Market Price Dynamics

Suppose for a commodity, demand function and supply function are as follows:

Q_d = α - βP ; α,β > 0 Q_s = -γ + δP ; γ,δ > 0

From last lecture, the equilibrium price is: P= (α +γ)/(β +δ)

If the initial price, P(0) = P, then, the market is in an equilibrium and, therefore, no need to analyze the movements of the price.

However, if P(0) ≠ P, we need to see the process from P(0) to P (if there is an equilibrium).

In this case, the price will change as time changes; so do Q_d and Q_s.

The interesting question is: will P(t) converge to P as t → ∞ ?

To answer the question, we need to know the movement of P(t).

In general, a price change depends on demand (Q_d) and supply (Q_s).

If Q_d > Q_s, then, P tends to increase; If Q_d < Q_s, then, P tends to decrease;

So, it can be assumed that dP/dt = λ (Q_d-Q_s); λ >0; λ represents coefficient of adjustment:

dP/dt = λ(α - βP + γ - δP) = λ(α +γ)- λ(β +δ)P or dP/dt + λ(β + δ)P = λ(α + γ)

Notice that the differential equation is of the form: dy/dt + ay = b; where: y(t) = P(t) and the solution is: P(t) = (P(0) – (α +γ)/(β +δ)) e- λ(β+δ)t + (α + γ)/(β + δ)

= (P(0) – P )e-kt + P ; k= λ(β +δ)

General form: dy/dt + u(t)y = w(t)

• If u(t) = a, constant, and w(t) =0, then the new form dy/dt + ay = 0 ; is called homogeneous differential equation.

• It can be written as: 1/y dy/dt = -a • General solution: y(t) = A e-at _;

• Particular solution: y(t) = y(0) e-at

The Non-homogeneous case

dy/dt + ay = b ; b ≠ 0 ; a ≠ 0

Solution: y(t) = y_c + y_p ; y_c : complementary function y_p : particular function

y_c the solution of its homogeneous differential equation, y_c = Ae-at

While y_p is obtained by assuming that y(t) = k therefore dy/dt =0 and then y_p = b/a ; finally, the solution y(t) = Ae-at _{+ b/a}

For t = 0, y(0) = A + b/a or A = y(0)-b/a Thus, y(t) = (y(0) – b/a) e-at + b/a.

(compare to the solution of P(t) on market price dynamics)

Back to earlier case:

dP/dt = λ(α - βP + γ - δP) = λ(α +γ)- λ(β +δ)P or dP/dt + λ(β + δ)P = λ(α + γ)

Notice that the differential equation is of the form: dy/dt + ay = b; where: y(t) = P(t) and the solution is: P(t) = (P(0) – (α +γ)/(β +δ)) e - λ(β+δ)t + (α + γ)/(β + δ)

= (P(0) – P )e -kt + P ; k= λ(β +δ)

The Dynamic Stability of Equilibrium

Will P(t) converge to P as t→ ∞ • See the following equation

P(t) = (P(0) – P )e-kt + P . . . (*)

From the equation, it can be seen that P(t) will converge to P since e-kt → 0 as t → ∞

because k= λ(β +δ) > 0.

Therefore, the equilibrium is dynamically stable.

There are 3 cases can happened from equation (*) 1 . If P(0) = P , then P(t) = P . This situation

represents the constant movement of P(t) at equilibrium price P

2 . If P(0) > P , then P(0) – P > 0. The price movement of P(t) approaches P from above. 3 . If P(0) < P , then P(0) – P < 0. The price

movement of P(t) approaches P from below.

In general, if deviation of P(t) and P equal zero or decreases as t increases, then it will reach an equilibrium that dynamically stable.

Examples

1. If dy/dt + 2 y = 6 , with initial condition y(0) =10; find the solution.

y(t) = (y₀- b/a) e-at + b/a ; b=6; a= 2 = (10-3)e-2t + 3

y(t) = 7 e-2t _{+ 3}

Verification: dy/dt + 2y = 7 (-2)e-2t + 2(7e-2t + 3) = 6

2. dy/dt + 4y = 0; initial condition y(0) = 1 obtain the solution.

y(t) = (y₀ – b/a) e-at _{+ b/a; b = 0; a = 4}

= (1 – 0)e-4t + 0 = e-4t

Verification: dy/dt + 4y = -4e-4t + 4e-4t = 0

Exercises

1. Find the solution of:

(a). dy/dt +4y = 12; y(0) = 2

y(t) = (y(0) – b/a) e-at + b/a ; b=12; a=4 = (2- 12/4) e-4t + 12/4

= - e-4t _{+ 3}

Verification: dy/dt + 4y = 4e-4et + 4(-e-4t + 3) = 12

(b). dy/dt – 2y = 0; y(0) = 9

y(t) = (y(0) – b/a) e-at + b/a; b=0; a=-2 = (9-0)e2t + 0 = 9e2t

verification: dy/dt-2y = (9)(2)e2t - 2(9e2t) = 0 (c). dy/dt + 10y = 15; y(0) = 0

y(t) = (y(0) – b/a) e-at + b/a; b = 15, a = 10 = (0 -15/10)e-10t _{+ 15/10 = -3/2 e}-10t _{+ 3/2}

verification:

dy/dt + 10y = -3/2(-10)e-10t + 10(- 3/2e-10t + 3/2) =15

(d). 2dy/dt + 4y = 6 ; y(0) = 3/2 or dy/dt + 2y = 3

y(t) = (y(0)-b/a)e-at + b/a ; b=3; a=2 = (3/2-3/2)e-2t _{+ 3/2 = 3/2}

verification: 2dy/dt + 4y = 2 (0) + 4(3/2) =6

(e). dy/dt +y = 4; y(0)=0

y(t) = (y(0)-b/a)e-at + b/a; b=4; a=1 = (0-4)e-t + 4 = -4e-t + 4

verification: 4e-t + (-4e-t + 4) = 4

(f). dy/dt – 5y = 0; y(0) = 6

y(t) = (y(0) - b/a) e-at + b/a; b = 0; a = -5 = (6-0)e5t _{– 0 = 6e}5t

verification: (6)(5)e5t – 5(6e5t) = 0

(g). dy/dt – 7y = 7; y(0)= 7

y(t) = (y(0) – b/a) e-at + b/a; b=7, a=7 = (7-1) e-7t _{+ 1 = 6e}-7t _{+ 1}

verification: (6) (-7)e-7t – 7 (6 e-7t + 1) = 7

dy/dt + u(t)y = w(t)

How to obtain the solution of y(t) ?

Homogeneous Case:

w(t) = 0

dy/dt + u(t)y = 0 or (1/y)(dy/dt) = -u(t) If we do integration on both sides:

∫ (1/y)(dy/dt) dt =∫ - u(t)dt or ∫dy/y = - ∫u(t) dt Ln y + c = -∫u(t) dt

Ln y = - c - ∫u(t) dt

y(t) = eLn y _{= e}-c _e-∫u(t) dt _{= Ae}-∫u(t) dt _{; A= e}-c

Ilustration:

Find the general solution of dy/dt + 3t2y = 0 From earlier discussion, y(t)= A e-∫u(t) dt ;

with ∫u(t) dt = ∫ 3t2 _{dt = t}3 _{+ c}

Thus, y(t) = Ae-t3 e-c = B e-t3 with B = Ae-c

Differential Equation with Variable Coefficient

Non-homogeneous case

w(t) ≠ 0

dy/dt + u(t) y = w(t)

Differential Equation with Variable Coefficient

Ilustration

Find the solution of dy/dt + 2ty = t u(t) = 2t ; w(t) = t

∫u (t) dt = ∫2t dt = t2 _{+ k; k: constant}

∫w(t) e∫u(t) dt _{dt = ∫t e}(t2 +k) _dt

= ek ∫ t et2dt = (ek/2) et2 + c ; c: constant. y(t) = e -∫u(t) dt (A +∫we ∫u dt dt)

= e –(t2 +k) ( A+ ek et2/2 + c)

= A e-k _e-t2 _{+ e}-t2 _e-k _ek _et2_{/2 +ce}-(t2+k)

= (A + c) e-k e-t2 +1/2

= B e-t2 +1/2; B=(A+c) e-k ; a constant

Another illustration

Find the solution of dy/dt + 4ty = 4t u(t) = 4t; w(t) = 4t

∫u(t) dt = 2t2

∫w(t)e∫u(t) dt _{dt = ∫4t e} 2t2 _dt

= ∫eν _{dν = e}ν _{= e}2t2 ; for ν = 2t2

y(t) = e-2t2 (A+e2t2) = Ae-2t2 + 1 ;

Differential Equation with Variable Coefficient

Exercises:

1. dy/dt + 5y = 15 ; u(t) = 5 ; w(t) = 15

•∫u dt = ∫5 dt = 5t

•∫we∫u dt _{dt = ∫15 e}5t _{dt = 3e}5t

y(t) = e-5t (A + 3e5t) = Ae-5t + 3

Verification using of constants u(t) and w(t) rule: y(t) = Ae-at _{+ b/a ; a= 5 ; b= 15}

= Ae-5t + 3

check: dy/dt + 5y = -5Ae-5t + 5(Ae-5t + 3 ) = 15

2. dy/dt +2ty = t; y(0) = 3_/

2; u(t) = 2t; w(t) = t

solution:∫u dt = ∫2t dt = t2

•∫we∫u dt _{= ∫te} t2 _{dt = ½ e}t2

•y(t) = e-t2 _{(A + ½ e}t2_{) = Ae}-t2 _{+ ½}

y(0) = A +1/₂ → A= y₀ -1/₂ check: dy/dt + 2ty = -Ae-t2 _{(2t) + 2t(Ae}-t2 _{+ ½ ) = t}

3. dy/dt + t2 y = 5t2 ; y(0) = 6; u(t) = t2 ; w(t) = 5t2 • ∫u dt = ∫ t2 _{dt = t}3_/3

• ∫we∫u dt _{dt = ∫ 5t}2 _e t3/3 _dt

for u = t3_{/3; then du = t}2 _dt

therefore, ∫5t2 _et3/3 dt = ∫ 5 eu _{du = 5e}u _{= 5e} t3/3

solution: y(t) = e-t3/3 (A + 5 e t3/3) = A e-t3/3 + 5 y(0) = A + 5 →A = y(0) - 5 = 1 y(t) = e-t3/3 + 5

check: dy/dt + t2 _{y = -t}2 _e-t3/3 _{+ t} 2 _(e –t3/3 _{+ 5) = 5t}2

4. 2 dy/dt + 12y + 2et = 0; y(0) = 6/7

atau dy/dt + 6y = -et _{; u(t) = 6; w(t) = -e}t

∫u dt = ∫ 6 dt = 6t

∫ we∫u dt _{dt = ∫-e}t _e6t dt = -∫ e7t _{dt = -e}7t_/7

Solution: y(t) = e-6t (A – e7t/₇) y(0) = (A-1) →A = y(0) +1/7 =

1. The changes of investment rate per year will have impacts on two things:

(i). Aggregate Demand (total) (ii). Production Capacity

2. The impact of demand from investment can be represented by:

dy/dt = ( dI/dt) (1/s)

s: marginal propensity to save

3. The impact of production capacity from investment can be represented by:

dk/dt = ρ dK/dt = ρ I k = ρK

k : production capacity

ρ: ratio between capacity and capital K: capital

4. In Domar Model, an equilibrium achieved when production capacity fulfilled.

This case happened when total demand (y(t)) equals to production capacity (k) at time t; or;

dy/dt = dk

5. How to obtain an investment pattern (I(t)) per year?

dy/dt = dI/dt (1/s); but dk/dt = ρI From these two equations:

(dI/dt) (1/s) = ρ I or (1/I)(dI/dt) = ρ s Integrate both sides:

∫(1/I) (dI/dt) dt = ∫ρ s dt Or ∫ dI/I = ∫ρ s dt

Ln I + c₁ = ρ s t + c₂ or Ln I = ρ s t + c

I = e (ρst + c)

I(t)= A eρst ; A= ec At t=0, I(0)=A e0 _{= A}

Then, I(t) = I(0) eρst ; I(0): initial investment

Interpretation: to maintain an equilibrium between production capacity and demand, rate of investment should grow at eρs. The higher the investment rate needed, the higher ρ and s required.

Exact Differential Equations

• If we have a two-variable function F(y,t), the total differential:

dF(y, t) = ( ∂F/∂y ) dy + ( ∂F/∂t ) dt • When dF(y, t) = 0,

(∂F/∂y) dy + ( ∂F/∂t ) dt = 0,

The form of this differential equation is called Exact Differential Equation since its left side is exactly the differential of the function F(y, t).

Exact Differential Equations

• For example F( y, t ) = y2 _{t + k; k: contstant}

The total differential: dF = 2y t dy + y2 dt, and the differential equation is in the form of:

2y t dy + y2 dt = 0 Or dy/dt + y2_{/2y t = 0}

Exact Differential Equations

• In general, the differential equation M dy + N dt = 0

is an exact differential equation if and only if there is a function F(y, t) with M = ∂F/∂y and N = ∂F/∂t

Since ∂2F/∂t∂y = ∂2F/∂y∂t,

it can be said that M dy + N dt = 0 if only if ∂M/∂t = ∂N/∂y

• Verify whether 2y t dy + y2 _{dt = 0 is an exact}

differential equation?

Check: M = 2y t; N= y2

∂M/∂t = 2y; ∂N/∂y = 2y Since, ∂M/∂t = ∂N/∂y = 2y;

therefore the DE is exact DE.

Exact Differential Equation

How to solve an Exact Differential Equation

Exact DE: M dy + N dt = 0 Solution: F( y, t ) = ∫ M dy + ψ(t) Example: (1). 2y t dy + y2 dt = 0 M = 2y t; N = y2 Solution: F(y, t) = ∫ 2y t dy + Ψ(t) = y2 t +ψ (t) How to obtain ψ (t) ?

Exact Differential Equation

∂F/∂t = y2+ψ' (t)

But ∂F/∂t = N = y2; therefore ψ ' (t) = 0 or (t) = k,

So, F ( y, t ) = y2 t + k

Thus, the solution of DE is:

Exact Differential Equation

(2). Find the following DE:

( t +2y ) dy + ( y + 3t2 _{) dt =0}

M = t + 2y; N = y + 3t2 ∂M/∂t = 1 = ∂N/∂y ;

so, ∂M/∂t = ∂N/∂y → exact DE F(y,t) = ∫ M dy + ψ(t)

= ∫ (t + 2y) dy + ψ(t) = yt + y2 + ψ(t)

Exact Differential Equation

But, N = ∂F/∂t = y + 3t2 _;

thus, ψ'(t) = 3t2 ; ψ(t) = t3

Therefore, F ( y, t ) = yt + y2 + t3 The solution of the exact DE is:

yt + y2 + t3 = c; c: constant

Verification: The total differential: ( ∂F/∂y ) dy + ( ∂F/∂t ) dt

= ( t + 2y ) dy + ( y + 3t2 _{) dt = 0}

Can a non-exact DE be transformed into an exact DE?

Exact Differential Equation

See he following example: (3). 2t dy + y dt = 0;

M = 2t; N = y

check: ∂M/∂t = 2; ∂N/∂y = 1 ; it means ∂M/∂t ≠ ∂N/∂y

Now, multiply the DE with y, then:

2t y dy + y2 dt = 0; is an exact DE (verify?) and its solution is:

y(t) = c t-0.5

In this case, y is a multiplication factor that can

transform a non exact DE to an exact DE; and y is called intergration factor.

1. 2y t3 _{dy + 3y}2 _t2 _{dt = 0 Apakah PD eksak?}

cari solusinya.

M = 2y t3 _{; N = 3y}2 _t2

∂M/∂t = 6y t2 ; ∂N/∂y = 6y t2 _;

berarti ∂M/∂t = ∂N/∂y, PD tersebut di atas merupakan PD eksak.

Solusi: F(y,t) = ∫M dy +ψ(t) = ∫2y t3 dy + ψ(t)

= y2 t3 + ψ(t), sehingga ∂F/∂t = 3y2 t2 + ψ'(t)

Exact Differential Equation

Padahal, N = ∂F/∂t = 3y2 _t2

Dengan demikian, ψ'(t) = 0 sehingga solusinya: F ( y, t ) = y2 t3 + k atau y2 t3 = c

Exact Differential Equation

2. 3y2t dy + (y3 + 2t) dt = 0

Apakah PD eksak ? cari solusinya: M = 3y2 _{t ; N = ( y}3 _{+ 2t)} ∂M/∂t = 3y2 = ∂N/∂y → PD eksak Solusi: F( y, t ) = ∫ M dy + ψ(t) = ∫3y2 t dy + ψ(t) = y3 t + ψ(t) ∂F/∂t = y3 +ψ'(t)

Exact Differential Equation

Sedangkan, N = ∂F/∂t = y3 + 2t; maka ψ'(t) = 2t

atau ψ(t) = t2

sehingga solusinya F ( y, t ) = y3 _{t + t}2 _{= c}

Exact Differential Equation

3. t(1+2y)dy + y (1+y)dt = 0

Apakah PD eksak? cari solusinya. M= t(1+2y); N= y (1+y) = (y+y2)

∂M/∂t = (1+2y) = ∂N/∂y; →PD eksak Solusi: F(y,t) = ∫ M dy + ψ(t)

= ∫ t (1 + 2y) dy + ψ(t) = t (y + y2) + ψ(t)

Exact Differential Equation

sedangkan N = ∂F/∂t = (y + y2_); berarti ψ'(t) = 0 atau ψ(t) = k solusi: F ( y, t ) = t (y + y2) + k atau t (y + y2_{) = c ; c: konstan} cek:

4. 2 (t3 _{+1) dy + 3y t}2 _{dt = 0}

Apakah PD eksak? solusi? M = 2(t3 _{+ 1) ; N = 3y t}2

∂M/∂t = 6t2 ≠ ∂N/∂y = 3t2  bukan PD eksak

Exact Differential Equation

Exact Differential Equation

Akan dicoba dicari faktor integrasinya

Bila PD tersebut di atas dikalikan dengan y, diperoleh:

2 y (t3 + 1) dy + 3 y2 t2 dt = 0 dengan M = 2y (t3 + 1) ; N = 3y2 t2

Solusi: F(y,t) = ∫ M dy + ψ(t)

= ∫ 2y (t3 + 1) dy + ψ(t)

= y2 _(t3 + 1) + ψ(t)

∂F/∂t = 3y2 _t2 + ψ'(t)

Sedangkan N = ∂F/∂t = 3y2 _t2

Dengan demikian, ψ'(t) = 0 atau ψ(t) = k

F( y, t ) = y2 (t3 + 1) + k atau y2 (t3 + 1) = c; c = konstan

Komentar:

Bagaimana mencari faktor integrasi?

5. 4y3 t dy + (2y4 + 3t) dt = 0 Apakah PD eksak? solusi? M = 4y3 t ; ∂M/∂t = 4y3 _t

N = 2y4 + 3t ∂N/∂y = 8y3; PD tidak eksak Sekarang dicari faktor integrasinya:

Bila PD diatas dikalikan dengan t, diperoleh: 4y3 t2 dy + (2y4 + 3t) t dt = 0

M = 4y3 t2 ; N = (2y4 + 3t) t ∂M/∂t = 8y3t = ∂N/∂y  PD eksak.

Solusi: F ( y, t ) = ∫ M dy + ψ(t) = ∫ 4y3 _t2 dy + ψ(t) = y4 t2 + ψ(t) ∂M/∂t = 2y4 t + ψ'(t) Sedangkan, N = ∂F/∂t = 2y4 _{t + 3t}2 Maka ψ'(t) = 3t2 ; ψ(t) = t3 _{+ k} F( y, t ) = y4 _t2 _{+ t}3 _{+ k atau y}4 _t2 _{+ t}3 _{= c}

Komentar: Bagaimana mencari faktor integrasi? Trial and Error?

Persamaan Diferensial Tdk Linier Orde 1

Degree Satu

PD Linear: (i). dy/dt dan y linier

(ii). tidak boleh ada perkalian y. (dy/dt) Dengan demikian meskipun dy/dt linier tetapi bila y berpangkat lebih besar dari satu, persamaannya menjadi tidak linier.

Secara umum, bentuk persamaannya: f(y,t) dy + g(y,t) dt = 0

atau dy/dt = h(y,t)

Ada 3 cara mencari solusinya

(i). Model PD eksak (sudah dipelajari) (ii). Model PD terpisah

(iii). Model tidak linier direduksi menjadi linier

Persamaan Diferensial Tdk Linier Orde 1

Degree Satu

PD dengan variabel terpisah

Bentuk umum: f(y,t) dy + g(y,t) dt = 0

Bila f(y,t) hanya merupakan fungsi dari y atau f(y) dan bila g(y,t) juga hanya merupakan fungsi dari t atau g(t), maka bentuk umum di atas berubah

menjadi

f(y) dy + g(t) dt = 0

PD ini disebut PD dengan variabel terpisah karena variabel y dan t muncul secara terpisah, mereka berada di ruas yang terpisah.

Contoh: (1). 3y2 dy – t dt = 0 atau 3y2 _{dy = t dt} ∫3y2 dy = ∫t dt y3 = t2/₂ + c Solusi: y(t) = (t2_/ 2 + c) 1/3

Contoh: (2). 2t dy + y dt = 0 dy/y + dt/2t = 0 ∫(1/y ) dy +∫(1/2t) dt = c Ln y + (1/2) Ln t + c atau Ln (yt1/2_{) = c} y t1/2 = ec = k y(t) = k t-1/2

Komentar:

Lihat lagi contoh 2:

2t dy + y dt = 0 atau 2t y dy + y2 dt = 0, merupakan PD Eksak dengan M = 2ty, N =y2_.

Solusi umum F(y,t) = ∫2yt dy + ψ (t) = y2 t + ψ(t)

∂F/∂t = y2 + ψ'(t)

Padahal: N = ∂F/∂t = y2 _;

berarti ψ' (t) = 0 atau ψ(t) = k₁

Solusi: F(y,t) = y2 _{t + k} 1 Atau y2 t + k₁ = c₁ y2 t = c y = k . t-1/2 Komentar:

Dengan metode PD terpisah maupun dengan metode PD eksak, solusi pada contoh no.2 mencapai hasil akhir yang sama.

PD yang dapat direduksi menjadi PD Linier

Bila PD dy/dt = h(y,t) dapat dinyatakan dalam bentuk tidak linier sebagai berikut:

dy/dt + Ry = Tym _{dengan R, T fungsi t}

dan m ≠0, m ≠1

maka PD tersebut selalu dapat direduksi menjadi PD linier.

Proses reduksi:

dy/dt + Ry = T ym ; persamaan Bernoulli y-m _{dy/dt + R y}1-m _{= T}

sederhanakan: z = y1-m dz/dt = dz/dy . dy/dt = (1-m) y-m . dy/dt (1-m)-1 _{dz/dt + Rz = T} dz +[(1-m) Rz - (1-m) T ] dt = 0 dz + (u z –wT) dt; u =(1-m)R ; w =(1-m)

Solusi: z(t) = e-∫ u(t) dt (A + ∫we ∫ u dt dt)

PD yang dapat direduksi menjadi PD Linier

Contoh:

1. cari solusi dari dy/dt + ty = 3 ty2 m = 2 ; z = y1-m ; R = t T = 3t

PD liniernya; dz + [(1- m) Rz – (1- m) T ] dt = 0 dz + [(1 – 2) tz - (1 – 2)(3t)] dt = 0

solusi z(t) = e-∫ u(t) dt (A + ∫we ∫ u dt dt) dengan u(t) = -t ; w(t) = -3t ∫ u(t) dt = ∫ -t dt = - t2_/ 2 ∫we ∫ u dt _{dt = - ∫ 3t e}- t2/2 _{dt = 3e}- t2/2 z(t) = e +t2/2 (A + 3e –et2/2) = A e t2/2 + 3 padahal, z(t) = y(t)-1 atau y(t) = z(t)-1 _{= (A + 3e} –t2/2₎-1

2. cari solusi dari dy/dt + (1/t) y = y3_.

m = 3 ; z = y-2 ; R = t-1 ; T = 1

PD liniernya; dz + [(1- m) Rz – (1- m) T ] dt = 0 dz + ( – 2t-1_{z + 2) dt = 0}

u(t) = – 2t-1 ; w(t) = -2 solusi z(t) = e-∫ u(t) dt (A + ∫we ∫ u dt dt) ∫ u(t) dt = - ∫ 2 dt/t = - 2 Ln t

∫we ∫ u dt _{dt = ∫- 2 e}-2 Ln t _dt = ∫- 2 t-2 dt = 2t-1 z(t) = e 2 Ln t (A + (2) t-1) = t2 (A + 2t-1) z(t) = At2 _{+ 2t} y(t) = z-1/2 = (At2 _{+ 2t)}-1/2

Model Pertumbuhan Solow

Q = f (K,L) ; K > 0 ; L > 0 K: Kapital; L: Labor; Q: Output Asumsi:

(i). f_K , f_L > 0

Artinya, output meningkat bila ada tambahan kapital maupun labor. (ii). f_KK < 0, f_LL < 0 ; diminishing return. (iii). f: CRTS

Q = L . f (K/L , 1) = L φ (k) ; k = K/L; φ (k) = f (K/L , 1)

Karena f_K = φ' (k),

maka f_KK = ∂/∂K (φ' (k)) = dφ' (k) / dk . ∂k/∂K = φ''(k) . 1/L

Asumsi Solow:

(i). dk/dt = sQ; sebagian dari Q di investasikan. s: Marginal Propensity to Save

(ii). (dL/dt) /L = λ; Labor tumbuh secara eksponensial atau

L = e λt

Model Pertumbuhan Solow secara lengkap. (1). Q = L f (K/L , 1) = L φ (k); k= K/L

(2). dk/dt = sQ

(3). (dL /dt) / L = λ

Model Pertumbuhan Solow

Bagaimana mencari solusi dari model tersebut? (2): dk/dt = s.Q = s . L . φ (k); k = K/L; sedangkan: K = k L, Akibatnya: dK/dt = dk/dt. L + dL/dt . k = L dk/dt + k λ L Berarti: s . L . φ (k) = L dk/dt + k λ L Atau s φ (k) = dk/dt + k λ Atau dk/dt = s φ (k) - k λ

Ini merupakan persamaan diferensial dalam k dengan parameter λ dan s.

Sebagai ilustrasi, bila Q = Kα L1-α

Q = Kα L1-α = L (K/L)α, sehingga φ (k) = kα dan persamaan diferensialnya menjadi:

dk/dt = s kα - λk atau dk/dt +λk = skα.

PD tersebut merupakan persamaan Bernoulli dengan R = λ ; T= s; m =α.

Model Pertumbuhan Solow

Bila z = k1-α , maka dz + [(1-α ) λz - [1-α )s] dt = 0 atau dz/dt + az = b; a = (1-α ) λ; b = (1-α )s solusinya: z(t) = (z(0) – s/λ ) e (1-α ) λt _{+ s/λ} atau k1-α = (k (0)1-α – s/λ ) e (1-α ) λt + s/λ k(0): nilai awal dari rasio Kapital dan labor.

Model Pertumbuhan Solow

Pada saat t → ∞ , k1-α _{→ s/λ atau k→ (s/λ)} (1/ (1-λ))

Artinya, rasio kapital dan tenaga kerja akan menca-pai konstan pada saat mencamenca-pai keseimbangan. Ni-lai keseimbangan ini tergantung pada MPS dan per-tumbuhan tenaga kerja λ .