Market Price Dynamics

Market Price Dynamics Suppose for a commodity, demand function and supply function are as follows: Qd = α - βP

; α,β > 0

Qs = -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.

Market Price Dynamics 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 Qd and Qs. The interesting question is: will P(t) converge to P as t → ∞ ?

Market Price Dynamics To answer the question, we need to know the movement of P(t). In general, a price change depends on demand (Qd) and supply (Qs). If Qd > Qs, then, P tends to increase; If Qd < Qs, then, P tends to decrease; So, it can be assumed that dP/dt = λ (Qd-Qs); 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)

Market Price Dynamics 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)

First Order Linear Differential Equations 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

First Order Linear Differential Equations The Non-homogeneous case dy/dt + ay = b ; b ≠ 0

; a≠0

Solution: y(t) = yc + yp ; yc : complementary function yp : particular function yc the solution of its homogeneous differential equation, yc = Ae-at

First Order Linear Differential Equations

While yp is obtained by assuming that y(t) = k therefore dy/dt =0 and then yp = 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)

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)

Market Price Dynamics 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.

Market Price Dynamics 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.

Market Price Dynamics

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.

First Order Linear Differential Equations 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

First Order Linear Differential Equations

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

First Order Linear Differential Equations 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

First Order Linear Differential Equations (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

First Order Linear Differential Equations (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

First Order Linear Differential Equations (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 = 6e5t

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

Differential Equation with Variable Coefficient

dy/dt + u(t)y = w(t) How to obtain the solution of y(t) ?

Differential Equation with Variable Coefficient 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 = t3 + 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) Solution : y(t) = e-∫u dt (A +∫ w e ∫u dt dt)


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

Differential Equation with Variable Coefficient ∫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

Differential Equation with Variable Coefficient 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 = eν = e2t2 ; for ν = 2t2 y(t) = e-2t2 (A+e2t2) = Ae-2t2 + 1 ;


Exercises: 1. dy/dt + 5y = 15 ; u(t) = 5 ; w(t) = 15 General solution: y(t) = e -∫u dt (A + ∫we∫u dt dt)

Differential Equation with Variable Coefficient •∫u dt = ∫5 dt = 5t •∫we∫u dt dt = ∫15 e5t dt = 3e5t 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 et2 •y(t) = e-t2 (A + ½ et2) = 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 = t3/3 • ∫we∫u dt dt = ∫ 5t2 e t3/3 dt for u = t3/3; then du = t2 dt therefore, ∫5t2 et3/3 dt = ∫ 5 eu du = 5eu = 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 = -t2 e-t3/3 + t 2 (e –t3/3 + 5) = 5t2


4.

2 dy/dt + 12y + 2et = 0; y(0) = 6/7 atau dy/dt + 6y = -et ; u(t) = 6; w(t) = -et ∫u dt = ∫ 6 dt = 6t ∫ we∫u dt dt = ∫-et e6t dt = -∫ e7t dt = -e7t/7 Solution: y(t) = e-6t (A – e7t/7) y(0) = (A-1) →A = y(0) +1/7 =

Domar Growth Model 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

Domar Growth Model

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

Domar Growth Model

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

Domar Growth Model

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

Analysis obtaining I(t) 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

Analysis obtaining I (t) 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 if only if

M dy + N dt = 0

∂M/∂t = ∂N/∂y

Exact Differential Equation • 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: y2 t = c; or y(t) = c t-0.5 ; c = constant

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) ∂F/∂t = y + 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

Therefore, the DE is a non exact DE.

Exact Differential Equation 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.

Exact Differential Equation 1.

2y t3 dy + 3y2 t2 dt = 0 Apakah PD eksak? cari solusinya. M = 2y t3 ; N = 3y2 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 Cek : diferensial totalnya: 2y t2 dy + 3y2 t2 dt = 0


3y2t dy + (y3 + 2t) dt = 0 Apakah PD eksak ? cari solusinya: M = 3y2 t ; N = ( y3 + 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 + t2 = c cek: diferensial totalnya: 3y2 t dy + ( y3 +2t) dt = 0


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) ∂F/∂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: diferensial totalnya: t (1 + 2y) dy + y ( 1+ y) dt = 0


2 (t3 +1) dy + 3y t2 dt = 0 Apakah PD eksak? solusi? M = 2(t3 + 1) ; N = 3y t2 ∂M/∂t = 6t2 ≠ ∂N/∂y = 3t2  bukan PD eksak

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 ∂M/∂t = 6y t2 = ∂N/∂y → PD eksak

Exact Differential Equation 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

Exact Differential Equation 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?


4y3 t dy + (2y4 + 3t) dt = 0 Apakah PD eksak? solusi? M = 4y3 t N = 2y4 + 3t

;

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

Exact Differential Equation 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 + 3t2

Maka ψ'(t) = 3t2 ; ψ(t) = t3 + k F( y, t ) = y4 t2 + t3 + k atau y4 t2 + t3 = 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.

Persamaan Diferensial Tdk Linier Orde 1 Degree Satu 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

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.

PD dengan variabel terpisah Contoh: (1). 3y2 dy – t dt = 0 atau

3y2 dy = t dt ∫3y2 dy = ∫t dt y 3 = t2 / 2 + c

Solusi:

y(t) = (t2/2 + c) 1/3

PD dengan variabel terpisah 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

PD dengan variabel terpisah 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

PD dengan variabel terpisah Solusi: F(y,t) = y2 t + k1 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 y1-m = T

PD yang dapat direduksi menjadi PD Linier 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 dz + ( -tz + 3t) dt = 0

PD yang dapat direduksi menjadi PD Linier 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

PD yang dapat direduksi menjadi PD Linier 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-1z + 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

PD yang dapat direduksi menjadi PD Linier ∫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). fK , fL > 0 Artinya, output meningkat bila ada tambahan kapital maupun labor. (ii). fKK < 0,

fLL < 0 ; diminishing return.

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

Model Pertumbuhan Solow Karena fK = φ' (k), maka fKK = ∂/∂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 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.

Model Pertumbuhan Solow 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 mencapai konstan pada saat mencapai keseimbangan. Nilai keseimbangan ini tergantung pada MPS dan pertumbuhan tenaga kerja λ .

Market Price Dynamics

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