exact differential equation

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 → ∞ ?


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)


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

First Order Linear Differential Equations

The Non-homogeneous case

dy/dt + ay = b ; b ≠ 0 ; a ≠ 0

Solution: y(t) = yc + yp ; yc : complementary functionyp : particular function

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


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)


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 equationP(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) = 1obtain 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) = 0y(t) = (y(0) – b/a) e-at + b/a; b = 15, a = 10

= (0 -15/10)e-10t + 15/10 = -3/2 e-10t + 3/2verification:

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


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


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



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

∫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 = 4tu(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)

•∫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 -1/2check: 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 =


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 Ik = ρKk : production capacityρ: ratio between capacity and capitalK: 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?

Domar Growth Model

dy/dt = dI/dt (1/s); but dk/dt = ρIFrom 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.

Analysis obtaining I (t)

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).


• 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


• In general, the differential equationM 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

Exact Differential EquationHow 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) ?


∂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 EquationBut, 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?


See he following example:

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

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

Therefore, the DE is a non exact DE.

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 + 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)



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


2. 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)


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

Exact Differential Equation3. 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)


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

4. 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 EquationAkan 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

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 = 0M = 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 + 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.

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 terpisahBentuk 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

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


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 + 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 tdan 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

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)



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

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-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


∫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 SolowQ = 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: CRTSQ = L . f (K/L , 1) = L φ (k) ;k = K/L; φ (k) = f (K/L , 1)

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 SolowBagaimana 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 LAtau 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.



Bila z = k1-α , maka dz + [(1-α ) λz - [1-α )s] dt = 0atau dz/dt + az = b;

a = (1-α ) λ;b = (1-α )s

solusinya: z(t) = (z(0) – s/λ ) e (1-α ) λt + s/λatau k1-α = (k (0)1-α – s/λ ) e (1-α ) λt + s/λ

k(0): nilai awal dari rasio Kapital dan labor.


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 mencapai keseimbangan. Ni-lai keseimbangan ini tergantung pada MPS dan per-tumbuhan tenaga kerja λ .

exact differential equation

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