and (d): The phase line is upward-sloping, and the equilibrium is accordingly dynamically unstable. (b) and (c): The phase line is downward-sloping, and the equilibrium is dynamically stable.

2.

104

(a) The phase line is upward-sloping for nonnegative y; the equilibrium y ¯ = 3 is dynamically unstable. (b) The phase line slopes upward from the point of origin, reaches a peak at the point

¡1 1 ¢ , 4 , 16

1 and then slopes downward thereafter. There are two equilibriums, y ¯ = 0 and y ¯ = 2 ; the former is dynamically unstable, but the latter is dynamically stable. 3.

dy dt

(a) An equilibrium can occur only when (b)

d dy

  ³ ´ dy = 2y − 8 = −2 dt  +2

= 0, i.e., only when

when

y

=3

when

y

=5

y

= 3, or

y

= 5.

Since this derivative measures the slope of the phase line, we can infer that the equilibrium at

y

= 3 is dynamically stable, but the equilibrium at

y

= 5 is dynamically unstable.

Exercise 15.7

1. Upon dividing by

k

throughout, the equation becomes ˙ k k

Since the

first

I

≡

s

φ( k ) k

term on the right is equal to

that:growth rate of 2.

=

K L

= growth rate of

dK = d Aeλt = Aλeλt = Beλt . dt dt

K

−

sKLQ L

−λ =

sQ K

=

growth rate of

Thus net investment,

K, K ˙

the equation above means

L

I,

obviously also grows at the rate

λ. 3. The assumption of linear homogeneity (constant returns to scale) is what enables us to focus on the capital-labor ratio. 4. (a) There is a single equilibrium y ¯ which lies between 1 and 3, and is dynamically stable.

105

(b) There are two equilibriums:

y¯1

(negative) is dynamically stable, and

dynamically unstable.

106

y¯2

(positive) is

107...