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)

Market Price Dynamics

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

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)

First Order Linear Differential Equations

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 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.

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.

Market Price Dynamics

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) = 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

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) = 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

First Order Linear Differential Equations

(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

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

First Order Linear Differential Equations

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

Differential Equation with Variable Coefficient

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

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)

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

Differential Equation with Variable Coefficient

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 ;

Differential Equation with Variable Coefficient

Differential Equation with Variable Coefficient

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

Differential Equation with Variable Coefficient

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

Differential Equation with Variable Coefficient

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

Differential Equation with Variable Coefficient

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

Differential Equation with Variable Coefficient

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 =

Differential Equation with Variable Coefficient

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).

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 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) ?

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 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?

Exact Differential Equation

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.

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

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

Exact Differential Equation

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)

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

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)

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

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 Equation

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

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?

Exact Differential Equation

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.

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?

Exact Differential Equation

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

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 dengan variabel terpisah

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)

PD yang dapat direduksi menjadi PD Linier

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

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

PD yang dapat direduksi menjadi PD Linier

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 Solow

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.

Model Pertumbuhan Solow

Model Pertumbuhan Solow

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.

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