ECO607 — Final Term Summary (Lectures 23–44)

📘 Lecture 23 — Various Economic Dynamics Using Differential Equations

📖 Overview: This lecture applies first-order linear differential equations to model the dynamic behavior of various economic variables, including the value of natural resources (oil wells and gold mines), net revenue from nonrenewable resources (Hotelling's Rule), bank deposits, monthly sales of new products, and population growth. It demonstrates how to derive time paths, assess dynamic stability, and solve for specific future values.

🗂️ Topics Covered

This lecture covers five distinct applications: dynamics of oil well value (Topic 094), dynamics of gold mine value (Topic 095), Hotelling's Rule for nonrenewable resources (Topic 096), future value of bank deposits (Topic 097), dynamics of monthly sales using a logistic-type model (Topic 098), and dynamics of population growth (Topic 099). Each application follows a systematic procedure: setting up the differential equation, identifying the case (homogenous vs. non-homogenous), finding the definite solution, and analyzing dynamic stability.

📝 Lecture Summary

TOPIC 094: DYNAMICS OF VALUE OF OIL WELL USING DIFFERENTIAL EQUATIONS

The initial value of an oil well is V(0) = $2700, and its value decays according to the differential equation dV/dt + 0.04 V(t) = 28. This models the depreciation of a natural resource asset over time.

🔑 Definition — Time Path: The function V(t) describing the value of a variable at any point in time. The general solution is y(t) = {y(0) - b/a} e^{-at} + b/a for the Non-Homogenous Case with a ≠ 0.

📐 Formula: For dV/dt + aV = b with a ≠ 0 and b ≠ 0, the definite solution is V(t) = {V(0) - b/a} e^{-at} + b/a. Substituting a = 0.04, b = 28, V(0) = 2700: V(t) = (2700 - 28/0.04) e^{-0.04t} + 28/0.04 = (2700 - 700) e^{-0.04t} + 700 = 2000 e^{-0.04t} + 700.

📌 Example: To find the future value at t = 10: V(10) = 2000 e^{-0.04*10} + 700 = 2000 e^{-0.4} + 700. Since e^{-0.4} ≈ 0.6703, V(10) = 2000(0.6703) + 700 = 1340.6 + 700 = $2040.6. The time path shows decay from $2700 to equilibrium V = 700*. 💡 Why this matters: This models the declining economic value of a finite resource as it is extracted over time.

TOPIC 095: DYNAMICS OF VALUE OF GOLD MINE USING DIFFERENTIAL EQUATIONS

The initial value of a gold mine is G(0) = $1,400, decaying at the rate dG/dt + 0.04 G(t) = 16. This is structurally identical to the oil well problem.

🔑 Definition — Dynamic Stability: A time path is dynamically stable if the deviation from equilibrium (complementary function) approaches zero as time goes to infinity (t → ∞).

📐 Formula: Using the same NHCa≠0 form with a = 0.04, b = 16: G(t) = {G(0) - b/a} e^{-at} + b/a = {1400 - 16/0.04} e^{-0.04t} + 16/0.04 = (1400 - 400) e^{-0.04t} + 400 = 1000 e^{-0.04t} + 400.

📌 Example: At t = 10: G(10) = 1000 e^{-0.4} + 400 = 1000(0.6703) + 400 = $1070.30. The complementary function G_c = 1000 e^{-0.04t} → 0 as t → ∞, so the time path converges to equilibrium G = 400*.

TOPIC 096: HOTELLING'S RULE USING DIFFERENTIAL EQUATIONS

Hotelling's Rule, named after economist Harold Hotelling, states that the net revenue π(t) from extracting a nonrenewable resource grows exactly at the rate of interest r.

🔑 Definition — Hotelling's Rule: The optimal extraction path occurs when the growth rate of net revenue equals the interest rate: π'(t)/π(t) = r, or equivalently, dπ/dt = r π(t).

📐 Formula: Rewriting as dπ/dt - rπ = 0 gives a = -r, b = 0, which is a Homogenous Case (HC) with a ≠ 0. The definite solution for HC is y(t) = {y(0)} e^{-at}. Substituting: π(t) = {π(0)} e^{-(-r)t} = {π(0)} e^{rt}. Wait — checking the substitution: The standard form is dy/dt + ay = 0 with y(t) = y(0) e^{-at}. Here dπ/dt - rπ = 0 means a = -r, so π(t) = π(0) e^{-(-r)t} = π(0) e^{rt}.

📌 Example: For Hotelling's Rule, π(t) = π(0) e^{rt}. However, the lecture presents the solution as π(t) = {π(0)} e^{-rt}, which represents a decay model. This discrepancy requires careful attention — the rule formalized states net revenue grows at rate r, suggesting π(t) = π(0) e^{rt}. The lecture solution with π(t) = π(0) e^{-rt} would show convergence to zero, not growth. Seek clarification on which formulation is correct for the exam. 💡 Why this matters: This rule is fundamental for understanding the economics of nonrenewable resource extraction and optimal depletion policies.

TOPIC 097: FUTURE VALUE OF INVESTMENT USING DIFFERENTIAL EQUATIONS

An initial deposit D(0) = $1,300 grows according to dD/dt - 0.05 D(t) = 15. Note the minus sign — rewriting gives dD/dt + (-0.05) D = 15, so a = -0.05 and b = 15.

🔑 Definition — Complementary Function (Deviation): The part of the time path y_c that represents the deviation from equilibrium. For dynamic stability, y_c → 0 as t → ∞.

📐 Formula: D(t) = {D(0) - b/a} e^{-at} + b/a. Substituting a = -0.05, b = 15: D(t) = {1300 - 15/(-0.05)} e^{-(-0.05)t} + 15/(-0.05) = {1300 + 300} e^{0.05t} - 300 = 1600 e^{0.05t} - 300.

📌 Example: Checking against the lecture: D(t) = (1300 - 300) e^{-0.05t} + 300 = 1000 e^{-0.05t} + 300. Wait — the lecture uses a = 0.05 (positive), but the given equation is dD/dt - 0.05D = 15, meaning a = -0.05. This is a critical inconsistency to verify. The table shows D_c = 1000 e^{-0.05t} → 0, converging to D = 300*.

TOPIC 098: DYNAMICS OF MONTHLY SALES USING DIFFERENTIAL EQUATIONS

Monthly sales S(t) of a new product satisfy dS/dt = k(M - S) with S(0) = C, where k, M, C are positive constants.

🔑 Definition — Saturation Level (M): The maximum possible monthly sales that the market can absorb — the equilibrium sales level S*.

📐 Formula: Rewriting dS/dt + kS = kM: a = k, b = kM. Using the NHCa≠0 solution: S(t) = {S(0) - b/a} e^{-at} + b/a = {C - kM/k} e^{-kt} + kM/k = (C - M) e^{-kt} + M.

📌 Example: If initial sales C = 0, then S(t) = -M e^{-kt} + M = M(1 - e^{-kt}). As t → ∞, e^{-kt} → 0, so S(t) → M. Dynamic stability holds regardless of whether C > M or C < M, since e^{-kt} → 0 for k > 0. 💡 Why this matters: This model (similar to the logistic or Bass diffusion model) predicts how new products penetrate markets over time.

TOPIC 099: DYNAMICS OF POPULATION USING DIFFERENTIAL EQUATIONS

A country's population N is currently 5 million and grows according to dN/dt = 0.015 N, where t is in years.

🔑 Definition — Dynamic Instability: A time path is dynamically unstable if the deviation from equilibrium grows without bound as t → ∞.

📐 Formula: dN/dt - 0.015N = 0 gives a = -0.015, b = 0 (Homogenous Case). Definite solution for HC: N(t) = N(0) e^{-at} = N(0) e^{-(-0.015)t} = 5,000,000 e^{0.015t}.

📌 Example — Doubling Time: When does N = 10,000,000? 10,000,000 = 5,000,000 e^{0.015T}. Dividing: 2 = e^{0.015T}. Taking natural logs: ln(2) = 0.015T. So T = ln(2)/0.015 = 0.693147/0.015 = 46.2 years. The positive exponent (0.015t) means N_c → ∞ as t → ∞, making this dynamically unstable. 💡 Why this matters: Policy-makers use such models to plan for infrastructure, healthcare, and educational needs based on population projections.

⭐ Key Takeaways

The most critical thing to remember is the standard form of a first-order linear differential equation: dy/dt + a y = b, and its three-case solution method: if b = 0 use y(t) = y(0) e^{-at} (Homogenous); if b ≠ 0 and a = 0 use y(t) = y(0) + b t; if b ≠ 0 and a ≠ 0 use y(t) = {y(0) - b/a} e^{-at} + b/a (Non-Homogenous). For dynamic stability, you check whether the exponent in the complementary function is negative — if yes, y_c → 0 and the time path converges to equilibrium (y* = b/a); if positive, the path diverges and is unstable. The equilibrium or steady-state value is always y = b/a* for the non-homogenous case with a ≠ 0. Be careful with sign of a — always rewrite the equation into dy/dt + a y = b form first.

🧠 Quick Revision Questions

1. An oil well with initial value $5000 decays according to dV/dt + 0.05V = 20. What is the equilibrium value V*, and is the time path dynamically stable or unstable?

2. For Hotelling's Rule, the equation is dπ/dt = rπ. What is the value of a and b when written in standard form dπ/dt + aπ = b?

3. A bank deposit grows according to dD/dt - 0.03D = 10 with D(0) = 2000. What are a and b in standard form, and what is the definite solution D(t)?

4. Monthly sales follow dS/dt = 0.1(500 - S) with S(0) = 100. How long will it take for sales to reach 90% of the saturation level?

5. A population grows as dN/dt = 0.02N with N(0) = 2 million. What is the doubling time? Is this system dynamically stable or unstable?

📘 Lecture 24 — VARIOUS ECONOMIC DYNAMICS USING DIFFERENTIAL EQUATIONS (CONTINUED)

📖 Overview: This lecture applies first-order linear differential equations to model the dynamic behavior of various economic and natural phenomena. It demonstrates how to solve, interpret, and assess the dynamic stability of time-paths for government spending, fuel consumption, unemployment, sales growth, ore mining, pollutant decay, and fish stock populations.

🗂️ Topics Covered

The lecture covers seven distinct applications: dynamics of government spending with a homogeneous differential equation, dynamics of fuel consumption with a non-homogeneous equation, dynamics of unemployment, limited growth models for sales volume, dynamics of ore mining with exponential growth, dynamics of atmospheric pollutant decay, and dynamics of fish stock growth toward a carrying capacity. Each application involves solving a differential equation, finding the definite time-path, and analyzing dynamic stability.

📝 Lecture Summary

TOPIC 100: DYNAMICS OF GOVERNMENT SPENDING USING DIFFERENTIAL EQUATIONS

Government expenditure, (G), satisfies the differential equation: (\frac{dG}{dt} = -0.05G), where (G) is measured in billions of dollars and time (t) is in years. The initial condition is (G(0) = 500).

This is a homogeneous case because (b = 0) and (a \neq 0). The standard form is (\frac{dy}{dt} + a y = b). Here, (\frac{dG}{dt} + 0.05G = 0), so (a = 0.05) and (b = 0).

🔑 Definition — Definite Solution of time-path of HC: (y(t) = {y(0)} e^{-a t})

Substituting values, the definite time-path for government expenditure is: [G(t) = (500)e^{-0.05t}]

Dynamic Stability: For dynamic stability, the deviation component ((G_c)) must fade to zero as (t \to \infty). Since the particular solution (G_p = 0), convergence should be toward zero. The expression (G(t) = (500)e^{-0.05t}) has a negative exponent, so it approaches zero as (t \to \infty). Therefore, the time-path of government expenditure converges toward equilibrium and is dynamically stable.

📐 Formula: (G(t) = (500)e^{-0.05t}) → Government spending decays exponentially from $500 billion. 📌 Example: To find government spending after 6 years: (G(6) = (500)e^{-0.05(6)} = (500)e^{-0.3} = $370.409) billion.

TOPIC 101: DYNAMICS OF FUEL CONSUMPTION USING DIFFERENTIAL EQUATIONS

Let (F(x)) denote the number of liters of fuel left in an aircraft’s fuel tank after flying (x) km. The fuel consumption satisfies: (F'(x) = -0.001F(x) - 8), with initial fuel (F(0) = 12000).

This is a non-homogeneous case because (b \neq 0) and (a \neq 0). Rewriting: (\frac{dF}{dx} + 0.001F = -8), so (a = 0.001), (b = -8).

🔑 Definition — Definite Solution of time-path of NHC with (a \neq 0): (y(t) = \left{ y(0) - \frac{b}{a} \right} e^{-a t} + \frac{b}{a})

Substituting values: [F(t) = \left{12000 - \frac{(-8)}{0.001}\right} e^{-(0.001)t} + \frac{(-8)}{0.001}] [F(t) = {12000 + 8000} e^{-(0.001)t} - 8000] [F(t) = 20000 e^{-0.001t} - 8000]

Dynamic Stability: The complementary function (deviation) is (F_c = 20000 e^{-0.001t}). As (t \to \infty), (F_c \to 0), so the time-path converges to equilibrium (F^* = 8000). The time-path is dynamically stable.

💡 Why this matters: The equilibrium value of 8000 liters represents the fuel remaining when the plane is empty of passengers but still has fuel for the aircraft's own weight.

TOPIC 102: DYNAMICS OF UNEMPLOYMENT USING DIFFERENTIAL EQUATIONS

Unemployment (U) (in thousands) in a small economy satisfies: (\frac{dU}{dt} = -3U + 180), with initial condition (U(0) = 150).

This is a non-homogeneous case with (a \neq 0). Rewriting: (\frac{dU}{dt} + 3U = 180), so (a = 3), (b = 180).

Using the definite solution formula: [U(t) = \left{150 - \frac{180}{3}\right} e^{-3t} + \frac{180}{3}] [U(t) = (150 - 60)e^{-3t} + 60] [U(t) = 90e^{-3t} + 60]

Dynamic Stability: The deviation (U_c = 90e^{-3t}) decays rapidly to zero as (t \to \infty). The time-path converges to equilibrium (U^* = 60) (thousand). The time-path is dynamically stable.

📌 Example: At (t = 5) years, unemployment is approximately (60.0000275) thousand (essentially at equilibrium).

TOPIC 103: LIMITED GROWTH USING DIFFERENTIAL EQUATIONS

This model represents limited growth, where a variable grows rapidly initially but levels off just short of a maximum limit. The differential equation is of the type (\frac{dy}{dt} = r(A - y)), where (A) is the maximum limit.

For a new printer, the rate of sales (Q) satisfies: (\frac{dQ}{dt} = 0.04(700 - Q)), with (Q(0) = 0).

Rewriting: (\frac{dQ}{dt} + 0.04Q = 28), so (a = 0.04), (b = 28).

Using the definite solution formula: [Q(t) = \left{0 - \frac{28}{0.04}\right} e^{-0.04t} + \frac{28}{0.04}] [Q(t) = (0 - 700)e^{-0.04t} + 700] [Q(t) = -700e^{-0.04t} + 700]

Dynamic Stability: The deviation (Q_c = -700e^{-0.04t}) is negative and decays to zero. The time-path rises from 0 and converges to equilibrium (Q^* = 700) printers. The time-path is dynamically stable.

📌 Example: After 100 years, (Q(100) = -12.8209 + 700 = 687.179) printers sold, approaching the maximum of 700.

TOPIC 104: DYNAMICS OF ORE MINING USING DIFFERENTIAL EQUATIONS

The rate of ore mining satisfies: (\frac{dM}{dt} = 0.02M), with initial reserves (M(0) = 400) million tons in 1989.

This is a homogeneous case. Rewriting: (\frac{dM}{dt} - 0.02M = 0), so (a = -0.02), (b = 0).

Using the definite solution for HC: [M(t) = {M(0)} e^{-a t}] [M(t) = (400)e^{-(-0.02)t}] [M(t) = (400)e^{0.02t}]

Dynamic Stability: The particular solution (M_p = 0), but here the exponent is positive. The deviation (M_c = 400e^{0.02t}) grows to infinity as (t \to \infty). Therefore, the time-path is dynamically unstable — ore reserves diverge away from equilibrium.

💡 Why this matters: A positive exponent indicates exponential growth of mining, which is unsustainable and will deplete finite reserves.

TOPIC 105: DYNAMICS OF POLLUTANT USING DIFFERENTIAL EQUATIONS

The amount of sulphur dioxide (P) (in tons) satisfies: (\frac{dP}{dt} = -0.2P), with initial release (P(0) = 50) tons.

This is a homogeneous case. Rewriting: (\frac{dP}{dt} + 0.2P = 0), so (a = 0.2), (b = 0).

Using the definite solution: [P(t) = (50)e^{-0.2t}]

Dynamic Stability: With a negative exponent, the deviation decays to zero. The time-path converges toward equilibrium (0 tons). The time-path is dynamically stable.

📌 Example: After 10 hours, (P(10) = 50e^{-2} = 6.767) tons of sulphur dioxide remaining.

TOPIC 106: DYNAMICS OF FISH STOCK USING DIFFERENTIAL EQUATIONS

The fish stock (F) (in thousands) in an inland lake satisfies: (\frac{dF}{dt} = 0.12(80 - F)), with initial condition (F(0) = 0).

Rewriting: (\frac{dF}{dt} + 0.12F = 9.6), so (a = 0.12), (b = 9.6).

Using the definite solution formula: [F(t) = \left{0 - \frac{9.6}{0.12}\right} e^{-0.12t} + \frac{9.6}{0.12}] [F(t) = (0 - 80)e^{-0.12t} + 80] [F(t) = -80e^{-0.12t} + 80]

Dynamic Stability: The deviation (F_c = -80e^{-0.12t}) decays to zero. The time-path rises from 0 and converges to equilibrium (F^* = 80) (thousand fish). The lake has a maximum carrying capacity of 80,000 fish.

📌 Example: To find years to reach 40,000 fish (i.e., (F = 40)): [40 = -80e^{-0.12t} + 80] [80e^{-0.12t} = 40] [e^{-0.12t} = 0.5] [t = \frac{\ln(0.5)}{-0.12} = 5.78 \text{ years}]

⭐ Key Takeaways

The critical exam concepts are: (1) Identifying whether a differential equation is homogeneous ((b=0)) or non-homogeneous ((b \neq 0)) determines which solution formula to use. (2) Dynamic stability depends entirely on the sign of the exponent in the complementary function — a negative exponent ((-a < 0)) means the deviation decays to zero and the time-path is stable, while a positive exponent means instability. (3) For non-homogeneous cases, the equilibrium value is given by (\frac{b}{a}). (4) Limited growth models ((dy/dt = r(A - y))) always have a stable equilibrium at (y^* = A), and the solution follows an exponential approach from below. (5) The specific applications (government spending, unemployment, fish stock, etc.) differ only in their numerical parameters, but the mathematical solution method is identical for equations with the same structure.

🧠 Quick Revision Questions

1. A differential equation of the form (dG/dt = -0.05G) is classified as which case, and what is the general solution formula?
2. For the fuel consumption model (F'(x) = -0.001F(x) - 8), what is the equilibrium value of remaining fuel, and how do you know the time-path is dynamically stable?
3. In the unemployment model (dU/dt = -3U + 180) with (U(0)=150), what is the value of the deviation component after 1 year?
4. What condition determines whether a first-order linear differential equation produces dynamic stability or instability in the homogeneous case?
5. For the fish stock model (dF/dt = 0.12(80-F)), what is the maximum carrying capacity, and how many years does it take to reach 40 thousand fish?

📘 Lecture 25 — EXACT DIFFERENTIAL EQUATIONS

📖 Overview: This lecture covers the solution of first-order differential equations with variable coefficients and variable terms, distinguishing between homogeneous and non-homogeneous cases. It then introduces exact differential equations as a method for solving nonlinear differential equations, including the exactness test and the use of integrating factors to convert inexact equations into exact ones.

🗂️ Topics Covered

This lecture begins by revisiting the standard form of first-order differential equations and classifying them into homogeneous and non-homogeneous cases based on the variable term. It then derives the general solution for the homogeneous case and presents the borrowed formula for the non-homogeneous case, followed by detailed numerical examples for both. The lecture transitions to nonlinear differential equations, introducing the concept of exact differential equations, developing the exactness test using Young's Theorem, and providing a 4-step solution procedure with a numerical example. Finally, it addresses inexact differential equations and explains both the trial-and-error and formula-based methods for finding integrating factors to convert them into exact equations.

📝 Lecture Summary

TOPIC 107: FIRST-ORDER DIFFERENTIAL EQUATIONS: VARIABLE TERM & VARIABLE COEFFICIENT: HOMOGENEOUS CASE

The standard form of a first-order differential equation with variable coefficients is given by dy/dt + u(t) · y = w(t) , where u(t) and w(t) are functions of time, not constants. This leads to two possible cases: Case I (Homogeneous) where u(t) ≠ 0, w(t) = 0, and Case II (Non-Homogeneous) where u(t) ≠ 0, w(t) ≠ 0.

For the homogeneous case, the equation becomes dy/dt + u(t)y = 0, which simplifies to dy/dt = -u(t)y. Separating variables and integrating with respect to t yields the general solution y(t) = Ae^(-∫u(t)dt) , where A is an arbitrary constant. This formula shows that the dependent variable y follows an exponential path determined by the integral of the variable coefficient u(t).

🔑 Definition — First-Order Differential Equation (Homogeneous): A differential equation of the form dy/dt + u(t)y = 0, where u(t) is a variable coefficient and the term w(t) equals zero.

📐 Formula: y(t) = Ae^(-∫u(t)dt) → The general solution where A is a constant and the exponent is the negative integral of the variable coefficient.

📌 Example: Find the general solution of dy/dt + 3t²y = 0. Here, u(t) = 3t² and w(t) = 0. Solving the exponent: ∫u(t)dt = ∫3t²dt = 3∫t²dt = 3(t³/3) + c = t³ + c. Substituting into the formula: y(t) = Ae^(-(t³ + c)) = A × e^(-t³) × e^(-c) = B × e^(-t³), where B = A × e^(-c) is another constant. Therefore, the general solution is y(t) = Be^(-t³) .

💡 Why this matters: The solution shows that when the variable coefficient is a power function of t, the time path of y is governed by an exponential function with a polynomial exponent.

TOPIC 108: FIRST-ORDER DIFFERENTIAL EQUATIONS: VARIABLE TERM & VARIABLE COEFFICIENT: NON-HOMOGENEOUS CASE

For the non-homogeneous case where u(t) ≠ 0 and w(t) ≠ 0, the standard form is dy/dt + u(t)y = w(t) . The general solution for this case is borrowed from the method of exact differential equations and is given by y(t) = e^(-∫u dt) {A + w ∫ e^(∫u dt) dt} . For simplicity, u and w are used to represent u(t) and w(t). A definite solution can be found by substituting initial conditions.

🔑 Definition — First-Order Differential Equation (Non-Homogeneous): A differential equation of the form dy/dt + u(t)y = w(t), where both u(t) and w(t) are non-zero variable functions.

📐 Formula: y(t) = e^(-∫u dt) {A + w ∫ e^(∫u dt) dt} → The general solution where A is an integration constant, and the solution involves an exponential factor multiplied by a term containing an integral.

📌 Example: Find the general solution of dy/dt + 2ty = t. Here, u(t) = 2t and w(t) = t. First, solve the exponent: ∫u(t)dt = ∫2t dt = t² + c₁. Substituting into the formula: y(t) = e^(-(t² + c₁)) {A + t ∫ e^(t² + c₁) dt} = e^(-t²)e^(-c₁) {A + e^(c₁) × t × ∫ e^(t²) dt}. Using the natural exponential function rule: ∫ e^(t²) dt = e^(t²)/(2t). Thus, y(t) = e^(-t²)e^(-c₁) {A + e^(c₁) × t × (e^(t²)/(2t)) + c₂}. Simplifying: y(t) = e^(-t²)e^(-c₁)A + e^(-c₁)e^(c₁) × (1/2) + e^(-t²)e^(-c₁)c₂ = Ae^(-c₁)e^(-t²) + (1/2) + ce^(-c₁)e^(-t²). Consolidating constants: y(t) = Be^(-t²) + 1/2, where B = (Ae^(-c₁) + c). The general solution is y(t) = Be^(-t²) + 1/2.

💡 Why this matters: The non-homogeneous case introduces a constant term in the solution, showing how the variable term w(t) affects the equilibrium value of the system.

TOPIC 109: NON-LINEAR DIFFERENTIAL EQUATION: EXACT DIFFERENTIAL EQUATIONS

Nonlinear differential equations can have the power of the dependent variable y greater than or other than 1. They can be solved using three methods: I- Exact Differential Equation Method, II- Separable Variables Method, and III- Bernoulli Equation Method. An exact differential equation can be either linear or nonlinear by itself.

Given a function of two variables F(y, t) = k, its total differential is d{F(y, t)} = d(k), which expands to (∂F/∂y)dy + (∂F/∂t)dt = 0. The function F(y, t) = k represents an Exact Differential Equation (EDE) as its total differential satisfies this equation. To develop an exactness test, we use Young's Theorem, which states that for a function F(y, t), the cross partial derivatives are equal: ∂/∂t(∂F/∂y) = ∂/∂y(∂F/∂t) . Substituting M = ∂F/∂y and N = ∂F/∂t, the exactness test becomes ∂M/∂t = ∂N/∂y. The exact differential equation form is (M)dy + (N)dt = 0.

🔑 Definition — Exact Differential Equation (EDE): A differential equation of the form (M)dy + (N)dt = 0 that satisfies the exactness test ∂M/∂t = ∂N/∂y, meaning it is the total differential of some function F(y, t) = k.

📐 Formula: ∂M/∂t = ∂N/∂y → The exactness test that confirms whether a differential equation is exact.

TOPIC 110: EXACT DIFFERENTIAL EQUATIONS: NUMERICAL

Given the differential equation 2yt dy + y² dt = 0, compare with standard form M dy + N dt = 0. Here, M = 2yt and N = y². Apply the exactness test: ∂M/∂t = ∂(2yt)/∂t = 2y, and ∂N/∂y = ∂(y²)/∂y = 2y. Since ∂M/∂t = ∂N/∂y (2y = 2y) , the equation is an Exact Differential Equation.

The 4-step procedure for solving an Exact Differential Equation is as follows: Step I: Write the preliminary function F(y, t) in terms of derivative and integral with respect to y: F(y, t) = ∫ M dy + ψ(t), where ψ(t) combines the integration constant and any terms containing t only. Substituting M = 2yt: F(y, t) = ∫ 2yt dy + ψ(t) = 2t ∫ y dy + ψ(t) = 2t(y²/2) + ψ(t) = y²t + ψ(t) . Step II: Differentiate F(y, t) with respect to t: ∂F/∂t = ∂(y²t + ψ(t))/∂t = y² + ψ'(t). Since ∂F/∂t = N, and N = y², we have y² = y² + ψ'(t), so ψ'(t) = 0. Step III: Integrate ψ'(t) to find ψ(t): ∫ ψ'(t) dt = ∫ 0 dt, so ψ(t) = k, where k is a constant. Step IV: Substitute the result from Step III into Step I: F(y, t) = y²t + k. Since d{F(y, t)} = 0 implies F(y, t) = c_F, we have c_F = y²t + k, or y²t = c_F - k = c'. This gives the implicit solution y²t = c'. Converting to explicit form: y² = c'/t, so y(t) = √(c'/t) = √c'/√t = c/t^(1/2) .

🔑 Definition — 4-Step Procedure for Exact DE: A systematic method for solving exact differential equations involving integration, differentiation, and substitution to find the function F(y, t).

📌 Example: Solve 2yt dy + y² dt = 0. M = 2yt, N = y². Step I: F(y,t) = ∫2yt dy + ψ(t) = y²t + ψ(t). Step II: ∂F/∂t = y² + ψ'(t) = N = y², so ψ'(t)=0. Step III: ψ(t)=k. Step IV: F(y,t) = y²t + k = c_F, so y²t = c'. Therefore, the general solution is y(t) = c/√t.

TOPIC 111: EXACT DIFFERENTIAL EQUATIONS: INTEGRATING FACTOR

If the exactness test fails, the equation is an Inexact Differential Equation: (M)dy + (N)dt = 0, where ∂M/∂t ≠ ∂N/∂y. To make it exact, we multiply the equation by an Integrating Factor (F_Int) . This changes the values of M and N while maintaining the equality. The integrating factor converts the left-hand side of the differential equation into an exact differential d{F(y, t)}.

Integrating factors can be found using two methods: I – Integrating Factor by Trial & Error Method: Experiment with different values (e.g., y, t, y², t², etc.) and select the one that makes the exactness test valid. II – Integrating Factor by Formula: Use two rules:

• Rule 1: If (1/N)(∂M/∂t - ∂N/∂y) = f(y) alone, then F_Int = e^(∫f(y)dy) .
• Rule 2: If (1/M)(∂N/∂y - ∂M/∂t) = g(t) alone, then F_Int = e^(∫g(t)dt) .

🔑 Definition — Inexact Differential Equation: A differential equation of the form (M)dy + (N)dt = 0 that does not satisfy the exactness test ∂M/∂t = ∂N/∂y.

🔑 Definition — Integrating Factor (F_Int): A multiplicative factor that converts an inexact differential equation into an exact differential equation, making it integrable.

📐 Formula (Rule 1): F_Int = e^(∫f(y)dy) → where f(y) = (1/N)(∂M/∂t - ∂N/∂y) 📐 Formula (Rule 2): F_Int = e^(∫g(t)dt) → where g(t) = (1/M)(∂N/∂y - ∂M/∂t)

📌 Example 1 (Trial & Error): Given 2t dy + y dt = 0, M = 2t, N = y. Exactness test: ∂M/∂t = 2, ∂N/∂y = 1, so 2 ≠ 1 (inexact). Multiply by y: (y)(2t dy + y dt) = 2yt dy + y² dt = 0. Now M = 2yt, N = y². Exactness test: ∂M/∂t = 2y, ∂N/∂y = 2y, so 2y = 2y (exact). Therefore, the integrating factor is F_Int = y.

📌 Example 2 (Formula): Given 5yt dy + (5y² + 8t) dt = 0, M = 5yt, N = 5y² + 8t. Exactness test: ∂M/∂t = 5y, ∂N/∂y = 10y, so 5y ≠ 10y (inexact). Apply Rule 1: (1/N)(∂M/∂t - ∂N/∂y) = (1/(5y²+8t))(5y - 10y) = -5y/(5y²+8t) ≠ f(y) alone. Apply Rule 2: (1/M)(∂N/∂y - ∂M/∂t) = (1/(5yt))(10y - 5y) = (5y)/(5yt) = 1/t = g(t) alone. Therefore, the integrating factor is F_Int = e^(∫(1/t)dt) = e^(ln|t|) = t.

💡 Why this matters: Once the integrating factor converts the inexact equation into an exact one, the 4-step procedure can be applied to find the solution, making a wide range of otherwise unsolvable differential equations integrable.

⭐ Key Takeaways

This lecture establishes that first-order differential equations with variable coefficients can be homogeneous (w(t)=0) or non-homogeneous (w(t)≠0), each with distinct solution formulas. The homogeneous case yields a solution of the form y(t) = Ae^(-∫u dt), while the non-homogeneous case requires y(t) = e^(-∫u dt){A + w∫e^(∫u dt) dt}. For nonlinear differential equations, the exact differential equation method is introduced, where the critical exactness test ∂M/∂t = ∂N/∂y determines if an equation is exact. The 4-step solution procedure systematically solves exact differential equations by integrating M, differentiating with respect to t, and substituting to find the implicit solution. Finally, when the exactness test fails, integrating factors found through trial-and-error or formula methods (Rule 1 for f(y) or Rule 2 for g(t)) can convert inexact equations into exact ones, after which the standard solution procedure applies.

🧠 Quick Revision Questions

1. What is the general solution formula for a first-order homogeneous differential equation with variable coefficients, and how is it derived?
2. For the non-homogeneous case dy/dt + 2ty = t, what is the general solution and how is the constant 1/2 obtained?
3. What is the exactness test for an exact differential equation, and which theorem justifies it?
4. Describe the 4-step procedure for solving an exact differential equation, using the example 2yt dy + y² dt = 0.
5. What are the two formula-based rules for finding an integrating factor, and when should each be applied?

📘 Lecture 26 — Economic Applications of Exact Differential Equations

📖 Overview: This lecture demonstrates how exact differential equations are applied to model key economic functions, including production functions, utility functions, and demand functions. It also introduces the separable variables method for solving non-linear differential equations and applies it to model capital dynamics over time.

🗂️ Topics Covered

The lecture covers four main applications: first, using exact differential equations to derive a Cobb-Douglas production function from given marginal products; second, using integrating factors to convert an inexact differential equation into an exact one for deriving a utility function; third, deriving a demand function from point elasticity using exact differential equations; and finally, introducing separable variables for solving non-linear differential equations, with an application to the dynamics of capital stock.

📝 Lecture Summary

TOPIC 112: PRODUCTION FUNCTION USING EXACT DIFFERENTIAL EQUATIONS

A firm uses L workers and K machines as inputs. The marginal products are given as MPₖ = 2(L/K)^(2/3) and MPₗ = 4(K/L)^(1/3). We must express the total product function assuming the firm stays on the same isoquant, and identify the type of production function.

The total differential of the production function Q(K,L) = k is dQ = 0, which gives (MPₖ)dK + (MPₗ)dL = 0. This yields the equation: {2(L/K)^(2/3)} dK + {4(K/L)^(1/3)} dL = 0.

🔑 Definition — Exact Differential Equation: A differential equation M dy + N dt = 0 is exact if ∂M/∂t = ∂N/∂y. Here, M = 2(L/K)^(2/3), N = 4(K/L)^(1/3), dy = dK, and dt = dL.

Exactness Test: ∂M/∂L = ∂/∂L [2(L/K)^(2/3)] = 2(1/K)^(2/3) * (2/3)L^(-1/3) = (4/3)(1/K)^(2/3)L^(-1/3) ∂N/∂K = ∂/∂K [4(K/L)^(1/3)] = 4(1/L)^(1/3) * (1/3)K^(-2/3) = (4/3)(1/L)^(1/3)K^(-2/3) Both equal (4/3)(1/K)^(2/3)(1/L)^(1/3), so the equation is exact.

Solution (4-Step Procedure): Step I: Write Q(K,L) = ∫ M dK + ψ(L). Q(K,L) = ∫ 2(L/K)^(2/3) dK + ψ(L) = 2L^(2/3) ∫ K^(-2/3) dK + ψ(L) = 2L^(2/3) * (K^(1/3))/(1/3) + ψ(L) = 6L^(2/3)K^(1/3) + ψ(L).

Step II: Differentiate w.r.t L and set equal to N. ∂Q/∂L = ∂/∂L [6L^(2/3)K^(1/3) + ψ(L)] = 6K^(1/3)*(2/3)L^(-1/3) + ψ'(L) = 4K^(1/3)L^(-1/3) + ψ'(L). Since ∂Q/∂L = N = 4(K/L)^(1/3) = 4K^(1/3)L^(-1/3), we get 4K^(1/3)L^(-1/3) + ψ'(L) = 4K^(1/3)L^(-1/3), so ψ'(L) = 0.

Step III: Integrate ψ'(L) to find ψ(L). ψ(L) = ∫ 0 dL = k (a constant).

Step IV: Substitute into Step I. Q(K,L) = 6L^(2/3)K^(1/3) + k. Since dQ = 0 implies Q is constant, we set Q(K,L) = c_Q, so c_Q = 6L^(2/3)K^(1/3) + k, which gives 6L^(2/3)K^(1/3) = c, where c = c_Q – k.

🔑 Definition — Cobb-Douglas Production Function: A production function of the form Q = A K^α L^β, where A is a constant and α, β are output elasticities.

The solution Q(K,L) = 6K^(1/3)L^(2/3) represents a Cobb-Douglas production function with constant returns to scale (sum of exponents = 1). The constant k must equal 0 to make economic sense, as output should be zero when either input is zero.

📐 Formula: Q(K,L) = 6 K^(1/3) L^(2/3) 📌 Example: If K=8 and L=27, then Q = 6 * (8)^(1/3) * (27)^(2/3) = 6 * 2 * 9 = 108 units of output.

💡 Why this matters: The solution demonstrates that exact differential equations can recover the underlying production function from its marginal products, confirming it is Cobb-Douglas.

TOPIC 113: UTILITY FUNCTION USING EXACT DIFFERENTIAL EQUATIONS

A consumer's utility function for two goods x and y is given by the differential equation 4xy dx + x² dy = 0. We must express the total utility function using integrating factors.

Comparing with standard form: M = 4xy, N = x², with dy = dx and dt = dy.

Exactness Test: ∂M/∂y = ∂/∂y(4xy) = 4x ∂N/∂x = ∂/∂x(x²) = 2x Since 4x ≠ 2x, the equation is inexact.

🔑 Definition — Integrating Factor: A function F_int that, when multiplied by an inexact differential equation, makes it exact.

Applying Rule 1: (1/N)(∂M/∂y – ∂N/∂x) = (1/x²)(4x – 2x) = (1/x²)(2x) = 2/x = f(x) (function of x only). The integrating factor is F_int = e^(∫ f(x) dx) = e^(∫ (2/x) dx) = e^(2 ln|x|) = x².

Multiplying the equation by x² gives: 4x³y dx + x⁴ dy = 0. Now M = 4x³y and N = x⁴. Exactness Test: ∂M/∂y = 4x³, ∂N/∂x = 4x³. The equation is now exact.

Solution (4-Step Procedure): Step I: U(x,y) = ∫ M dx + ψ(y) = ∫ 4x³y dx + ψ(y) = x⁴y + ψ(y).

Step II: ∂U/∂y = x⁴ + ψ'(y). Since ∂U/∂y = N = x⁴, we get ψ'(y) = 0.

Step III: ψ(y) = ∫ 0 dy = k.

Step IV: U(x,y) = x⁴y + k. Setting U = c_U gives x⁴y = c.

📐 Formula: U(x,y) = x⁴y – c (where c is absorbed constant) 📌 Example: If a consumer consumes x=2 units and y=5 units, utility is U = 2⁴ * 5 = 16 * 5 = 80 utils.

💡 Why this matters: This shows how integrating factors convert a non-exact utility differential equation into an exact one, allowing recovery of the underlying utility function.

TOPIC 114: POINT ELASTICITY FORMULA USING EXACT DIFFERENTIAL EQUATIONS

A firm faces point elasticity of demand E_d = -2. Use exact differential equations to find the demand function.

From the elasticity formula: (dQ_d/dP) * (P/Q_d) = -2, which rearranges to (1/Q_d) dQ_d + (2/P) dP = 0.

Comparing with standard form: M = 1/Q_d, N = 2/P, with dy = dQ_d and dt = dP.

Exactness Test: ∂M/∂P = ∂/∂P (1/Q_d) = 0 ∂N/∂Q_d = ∂/∂Q_d (2/P) = 0 The equation is exact.

Solution (4-Step Procedure): Step I: F(Q_d, P) = ∫ M dQ_d + ψ(P) = ∫ (1/Q_d) dQ_d + ψ(P) = ln|Q_d| + ψ(P).

Step II: ∂F/∂P = ψ'(P). Since ∂F/∂P = N = 2/P, we get ψ'(P) = 2/P.

Step III: ψ(P) = ∫ (2/P) dP = 2 ln|P| + c₁.

Step IV: F(Q_d, P) = ln|Q_d| + 2 ln|P| + c₁. Setting F = c_D gives ln|Q_d * P²| = c₂, so Q_d * P² = c, or Q_d = c/P².

📐 Formula: Q_d = c / P² 📌 Example: If the constant c = 100 and price P = 5, quantity demanded is Q_d = 100/25 = 4 units.

💡 Why this matters: This demonstrates how point elasticity information can be used with exact differential equations to derive the functional form of the demand curve.

TOPIC 115: NON-LINEAR DIFFERENTIAL EQUATION: SEPARABLE VARIABLES

🔑 Definition — Separable Variables: A differential equation f(y,t) dy + g(y,t) dt = 0 is separable if it can be written as f(y) dy + g(t) dt = 0, where y and t are separated.

Example #1: Solve 3y² dy – t dt = 0. Rearranged: 3y² dy = t dt. Integrating: ∫ 3y² dy = ∫ t dt → y³ + c₁ = t²/2 + c₂ → y³ = t²/2 + c' → y = (t²/2 + c')^(1/3).

Example #2: Solve 2t dy + y dt = 0. Rearranged: 2t dy = -y dt → dy/y = -(1/2)(dt/t). Integrating: ln|y| = -(1/2) ln|t| + c → ln|y| + ln|t^(1/2)| = c → ln|y * t^(1/2)| = c → y = c * t^(-1/2).

📐 Formula: For separable equations, integrate both sides: ∫ f(y) dy = -∫ g(t) dt 📌 Example: For 2t dy + y dt = 0 with initial condition y(1)=4, the solution is y = 4/√t.

TOPIC 116: DYNAMICS OF CAPITAL USING SEPARABLE VARIABLES METHOD

Given: X = √K √L (Cobb-Douglas production), dK/dt = 0.4X (investment proportional to output), L = e^(0.04t) (exponential labor growth). Derive a differential equation for K(t) and solve given K(0) = 10,000.

Substituting X and L into dK/dt: dK/dt = 0.4 * √K * √(e^(0.04t)) = 0.4 * √K * e^(0.02t)

Rearranging for separation of variables: K^(-1/2) dK = 0.4 e^(0.02t) dt

Integrating both sides: ∫ K^(-1/2) dK = ∫ 0.4 e^(0.02t) dt 2K^(1/2) = (0.4/0.02) e^(0.02t) + c = 20 e^(0.02t) + c √K(t) = 10 e^(0.02t) + c₂

Applying initial condition K(0) = 10,000: √10000 = 10 e⁰ + c₂ → 100 = 10 + c₂ → c₂ = 90

Solution: √K(t) = 10 e^(0.02t) + 90, so K(t) = [10(e^(0.02t) + 9)]² = 100(e^(0.02t) + 9)²

📐 Formula: K(t) = 100(e^(0.02t) + 9)² 📌 Example: At t=0, K(0) = 100(1+9)² = 100(100) = 10,000. At t=50, K(50) = 100(e¹ + 9)² = 100(2.718 + 9)² ≈ 100(11.718)² ≈ 13,732.

💡 Why this matters: This example shows how separable variables solve a non-linear differential equation modeling capital accumulation over time, incorporating production technology and labor growth.

⭐ Key Takeaways

The lecture demonstrates four critical economic applications of differential equations. First, exact differential equations can recover a Cobb-Douglas production function from its marginal products, with the 4-step procedure verifying exactness and yielding Q = 6K^(1/3)L^(2/3). Second, inexact differential equations require integrating factors (such as x² for utility functions) to become exact, allowing recovery of utility functions. Third, point elasticity information can directly derive demand functions in the form Q_d = c/P² using exact differential equations. Finally, separable variables provide a straightforward method for solving non-linear differential equations, powerfully applied to model the time-path of capital stock K(t) = 100(e^(0.02t) + 9)² under Cobb-Douglas production and exponential labor growth.

🧠 Quick Revision Questions

1. What are the two conditions that must be verified to determine if a differential equation M dy + N dt = 0 is exact, and how is the exactness test performed?
2. In Topic 112, why must the constant k be set to zero in the production function solution, and what economic paradox would arise otherwise?
3. How do you determine the integrating factor for an inexact differential equation, and what was the integrating factor used in Topic 113?
4. What is the key characteristic of a separable differential equation, and how does the solution method differ from the 4-step procedure used for exact equations?
5. In the capital dynamics problem, explain how the production function, investment equation, and labor growth function combine to form a single separable differential equation for capital.

📘 Lecture 27 — Economic Applications of Exact Differential Equations (Continued)

📖 Overview: This lecture continues exploring economic applications of differential equations, focusing on nonlinear Bernoulli equations and their solutions. It demonstrates how Bernoulli equations can be linearized and solved, applies this technique to find average cost functions, introduces qualitative analysis through phase diagrams, and applies both qualitative and quantitative methods to the Solow Growth Model.

🗂️ Topics Covered

The lecture covers six main topics: understanding Bernoulli's nonlinear differential equation structure and its reduction to linear form; solving a Bernoulli equation numerically with a specific example; finding an average cost function using Bernoulli equation techniques; introducing phase diagrams for qualitative analysis of autonomous differential equations; conducting qualitative analysis of the Solow Growth Model; and performing quantitative analysis of the Solow Growth Model using a Cobb-Douglas production function to find the steady-state equilibrium.

📝 Lecture Summary

TOPIC 117: NON-LINEAR DIFFERENTIAL EQUATION: UNDERSTANDING BERNOULLI’S EQUATION

A Bernoulli Differential Equation is a specific nonlinear differential equation attributed to Jacob Bernoulli. Its standard form is: dy/dt + R y = T y^m, where R and T are functions of t, and m is a power not equal to 0 or 1. When m=0, the equation becomes linear (dy/dt + R y = T). When m=1, it becomes dy/dt + (R - T)y = 0, also linear. The nonlinearity arises from the y^m term.

The solution method involves reducing the Bernoulli equation to a linear differential equation. By dividing through by y^m, we get: y^(-m) dy/dt + R y^(1-m) = T. Through substitutions: Substitution-I sets z = y^(1-m), replacing the nonlinear y term with a linear z term. Substitution-II applies the chain rule since z depends on y which depends on t: dz/dt = (dz/dy) × (dy/dt). Differentiating z = y^(1-m) with respect to y gives dz/dy = (1 - m) × y^(-m). Substituting into the chain rule yields: (1/(1-m)) × dz/dt = y^(-m) × dy/dt.

🔑 Definition — Bernoulli Differential Equation: A nonlinear differential equation of the form dy/dt + R(t) y = T(t) y^m, where m ≠ 0 and m ≠ 1. 📐 Formula: dz/dt + {(1-m) × (R z - T)} dt = 0 → This is the linearized form after substitution, where z = y^(1-m).

TOPIC 118: NON-LINEAR DIFFERENTIAL EQUATION: BERNOULLI’S EQUATION NUMERICALLY

Solve the equation dy/dt + t y = 3t y^2. Comparing with standard form: R = t, T = 3t, m = 2. Using substitution: z = y^(1-2) = y^(-1). The linearized form is dz + {(1-m) × (R z - T)} dt = 0. Substituting values: dz + {(1-2) × (t z - 3t)} dt = 0, which simplifies to dz + (-t z + 3t) dt = 0. Converting to standard FODE form: dz/dt + (-t) z = -3t.

Using the general solution for FODE with variable coefficients: z(t) = e^(-∫ u dt) (A + ∫ e^(∫ u dt) w dt). Here, u(t) = -t and w(t) = -3t. ∫ u dt = -t²/2. Then z(t) = e^(t²/2) (A + 3 ∫ (-t) e^(-t²/2) dt). Noting that f(x) = -t²/2 and f'(x) = -t, the integral becomes ∫ f'(x) e^(f(x)) dx = e^(f(x)). Thus, z(t) = e^(t²/2) (A + 3 e^(-t²/2)) = A e^(t²/2) + 3.

Since z = y^(-1), we restore y: y^(-1) = A e^(t²/2) + 3, so y(t) = 1/(A e^(t²/2) + 3).

📌 Example: Given dy/dt + t y = 3t y^2, with R=t, T=3t, m=2, the solution is y(t) = 1/(A e^(t²/2) + 3).

TOPIC 119: AVERAGE COST FUNCTION USING EXACT DIFFERENTIAL EQUATIONS

Given: d(AC)/dQ + (1/Q) AC = (AC)³. This is a Bernoulli equation where y = AC, t = Q, R = 1/Q, T = 1, m = 3. Using substitution: z = (AC)^(1-3) = (AC)^(-2). The linearized form is dz + {(1-m) × (R z - T)} dQ = 0. Substituting: dz + [(1-3) × {(1/Q)z - 1}] dQ = 0, simplifying to dz - 2(1/Q)z dQ - 2 dQ = 0. Converting to standard FODE form: dz/dQ + (-2/Q)z = 2, where u(Q) = -2/Q and w(Q) = 2.

∫ u(Q) dQ = ∫ (-2/Q) dQ = -2 ln|Q|. Using the general solution: z(Q) = e^(-(-2 ln|Q|)) (A + ∫ e^(-2 ln|Q|) × 2 dQ) = e^(2 ln|Q|) (A + 2 ∫ e^(-2 ln|Q|) dQ) = Q² (A + 2 ∫ Q⁻² dQ) = Q²(A + 2 × (-Q⁻¹)) = A Q² - 2Q.

Restoring AC: (AC)^(-2) = A Q² - 2Q, so AC(Q) = 1/(A Q² - 2Q)^(1/2). This solution shows a negative relationship between average cost and output (AC declines as Q increases), suggesting that AC is predominantly fixed cost, characteristic of natural monopoly.

📐 Formula: AC(Q) = 1/(A Q² - 2Q)^(1/2) → The general solution showing declining average cost with increasing output.

💡 Why this matters: This mathematical result confirms the economic theory of natural monopoly, where high fixed costs lead to declining average costs over the range of output, a situation where a single firm can serve the entire market most efficiently.

TOPIC 120: QUALITATIVE ANALYSIS: PHASE DIAGRAM

A Phase Diagram is a qualitative tool for analyzing the dynamic properties of autonomous differential equations. An autonomous differential equation has the form dy/dt = f(y), where the derivative depends only on y, not on t. The phase diagram plots dy/dt against y on a single axis (the phase line), showing the direction of movement of y over time.

For dy/dt = y - 7: The x-intercept (where dy/dt = 0) occurs at y = 7. Before the intercept (y < 7), dy/dt < 0, so y decreases (moves leftwards). After the intercept (y > 7), dy/dt > 0, so y increases (moves rightwards). This shows repulsion from the equilibrium y=7, indicating divergence and dynamic instability (a Repellor).

For dy/dt = 4 - y/2: The x-intercept occurs at y = 8. Before the intercept (y < 8), dy/dt > 0, so y increases (moves rightwards). After the intercept (y > 8), dy/dt < 0, so y decreases (moves leftwards). This shows attraction towards the equilibrium y=8, indicating convergence and dynamic stability (an Attractor or Steady-State Equilibrium).

Verification using the FODE solution: For dy/dt + a y = b, convergence occurs if a > 0, divergence if a < 0. In the first example (dy/dt = y - 7), rewriting as dy/dt + (-1)y = -7, a = -1 < 0, confirming divergence. In the second example (dy/dt = 4 - y/2), rewriting as dy/dt + (1/2)y = 4, a = 1/2 > 0, confirming convergence.

🔑 Definition — Phase Diagram: A graphical representation showing the qualitative behavior of an autonomous differential equation by plotting dy/dt against y, indicating whether the system converges to or diverges from equilibrium. 🔑 Definition — Repellor: An equilibrium point where the system moves away from it (divergent, unstable). 🔑 Definition — Attractor: An equilibrium point where the system moves toward it (convergent, stable, steady-state).

TOPIC 121: SOLOW GROWTH MODEL: QUALITATIVE ANALYSIS

The Solow Growth Model (attributed to Robert M. Solow) analyzes the qualitative behavior of the capital-labor ratio k. The fundamental equation is: dk/dt = s φ(k) - λ k, where s is the MPS (marginal propensity to save), φ(k) is the production function, and λ represents depreciation of capital. The term s φ(k) represents investment, and λ k represents depreciation.

Three cases occur: (1) If dk/dt > 0, then s φ(k) > λ k, meaning investment exceeds depreciation, so k increases (moves rightwards towards k*). (2) If dk/dt = 0, then s φ(k) = λ k, meaning the system is at steady-state equilibrium k*. (3) If dk/dt < 0, then s φ(k) < λ k, meaning depreciation exceeds investment, so k decreases (moves leftwards towards k*). In all cases, the system converges to the steady-state equilibrium k*, showing dynamic stability.

The phase diagram for the Solow model shows that before the x-intercept (k*), the phase line is above the axis (dk/dt > 0), and after k*, the phase line is below the axis (dk/dt < 0). This indicates attraction towards k*, confirming that Solow's model predicts convergence to a stable, steady-state equilibrium.

📐 Formula: dk/dt = s φ(k) - λ k → Rate of change of capital-labor ratio equals investment minus capital depreciation.

TOPIC 122: SOLOW GROWTH MODEL: QUANTITATIVE ANALYSIS

For quantitative analysis, a linearly homogeneous Cobb-Douglas production function is assumed: Q = K^α L^(1-α). Expressing in per-capita terms: φ(k) = k^α. Substituting into the Solow equation: dk/dt = s k^α - λ k, or dk/dt + λ k = s k^α. This is a Bernoulli equation where y = k, R = λ, T = s, m = α (with α ≠ 1).

Using substitution: z = k^(1-α). The linearized form becomes dz/dt + (1-α)λ z = (1-α)s. This is comparable to the standard FODE form dy/dt + a y = b, where a = (1-α)λ and b = (1-α)s.

The definite solution is: k^(1-α) = [{k(0)}^(1-α) - s/λ] e^(-(1-α)λ t) + s/λ. Here, the term {k(0)}^(1-α) - s/λ is the arbitrary constant based on initial condition. The exponential expression e^(-(1-α)λ t) determines dynamic stability. As t → ∞, e^(-(1-α)λ t) → 0, so k^(1-α) → s/λ, which means k(t) → (s/λ)^(1/(1-α)).

The exponent (1-α)λ > 0 because 0 < α < 1 and λ > 0. This ensures the exponential decay term diminishes to zero over time, leading to convergence. The steady-state capital-labor ratio k* = (s/λ)^(1/(1-α)), which varies directly with the savings rate s and inversely with the depreciation rate λ.

📌 Example: With k(0)=100, s=0.2, α=0.7, λ=0.025, the steady-state k* = s/λ = 0.2/0.025 = 8. The exponential decay term causes k(t) to converge to 8 as t increases.

🔑 Definition — Steady-State Equilibrium (k)**: The long-run equilibrium value of the capital-labor ratio in the Solow Growth Model, given by k = (s/λ)^(1/(1-α)).

💡 Why this matters: The Solow model predicts that economies with different initial capital-labor ratios will converge to the same steady-state level, implying that poorer economies (with lower k) will grow faster than richer economies, converging to a common equilibrium determined by savings and depreciation rates.

⭐ Key Takeaways

This lecture establishes four critical concepts. First, Bernoulli differential equations (dy/dt + R y = T y^m) can be linearized through substitution z = y^(1-m), transforming nonlinear economic dynamics into solvable linear forms. Second, phase diagrams provide powerful qualitative analysis of autonomous differential equations by showing whether equilibria are attractors (convergent, stable) or repellors (divergent, unstable). Third, the Solow Growth Model's fundamental equation dk/dt = s φ(k) - λ k predicts convergence to a steady-state capital-labor ratio, demonstrated both qualitatively through phase diagrams and quantitatively through the solved Bernoulli equation. Fourth, the steady-state equilibrium k* = (s/λ)^(1/(1-α)) depends directly on savings and inversely on depreciation, showing the long-run determinants of capital accumulation.

🧠 Quick Revision Questions

1. What is the standard form of a Bernoulli differential equation, and what condition must m satisfy for the equation to be nonlinear?
2. Explain the two substitutions (Substitution-I and Substitution-II) used to linearize a Bernoulli equation.
3. In the average cost problem (d(AC)/dQ + (1/Q) AC = (AC)³), what does the solution AC(Q) = 1/(A Q² - 2Q)^(1/2) imply about the relationship between average cost and output?
4. Using a phase diagram, explain the difference between an attractor and a repellor equilibrium. How does the sign of dy/dt on either side of the equilibrium determine stability?
5. In the Solow Growth Model, what is the steady-state capital-labor ratio k* and how is it derived from the equation dk/dt = s k^α - λ k? What conditions ensure dynamic stability?

📘 Lecture 28 — SECOND ORDER DIFFERENTIAL EQUATIONS

📖 Overview: This lecture introduces higher-order differential equations, focusing on the second-order case (SODE). It explains how to find the particular integral (equilibrium) and complementary function (deviation from equilibrium) to form the general solution. The lecture covers the classification of SODE into homogeneous and non-homogeneous cases, their subcases, and the method for solving with distinct real roots.

🗂️ Topics Covered

The lecture begins by defining the general nth-order linear differential equation with constant coefficients. It then narrows down to the second-order differential equation (SODE) and its two main categories: homogeneous and non-homogeneous cases. Within the non-homogeneous case, three subcases are explored based on the coefficients a₁ and a₂, each with its own formula for the particular integral. Finally, the concept of the complementary function is derived using an exponential trial solution, leading to the characteristic equation and its roots. The case of distinct real roots is solved with a numerical example.

📝 Lecture Summary

Topic 123: Higher (Second) Order Differential Equation

This section introduces the concept of rate of change of the rate of change in economic variables, which leads to second-order differential equations. It generalizes this to the nth-order linear differential equation.

🔑 Definition — Linear Differential Equation with nth Order: An equation of the form: dⁿy/dtⁿ + a₁ dⁿ⁻¹y/dtⁿ⁻¹ + a₂ dⁿ⁻²y/dtⁿ⁻² + ⋯ + aₙ₋₁ dy/dt + aₙ y = b where the a's are constant coefficients, b is a constant term, and the dependent variable is y. The order is n (due to the highest derivative), and the degree is linear/first degree (due to the highest power 1 of the nth derivative). This course only considers linear equations with constant coefficients and no cross-terms (products of y and its derivatives).

For n=2, this simplifies to the Standard Form of Second-Order Differential Equation (SODE): d²y/dt² + a₁ dy/dt + a₂ y = b or y″(t) + a₁ y′(t) + a₂ y(t) = b

Depending on the value of b, there are two main cases:

• Homogeneous Case (b = 0): The particular solution yₚ = 0.
• Non-Homogeneous Case (b ≠ 0): The particular solution yₚ ≠ 0.

Further sub-cases for each are identified based on the values of a₁ and a₂.

Topic 124: Second-Order Differential Equation: Particular Integral

This section defines the particular integral (yₚ) as any particular solution of the non-homogeneous equation that contains no arbitrary constants. It represents the equilibrium value of y(t). The time path is y(t) = y꜀ + yₚ, where y꜀ is the deviation from equilibrium.

The lecture derives formulas for yₚ for three non-homogeneous subcases:

• Subcase I: If a₂ ≠ 0, assume a constant solution y = k.

  • y = k → y′(t) = 0 → y″(t) = 0
  • Substituting into y″(t) + a₁y′(t) + a₂y(t) = b gives a₂k = b → k = b/a₂.
  • 📐 Formula: yₚ = b/a₂
  • 💡 Why this matters: This gives a constant equilibrium value.

• Subcase II: If a₂ = 0 and a₁ ≠ 0, the formula from Subcase I fails (divide by zero). Assume y = kt.

  • y = kt → y′(t) = k → y″(t) = 0
  • Substituting into y″(t) + a₁y′(t) + 0·y(t) = b gives a₁k = b → k = b/a₁.
  • 📐 Formula: yₚ = (b/a₁) t
  • 💡 Why this matters: This gives a non-constant equilibrium value that changes over time.

• Subcase III: If a₂ = 0 and a₁ = 0, both previous formulas fail. Assume y = kt².

  • y = kt² → y′(t) = 2kt → y″(t) = 2k
  • Substituting into y″(t) + 0·y′(t) + 0·y(t) = b gives 2k = b → k = b/2.
  • 📐 Formula: yₚ = (b/2) t²
  • 💡 Why this matters: This gives a quadratic equilibrium value over time.

Topic 125: Second-Order Differential Equation: Particular Integral Numerical

This section provides numerical examples for each non-homogeneous subcase.

Non-Homogeneous Subcase I: y″(t) + y′(t) - 2y = -10

• Here, a₁ = 1, a₂ = -2, b = -10. Since a₂ ≠ 0, Subcase I applies.
• 📐 Formula: yₚ = b/a₂
• 📌 Example: yₚ = (-10)/(-2) = 5

Non-Homogeneous Subcase II: y″(t) + y′(t) = -10

• Here, a₁ = 1, a₂ = 0, b = -10. Since a₂ = 0 and a₁ ≠ 0, Subcase II applies.
• 📐 Formula: yₚ = (b/a₁) t
• 📌 Example: yₚ = (-10/1) t = -10t

Non-Homogeneous Subcase III: y″(t) = -10

• Here, a₁ = 0, a₂ = 0, b = -10. Since both a₂ = 0 and a₁ = 0, Subcase III applies.
• 📐 Formula: yₚ = (b/2) t²
• 📌 Example: yₚ = (-10/2) t² = -5t²

The section concludes by noting that in the homogeneous case (b = 0), all subcases yield a particular solution of yₚ = 0.

Topic 126: Second-Order Differential Equation: Standard Form & Complementary Function

This section defines the complementary function (y꜀) as the general solution of the reduced (homogeneous) equation y″(t) + a₁y′(t) + a₂y(t) = 0. It represents the deviation of the time path y(t) from the equilibrium value at every point in time. For dynamic stability, the y꜀ term should disappear over time (y꜀ → 0), so y(t)→ yₚ.

To derive y꜀, the lecture uses an exponential trial solution y(t) = Aeʳᵗ, as this is the only function whose derivatives are proportional to itself, allowing the homogeneous equation to be reduced to zero.

• For y(t) = Aeʳᵗ, we have:
  • y′(t) = rAeʳᵗ
  • y″(t) = r²Aeʳᵗ
• Substituting into y″(t) + a₁y′(t) + a₂y(t) = 0 gives: r²Aeʳᵗ + a₁rAeʳᵗ + a₂Aeʳᵗ = 0 Aeʳᵗ(r² + a₁r + a₂) = 0 Since Aeʳᵗ ≠ 0, the characteristic equation is: r² + a₁r + a₂ = 0

🔑 Definition — Characteristic Equation: The quadratic equation r² + a₁r + a₂ = 0 derived from the homogeneous SODE.

Using the quadratic formula, the roots (r₁, r₂) are: 📐 Formula: r = [-a₁ ± √(a₁² - 4a₂)] / 2

• r₁ = [-a₁ + √(a₁² - 4a₂)] / 2 and r₂ = [-a₁ - √(a₁² - 4a₂)] / 2

Two relationships to verify the roots:

• Relationship I: Sum of roots r₁ + r₂ = -a₁
• Relationship II: Product of roots r₁ × r₂ = a₂

Topic 127: Second-Order Differential Equation: Distinct Real Roots

This section examines the case where the discriminant (a₁² - 4a₂) > 0, leading to two distinct real roots (r₁ ≠ r₂).

The complementary function for this case is: 📐 Formula: y꜀ = A₁e^(r₁t) + A₂e^(r₂t) where A₁ and A₂ are arbitrary constants.

Numerical Example: Solve y″(t) + y′(t) - 2y = -10 with initial conditions y(0) = 12 and y′(0) = -2.

1. Identify coefficients: a₁ = 1, a₂ = -2, b = -10.
2. Find Particular Integral: Since a₂ ≠ 0, use yₚ = b/a₂ = (-10)/(-2) = 5.
3. Find Characteristic Roots: r = [-1 ± √(1² - 4(-2))] / 2 = [-1 ± √9] / 2.
   • r₁ = (-1 + 3)/2 = 1
   • r₂ = (-1 - 3)/2 = -2
4. Form Complementary Function: y꜀ = A₁e¹ᵗ + A₂e⁻²ᵗ.
5. General Solution: y(t) = y꜀ + yₚ = A₁eᵗ + A₂e⁻²ᵗ + 5.
6. Apply Initial Condition 1 (y(0) = 12): 12 = A₁e⁰ + A₂e⁰ + 5 → 12 = A₁ + A₂ + 5 → A₁ + A₂ = 7 (Equation 1)
7. Apply Initial Condition 2 (y′(0) = -2):
   • First find derivative: y′(t) = A₁eᵗ - 2A₂e⁻²ᵗ.
   • At t=0: -2 = A₁e⁰ - 2A₂e⁰ → -2 = A₁ - 2A₂ (Equation 2)
8. Solve for Constants:
   • From Equation 1: A₁ = 7 - A₂.
   • Substitute into Equation 2: -2 = (7 - A₂) - 2A₂ → -2 = 7 - 3A₂ → 3A₂ = 9 → A₂ = 3.
   • Then A₁ = 7 - 3 = 4.
9. Definite Solution: y(t) = 4eᵗ + 3e⁻²ᵗ + 5.

⭐ Key Takeaways

The most critical concept is that the solution to a second-order differential equation has two components: the particular integral (yₚ) representing equilibrium and the complementary function (y꜀) representing deviations from it. You must know how to classify a SODE as homogeneous or non-homogeneous, then further classify it based on its coefficients to apply the correct formula for yₚ. The complementary function is found by solving the characteristic equation (r² + a₁r + a₂ = 0) derived from an exponential trial solution. For the distinct real roots case, the complementary function is A₁e^(r₁t) + A₂e^(r₂t), and the arbitrary constants A₁ and A₂ are determined using initial conditions.

🧠 Quick Revision Questions

1. What is the general form of a second-order linear differential equation?
2. How do you determine which subcase (I, II, or III) to use for finding the particular integral?
3. What is the characteristic equation for a SODE, and how is it derived?
4. Describe the three possible natures of the roots of the characteristic equation based on the discriminant.
5. In the numerical example with y″(t) + y′(t) - 2y = -10, how were the constants A₁ and A₂ in the complementary function solved?

📘 Lecture 29 — SECOND ORDER DIFFERENTIAL EQUATIONS (CONTINUED 1)

📖 Overview: This lecture completes the analysis of second-order differential equations by examining the remaining two cases of characteristic roots: repeated real roots and complex roots. Understanding these cases is critical for modeling economic systems where equilibrium paths exhibit different dynamic behaviors, such as steady convergence or oscillatory adjustment.

🗂️ Topics Covered

This lecture covers three main topics: solving second-order differential equations with repeated real roots (Case II), solving second-order differential equations with complex roots (Case III), a summary table of all complementary function forms for the three root cases, and a detailed analysis of dynamic stability using the distinct real roots case as an example, demonstrating when solutions converge to or diverge from equilibrium.

📝 Lecture Summary

SECOND-ORDER DIFFERENTIAL EQUATION: REPEATED REAL ROOTS

For a second-order differential equation ( y''(t) + a_1 y'(t) + a_2 y = b ), the characteristic roots are found using: [ r = \frac{-a_1 \pm \sqrt{a_1^2 - 4a_2}}{2} ]

The discriminant ( (a_1^2 - 4a_2) ) determines the nature of the roots. When ( a_1^2 - 4a_2 = 0 ), we have repeated real roots. In this case, both roots are equal: ( r_1 = r_2 = r = -\frac{a_1}{2} ).

The complementary function for repeated real roots is: [ y_c = A_3 e^{rt} + A_4 t e^{rt} ] The second term ( A_4 t e^{rt} ) is needed because the two exponential terms would otherwise be identical, so we multiply by ( t ) to obtain a second independent solution.

🔑 Definition — Repeated Real Roots: Characteristic roots where the discriminant equals zero (( a_1^2 = 4a_2 )), resulting in two identical real roots ( r_1 = r_2 ).

📐 Formula: General solution is ( y(t) = y_c + y_p = A_3 e^{rt} + A_4 t e^{rt} + y_p )

📌 Example: Solve ( y''(t) + 6y'(t) + 9y = 27 ), given ( y(0) = 5 ) and ( y'(0) = -5 ).

Step 1: Compare to standard form. Here ( a_1 = 6, a_2 = 9, b = 27 ).

Step 2: Check discriminant. ( a_1^2 - 4a_2 = 36 - 36 = 0 ). This is Repeated Real Roots.

Step 3: Calculate root. ( r = -\frac{a_1}{2} = -\frac{6}{2} = -3 ).

Step 4: Complementary function. ( y_c = A_3 e^{-3t} + A_4 t e^{-3t} ).

Step 5: Find particular integral. Since ( b = 27 ) is constant, try ( y_p = k ). Substituting: ( 0 + 0 + 9k = 27 \Rightarrow k = 3 ). So ( y_p = 3 ).

Step 6: General solution. ( y(t) = A_3 e^{-3t} + A_4 t e^{-3t} + 3 ).

Step 7: Use initial condition ( y(0) = 5 ). ( y(0) = A_3(1) + 0 + 3 = 5 \Rightarrow A_3 = 2 ).

Step 8: Derivative. ( y'(t) = -3A_3 e^{-3t} + A_4 e^{-3t} - 3A_4 t e^{-3t} ). Substituting ( A_3 = 2 ): ( y'(t) = -6 e^{-3t} + A_4 e^{-3t} - 3A_4 t e^{-3t} ).

Step 9: Use ( y'(0) = -5 ). ( y'(0) = -6(1) + A_4(1) - 0 = -5 \Rightarrow -6 + A_4 = -5 \Rightarrow A_4 = 1 ).

Step 10: Definite solution. ( y(t) = 2e^{-3t} + t e^{-3t} + 3 ).

Interpretation: The definite solution ( y(t) = 2e^{-3t} + t e^{-3t} + 3 ) converges to equilibrium at ( y_p = 3 ) as ( t \to \infty ) because ( e^{-3t} ) decays exponentially, but the term ( t e^{-3t} ) initially increases before decaying. 💡 Why this matters: This "overshoot" pattern can model economic variables that initially move away from equilibrium before converging.

SECOND-ORDER DIFFERENTIAL EQUATION: COMPLEX ROOTS

When the discriminant ( a_1^2 - 4a_2 < 0 ), we have complex roots. The roots are conjugate complex numbers of the form ( r_1 = h + vi ) and ( r_2 = h - vi ), where ( i = \sqrt{-1} ).

The real part is ( h = -\frac{a_1}{2} ) and the imaginary part is ( v = \frac{\sqrt{4a_2 - a_1^2}}{2} ).

The complementary function for complex roots can be expressed using trigonometric functions via Euler's formula: [ y_c = e^{ht} (A_5 \cos vt + A_6 \sin vt) ]

🔑 Definition — Complex (Conjugate) Roots: Characteristic roots where the discriminant is negative (( a_1^2 < 4a_2 )), resulting in two roots of the form ( h \pm vi ) that are complex conjugates.

📐 Formula: Complementary function ( y_c = e^{ht} (A_5 \cos vt + A_6 \sin vt) ), where ( h = -\frac{a_1}{2} ) and ( v = \frac{\sqrt{4a_2 - a_1^2}}{2} ).

📌 Example: Solve ( y''(t) + 2y'(t) + 17y = 34 ), given ( y(0) = 3 ) and ( y'(0) = 11 ).

Step 1: Identify parameters. ( a_1 = 2, a_2 = 17, b = 34 ).

Step 2: Check discriminant. ( a_1^2 - 4a_2 = 4 - 68 = -64 < 0 ). This is Complex Roots.

Step 3: Compute ( h ) and ( v ). ( h = -\frac{2}{2} = -1 ). ( v = \frac{\sqrt{68 - 4}}{2} = \frac{\sqrt{64}}{2} = \frac{8}{2} = 4 ). So roots are ( r_1 = -1 + 4i ) and ( r_2 = -1 - 4i ).

Step 4: Complementary function. ( y_c = e^{-t} (A_5 \cos 4t + A_6 \sin 4t) ).

Step 5: Find particular integral. Try ( y_p = k ). Substituting: ( 0 + 0 + 17k = 34 \Rightarrow k = 2 ). So ( y_p = 2 ).

Step 6: General solution. ( y(t) = e^{-t} (A_5 \cos 4t + A_6 \sin 4t) + 2 ).

Step 7: Use ( y(0) = 3 ). ( y(0) = e^0 (A_5 \cos 0 + A_6 \sin 0) + 2 = A_5(1) + A_6(0) + 2 = 3 \Rightarrow A_5 = 1 ).

Step 8: Derivative. Using product rule: ( y'(t) = e^{-t}(-A_5 \cos 4t - A_6 \sin 4t) + e^{-t}(-4A_5 \sin 4t + 4A_6 \cos 4t) ).

Step 9: Use ( y'(0) = 11 ). With ( A_5 = 1 ): ( y'(0) = (1)(-1(1) - A_6(0)) + (1)(-4(1)(0) + 4A_6(1)) = -1 + 4A_6 = 11 \Rightarrow 4A_6 = 12 \Rightarrow A_6 = 3 ).

Step 10: Definite solution. ( y(t) = e^{-t} (\cos 4t + 3 \sin 4t) + 2 ).

Interpretation: The solution ( y(t) = e^{-t}(\cos 4t + 3 \sin 4t) + 2 ) exhibits damped oscillations — it oscillates (due to the sine and cosine terms) but the amplitude decays to zero (due to the ( e^{-t} ) term), converging to equilibrium ( y_p = 2 ). The Argand diagram graphically represents a complex number ( h + vi ) as a point in a 2D plane with coordinates ( (h, v) ) and distance from origin ( R = \sqrt{h^2 + v^2} ). 💡 Why this matters: Complex roots model economic variables that oscillate around equilibrium before settling, like exchange rates or business cycles.

SECOND-ORDER DIFFERENTIAL EQUATION: SUMMARY OF COMPLEMENTARY FUNCTION

The summary table for all three cases of the complementary function ( y_c ) for ( y'' + a_1 y' + a_2 y = b ) is:

SECOND-ORDER DIFFERENTIAL EQUATION: DYNAMIC STABILITY OF DISTINCT REAL ROOTS

Dynamic stability refers to whether the solution ( y(t) ) converges to the equilibrium (particular integral ( y_p )) as time ( t \to \infty ). The deviation from equilibrium is ( y_c = A_1 e^{r_1 t} + A_2 e^{r_2 t} ). Stability depends entirely on the sign of the characteristic roots.

• If both roots ( r_1 ) and ( r_2 ) are negative (( < 0 )), then both exponential terms ( e^{r_1 t} ) and ( e^{r_2 t} ) undergo exponential decay and approach zero as ( t \to \infty ). Thus, ( y_c \to 0 ) and ( y(t) \to y_p ). This is dynamically stable (convergence).
• If either root is positive (( > 0 )), that exponential term grows without bound as ( t \to \infty ), so ( y_c \to \infty ) and ( y(t) ) diverges from equilibrium. This is dynamically unstable (divergence).

📌 Example (Numerical Confirmation):

The numerical table in the lecture confirms this: At ( t = 0 ), all three cases start at ( y_c = 7 ). By ( t = 6 ), Case 1's deviation explodes to ( \approx 489,878 ), Case 2's to ( \approx 1614 ), but Case 3's deviation shrinks to ( \approx 0.01 ), confirming convergence. Hence, dynamic stability requires BOTH roots to be negative: ( (r_1 \text{ and } r_2) < 0 ).

⭐ Key Takeaways

The most critical concepts from this lecture are: (1) When the discriminant ( a_1^2 - 4a_2 = 0 ), the solution has repeated real roots and the complementary function includes a ( t e^{rt} ) term. (2) When the discriminant is negative, the solution has complex conjugate roots ( h \pm vi ), leading to oscillatory behavior captured by ( e^{ht}(A_5 \cos vt + A_6 \sin vt) ). (3) The particular integral ( y_p ) for a constant ( b ) is found by assuming a constant solution ( y = k ). (4) Dynamic stability for distinct real roots requires that both characteristic roots be negative; if either root is positive, the system diverges from equilibrium. (5) For repeated roots, stability requires the single root ( r < 0 ); for complex roots, stability requires the real part ( h < 0 ) (damped oscillations).

🧠 Quick Revision Questions

1. What condition on the discriminant ( a_1^2 - 4a_2 ) leads to repeated real roots, and what is the corresponding formula for the complementary function?
2. For the differential equation ( y'' + 2y' + 17y = 34 ), what are the values of ( h ) and ( v ) in the complementary function ( e^{ht}(A_5 \cos vt + A_6 \sin vt) )?
3. Explain why the term ( A_4 t e^{rt} ) is necessary in the complementary function for repeated real roots.
4. State the condition for dynamic stability in the case of distinct real roots. What happens to the deviation from equilibrium if one root is positive and the other negative?
5. In the example ( y(t) = 4e^{-t} + 3e^{-2t} + 5 ), what is the value of the deviation from equilibrium ( y_c ) as ( t \to 6 )? (Refer to the numerical table.)

📘 Lecture 30 — SECOND ORDER DIFFERENTIAL EQUATIONS (CONTINUED 2)

📖 Overview: This lecture completes the analysis of dynamic stability in second-order differential equations, examining the conditions under which repeated real roots and complex roots lead to convergence or divergence from equilibrium. It also introduces alternative forms for representing complex roots—Cartesian, Polar, and Exponential—and explains how to convert between them, which is essential for solving and interpreting dynamic economic models.

🗂️ Topics Covered

This lecture covers four main topics: dynamic stability analysis for the case of repeated real roots, where the role of a negative root in ensuring convergence is confirmed; dynamic stability analysis for complex roots, where the real part of the root determines convergence or divergence through a multiplier effect; the three alternative forms of representing complex roots (Cartesian, Polar, and Exponential) along with De Moivre's theorem for raising them to powers; and interconversion among these three forms using basic trigonometric relationships and the Pythagorean theorem.

📝 Lecture Summary

TOPIC 132: SECOND-ORDER DIFFERENTIAL EQUATION: DYNAMIC STABILITY OF REPEATED REAL ROOTS

The general solution for a second-order differential equation with repeated real roots is given by:
풚(풕) = 푨₃ 풆^(풓풕) + 푨₄ 풕 풆^(풓풕) + 풚ₚ

Here, 푨₃ and 푨₄ are constants determined by initial conditions, and 풕 is the independent variable spanning (0, ∞). The complementary function (풚𝒄 = 푨₃ 풆^(풓풕) + 푨₄ 풕 풆^(풓풕)) represents the deviation from equilibrium. Dynamic stability depends entirely on the root (풓).

If 풓 is negative (-ve), both exponential terms undergo exponential decay. Although the second term (푨₄ 풕 풆^(풓풕)) has an additional factor of 풕, this does not reverse the decaying dynamics. As 풕 → ∞, the exponential decay dominates the linear multiplier, and the sum of the two exponential terms fades to zero, meaning 풚𝒄 → 0 and the system converges to equilibrium.

🔑 Definition — Exponential Decay: A process where a quantity decreases at a rate proportional to its current value, causing it to approach zero as time increases.

📐 Formula: 풚(풕) = 푨₃ 풆^(풓풕) + 푨₄ 풕 풆^(풓풕) + 풚ₚ
→ The deviation from equilibrium (풚𝒄 = 푨₃ 풆^(풓풕) + 푨₄ 풕 풆^(풓풕)) will converge to zero if and only if the root 풓 is negative (풓 < 0).

📌 Example:
Two cases are compared numerically.
Case 1 (positive root): 풚(풕) = 2풆^(³ᵗ) + 풕풆^(³ᵗ) + 3
At t=0: 풚𝒄 = 2+0 = 2. At t=1: 2풆³ ≈ 40.2, 풕풆³ ≈ 20.1, sum ≈ 60.3. At t=5: 2풆¹⁵ ≈ 6,538,034.7, 풕풆¹⁵ ≈ 16,345,086.9, sum ≈ 22,883,121.6. Both exponential terms increase → deviation increases → dynamically unstable (divergence).

Case 2 (negative root): 풚(풕) = 2풆^(⁻³ᵗ) + 풕풆^(⁻³ᵗ) + 3
At t=0: 풚𝒄 = 2+0 = 2. At t=1: 2풆⁻³ ≈ 0.0996, 풕풆⁻³ ≈ 0.0498, sum ≈ 0.1494. At t=5: 2풆⁻¹⁵ ≈ 0.0000006, 풕풆⁻¹⁵ ≈ 0.000002, sum ≈ 0.000002. Both exponential terms decrease → deviation decreases → dynamically stable (convergence).

💡 Why this matters: This confirms numerically that for dynamic stability with repeated real roots, the root must be negative, 풓 < 0.

TOPIC 133: SECOND-ORDER DIFFERENTIAL EQUATION: DYNAMIC STABILITY OF COMPLEX ROOTS

The general solution for a complex roots case is:
풚(풕) = 풆^(풉풕) (푨₅ cos(풖풕) + 푨₆ sin(풖풕)) + 풚ₚ

The deviation from equilibrium is 풚𝒄 = 풆^(풉풕) (푨₅ cos(풖풕) + 푨₆ sin(풖풕)). The multiplicand (푨₅ cos(풖풕) + 푨₆ sin(풖풕)) produces a wave-like graph with uniform amplitude above and below equilibrium, meaning by itself it cannot cause convergence. The multiplier 풆^(풉풕) determines dynamic stability. If 풉 is negative, then 풆^(풉풕) → 0 as 풕 → ∞, mitigating the amplitude and causing damped fluctuation and convergence to equilibrium.

🔑 Definition — Damped Fluctuation: Oscillatory behavior around an equilibrium where the amplitude of oscillation decreases over time due to a negative exponential multiplier, leading to convergence.

📐 Formula: 풚(풕) = 풆^(풉풕) (푨₅ cos(풖풕) + 푨₆ sin(풖풕)) + 풚ₚ
→ For dynamic stability, the real part of the complex root (풉) must be negative (풉 < 0). This ensures the multiplier 풆^(풉풕) decays to zero.

📌 Example:
Three cases of 풚(풕) = 풆^(풉풕)(cos(4풕) + 3 sin(4풕)) + 2 are compared.
Case 1 (positive h, h=1): 풆^(¹·¹ᵗ) grows explosively. At t=0: 풚𝒄 = 1. At t=3: amplitude ~20.1. At t=9: amplitude ~8103.1. This produces explosive fluctuation and divergence → dynamically unstable.

Case 2 (zero h, h=0): 풆^(⁰·ᵗ) = 1, so 풚𝒄 = (cos(4풕) + 3 sin(4풕)). The amplitude remains uniform over time. At t=0: 1. At t=3: -0.8. At t=9: -3.1. This produces uniform fluctuation, neither diverging nor converging → unstable (not dynamically stable).

Case 3 (negative h, h=-1): 풆^(⁻¹·ᵗ) decays. At t=0: 풚𝒄 = 1. At t=3: ~0.0498 × (-0.8) ≈ -0.0381. At t=9: ~0.0001 × (-3.1) ≈ -0.0004. This produces damped fluctuation and convergence → dynamically stable.

💡 Why this matters: This numerically confirms that for dynamic stability with complex roots, the real part of the root (풉) must be negative: 풉 < 0.

TOPIC 134: ALTERNATIVE FORMS OF COMPLEX ROOTS

Complex roots can be expressed in three alternative forms. The Cartesian Form is (풉 ± 풖풊), based on Quadrant I of cartesian coordinates with 풉 on the x-axis and 풖 on the y-axis. The Polar Form (also called Triangular Form) is {푹(cos 휽 ± 풊 sin 휽)}, based on polar functions with angles (휽). The Exponential Form is (푹풆^(±풊휽)), based on the exponential function.

De Moivre's Theorem states that if a complex number has a power 풏 other than 1, all three forms can be rewritten:

• Cartesian Form with 풏 ≠ 1: Simply raise (풉 ± 풖풊) to the power 풏.
• Exponential Form with 풏 ≠ 1: (푹 풆^(±풊휽))^풏 = 푹ⁿ 풆^(±풊풏휽). Here 푹 is raised to 푹ⁿ and 풆^(±풊휽) is raised to 풆^(±풊(풏휽)).
• Polar Form with 풏 ≠ 1: Using the Exponential form observation, 푹 is raised to 푹ⁿ and 휽 becomes 풏휽: 푹ⁿ {cos(풏휽) ± 풊 sin(풏휽)}.

🔑 Definition — De Moivre's Theorem: A formula for raising complex numbers to any integer power 푛, such that (풉 ± 풊풖)^푛 = 푹^푛 [cos(풏휽) ± 풊 sin(풏휽)] = 푹^푛 풆^(±풊풏휽).

📐 Formula:
(풉 ± 풖풊)^풏 = 푹ⁿ풆^(±풊풏휽) = 푹ⁿ{cos(풏휽) ± 풊 sin(풏휽)}
→ This provides a universal rule for raising any form of a complex number to a power.

TOPIC 135: ALTERNATIVE FORMS OF COMPLEX ROOTS: INTERCONVERSION

Interconversion among the three forms uses basic trigonometric ratios and the Pythagorean Theorem. For a vector in the complex plane with base 풉, perpendicular 풖, and hypotenuse 푹:

• sin 휽 = 풖/푹, so 풖 = 푹 sin 휽 and 휽 = sin⁻¹(풖/푹)
• cos 휽 = 풉/푹, so 풉 = 푹 cos 휽 and 휽 = cos⁻¹(풉/푹)
• Pythagorean Theorem: 푹² = 풉² + 풖², so 푹 = √(풉² + 풖²)

📌 Example 6 (p-520): Find the Cartesian form of the Exponential form 5풆^(³·풊·흅/²).

Solution: From the standard form 푹풆^(±풊휽), we have 푹 = 5 and 휽 = ³흅/².
Using 풖 = 푹 sin 휽 and 풉 = 푹 cos 휽:
풉 = 5 cos(³흅/²) = 5 × (0) = 0
풖 = 5 sin(³흅/²) = 5 × (-1) = -5
Cartesian Form: 풉 ± 풖풊 = 0 ± (-5)풊 = -5풊
Polar Form: {5[cos(³흅/²) ± 풊 sin(³흅/²)]}

📌 Example 7 (p-520): Find Polar and Exponential forms of the Cartesian form (1 + √3 풊).

Solution: From the standard form (풉 ± 풖풊), we have 풉 = 1 and 풖 = √3.
Using 푹 = √(풉² + 풖²): 푹 = √(1² + (√3)²) = √(1 + 3) = 2
Using 휽 = sin⁻¹(풖/푹) = sin⁻¹(√3/2) = 60° (or 흅/3)
Polar Form: {2(cos 60° ± 풊 sin 60°)}
Exponential Form: 2풆^(±풊(흅/³))

💡 Why this matters: Interconversion among the three forms is essential for solving differential equations as it allows choosing the most convenient form for analysis—Cartesian for direct algebraic manipulation, Polar for visualizing magnitude and angle, and Exponential for handling powers and derivatives.

⭐ Key Takeaways

For a second-order differential equation, dynamic stability requires the repeated real root to be negative (풓 < 0) so the complementary function decays to zero despite the presence of a time multiplier. For complex roots, stability requires the real part of the root (풉) to be negative (풉 < 0), causing damped fluctuation toward equilibrium—a positive 풉 leads to explosive fluctuation and divergence. Complex roots can be represented in three interconvertible forms: Cartesian (풉 ± 풖풊), Polar {푹(cos 휽 ± 풊 sin 휽)}, and Exponential (푹풆^(±풊휽)), with De Moivre's theorem providing the formula for raising any of these forms to a power 풏: 푹ⁿ{cos(풏휽) ± 풊 sin(풏휽)}. Mastery of interconversion using trigonometric ratios and the Pythagorean theorem enables seamless shifting between forms for effective analysis and solution of dynamic economic models.

🧠 Quick Revision Questions

1. For the repeated real roots case, why does the term 푨₄ 풕 풆^(풓풕) not prevent convergence even though it has a time multiplier, when 풓 is negative?
2. In the complex roots case, what determines whether fluctuation is damped, uniform, or explosive, and what condition must be met for dynamic stability?
3. Write the three alternative forms (Cartesian, Polar, Exponential) for the complex number 풓 = −1 + √3 풊, using the conversions provided.
4. State De Moivre's theorem for raising a complex number to power 풏 and explain how it applies to each of the three forms.
5. If the exponential form of a complex root is 3풆^(⁻풊흅/2), what is its Cartesian form?

📘 Lecture 31 — A Market Model with Price Expectations

📖 Overview: This lecture extends the basic market model by incorporating price expectations into the demand function. It introduces a framework where current quantity demanded depends not only on current price but also on price trends (first and second derivatives of price over time). The lecture derives the general solution for the time path of price under market-clearing conditions and analyzes dynamic stability based on key parameters.

🗂️ Topics Covered

Lecture 31 covers the framework and rationale for a market model with price expectations, including how price trends are measured mathematically and how they augment the demand function. It then provides the general solution to the second-order differential equation that arises from the market-clearing condition. The lecture proceeds to examine the possible cases of complementary functions (distinct real, repeated real, and complex roots) based on the discriminant and analyzes the conditions for dynamic stability. Finally, two detailed numerical examples are solved to demonstrate the application of the model and the determination of the time path P(t).

📝 Lecture Summary

Topic 136: A Market Model with Price Expectations: Framework

In a simple market model, current quantity demanded (Qd) and current quantity supplied (Qs) depend only on current price (P). However, in real life, the trend of price changes can affect Qd because trends help form price expectations. These expectations influence current buying decisions.

The trend is understood from two aspects:

1. Rate of change of price: Is price increasing or decreasing over time? (dP/dt)
2. Slope of the rate of change: Is the rate of change itself increasing or decreasing over time? (d²P/dt²)

Mathematically, the trend is measured using first and second-order derivatives. The new demand and supply functions are expressed as:

Qd = f( P(t) , dP(t)/dt , d²P(t)/dt² ) Qs = f( P(t) , dP(t)/dt , d²P(t)/dt² )

This means current quantities depend on current price and the price-trend. The general equations for the price-expectations-augmented functions are:

Qd = a - bP + mP' + nP'' (where a, b > 0) Qs = -γ + δP + uP' + wP'' (where γ, δ > 0)

Here, a, b, γ, δ are conventional parameters, while m, n, u, w are parameters generated by terms of price expectations.

The lecture assumes price expectations exist in the demand function only and not in the supply function (so u = 0, w = 0). Effect of m and n on demand: If m > 0, Qd increases because buyers expect prices to rise and buy now. If n > 0, Qd increases even faster as the rate of price increase is accelerating.

Topic 137: A Market Model with Price Expectations: Solution

Assuming market-clearing at all points in time (Qd = Qs), the equation is: a - bP + mP' + nP'' = -γ + δP

Rearranging and normalizing by dividing by n gives the standard second-order linear differential equation form: P'' + a₁P' + a₂P = b

Where:

• a₁ = m/n
• a₂ = -(b+δ)/n
• b = -(a+γ)/n

The particular solution (Pp) is found using Case I (since a₂ ≠ 0): Pp = b / a₂ = (a+γ) / (b+δ)

Since all parameters are positive constants, Pp is also a constant value. This implies the model has a stationary equilibrium.

Topic 138: A Market Model with Price Expectations: Possible Cases

The nature of the complementary function (Pc) depends on the discriminant, leading to three cases:

1. Distinct Real Roots (a₁² > 4a₂): For this, the condition is (m/n)² > -4(b+δ)/n.

   • Pc = A₁e^(r₁t) + A₂e^(r₂t)
   • The general solution is P(t) = A₁e^(r₁t) + A₂e^(r₂t) + (a+γ)/(b+δ).

2. Repeated Real Roots (a₁² = 4a₂): For this, the condition is (m/n)² = -4(b+δ)/n.

   • Pc = A₃e^(rt) + A₄te^(rt), where r = -m/(2n).
   • The general solution is P(t) = A₃e^(-mt/(2n)) + A₄te^(-mt/(2n)) + (a+γ)/(b+δ).

3. Complex Roots (a₁² < 4a₂): For this, the condition is (m/n)² < -4(b+δ)/n.

   • Pc = e^(ht) (A₅ cos ut + A₆ sin ut), where h = -a₁/2 = -m/(2n) and u = (1/2)√(4a₂ - a₁²).
   • The general solution is P(t) = e^(-mt/(2n)) (A₅ cos ut + A₆ sin ut) + (a+γ)/(b+δ).

Parametric Interpretation for Dynamic Stability:

• The parameter n plays a pivotal role.
• If n > 0: Only Distinct Real Roots can exist (r₁ > 0, r₂ < 0), making the system dynamically unstable. It can only be stable if A₁ = 0.
• If n < 0: All three cases are possible. Dynamic stability is ensured if m < 0 (the price coefficient on the first derivative) in any of these cases.

💡 Why this matters: The sign of n (the coefficient for the second derivative of price in the demand function) is the primary determinant of dynamic stability in this model.

Topic 139: A Market Model with Price Expectations: Numerical Examples

Example 1 (p. 530) Given: Qd = 42 - 4P - 4P' + P'', Qs = -6 + 8P, with P(0)=6, P'(0)=4.

• Comparison with standard form: a=42, b=4, γ=6, δ=8, m=-4, n=1, u=0, w=0.
• Since n=1 > 0, only distinct real roots are possible, and dynamic instability is present unless A₁=0.
• Roots are found: r₁=6, r₂=-2. The particular solution Pp = (42+6)/(4+8) = 48/12 = 4.
• The general solution is P(t) = A₁e^(6t) + A₂e^(-2t) + 4.
• Using initial conditions to find A₁ and A₂:
  • P(0)=6 => 6 = A₁ + A₂ + 4 => A₁ + A₂ = 2.
  • P'(0)=4 => 4 = 6A₁ - 2A₂.
  • Solving: A₁=1, A₂=1.
• The definite solution is P(t) = e^(6t) + e^(-2t) + 4.
• Interpretation: Since n>0 and A₁≠0, the path is dynamically unstable due to the positive root r₁=6.

Example 2 (p. 531) Given: Qd = 40 - 2P - 2P' - P'', Qs = -5 + 3P, with P(0)=12, P'(0)=1.

• Comparison: a=40, b=2, γ=5, δ=3, m=-2, n=-1, u=0, w=0.
• Since n=-1 < 0 and m=-2 < 0, dynamic stability is present.
• Test for nature of roots: (m/n)² = 4, -4(b+δ)/n = 20. Since 4 < 20, complex roots exist.
• Roots are: h = -1, u = 2. Particular solution Pp = 9.
• General solution: P(t) = e^(-t)(A₅ cos 2t + A₆ sin 2t) + 9.
• Using initial conditions:
  • P(0)=12 => 12 = A₅ + 9 => A₅ = 3.
  • P'(0)=1 => 1 = -A₅ + 2A₆ => 1 = -3 + 2A₆ => A₆ = 2.
• The definite solution is P(t) = e^(-t)(3 cos 2t + 2 sin 2t) + 9.
• Interpretation: Since n<0 and m<0, dynamic stability is present. The time path is a dampened fluctuation.

⭐ Key Takeaways

The core of this lecture is understanding how price expectations, modeled through the first (dP/dt) and second (d²P/dt²) derivatives of price, modify the standard demand function. The resulting market model leads to a second-order linear differential equation for the time path of price. The dynamic stability of this model is critically determined by the signs of n and m: when n > 0, the system is generally unstable, while when n < 0 and m < 0, it is dynamically stable. The solution for the time path P(t) involves a particular solution representing the stationary equilibrium and a complementary function whose form (exponential or sinusoidal) depends on the nature of the characteristic roots. The two numerical examples illustrate how to derive the definite time path and interpret its dynamic properties based on parameter values.

🧠 Quick Revision Questions

1. How do price expectations change the standard demand function in this model?
2. What are the two aspects of a price trend considered in this framework?
3. Write the standard form of the second-order differential equation obtained from the market-clearing condition.
4. For dynamic stability, what must be true about the signs of parameters n and m?
5. In Example 1, why is the price time path dynamically unstable?

📘 Lecture 32 — PHILLIPS CURVE: THE PHILLIPS RELATION

📖 Overview: This lecture explores the Phillips Curve, a fundamental macroeconomic model analyzing the interaction between inflation and unemployment. It progresses from the original relation through the expectations-augmented version, culminating in a dynamic model that uses second-order differential equations to derive time paths for expected inflation, actual inflation, and unemployment, revealing the long-run neutrality of monetary policy.

🗂️ Topics Covered

The lecture begins with the original and adapted Phillips relations, linking wage growth and inflation to unemployment. It then introduces the expectations-augmented Phillips relation incorporating inflation expectations and adaptive expectations formation. A feedback mechanism from inflation to unemployment via monetary policy is established. The core of the lecture derives the time path of expected inflation using a second-order differential equation, followed by numeric examples and derivations of time paths for actual inflation and unemployment, culminating in the concept of the natural rate and the long-run Phillips curve.

📝 Lecture Summary

TOPIC 140: PHILLIPS CURVE: THE PHILLIPS RELATION

The Phillips relation models the negative empirical relationship between money wage growth (풘) and the unemployment rate (푼). The Original Phillips Relation is 풘 = 풇(푼) where 풇′(푼) < 0. An Adapted Phillips Relation substitutes the inflation rate (풑) for wage growth, based on the idea that higher wages increase production costs, leading to cost-push inflation: 풑 = 풘 − 푻, where 푻 is the growth rate of productivity. Assuming a linear form for the original relation (풘 = 풂 − 흰푼), the adapted relation becomes 풑 = 풂 − 흰푼 − 푻, meaning both higher productivity and higher unemployment negatively affect the inflation rate.

🔑 Definition — Original Phillips Relation: An inverse empirical relationship between the rate of growth of money wages and the rate of unemployment.

TOPIC 141: PHILLIPS CURVE: EXPECTATIONS AUGMENTED PHILLIPS RELATION

The expectations-augmented Phillips relation incorporates expected inflation (흅) into wage demands: 풘 = 풇(푼) + 품흅, where 0 < 품 ≤ 1. The adapted version becomes 풑 = 풂 − 푻 − 흰푼 + 품흅. Adaptive Expectations form the basis for how expectations evolve; here, the change in expected inflation is proportional to the discrepancy between actual and expected inflation: 풅흅/풅풕 = 풋(풑 − 흅), where 0 < 풋 ≤ 1. If actual inflation exceeds expected (풑 > 흅), expectations are revised upward; if 풑 = 흅, no revision occurs; if 풑 < 흅, downward revision is needed.

🔑 Definition — Adaptive Expectations Hypthesis: The hypothesis that forecasts of future values are made using only past values of the variable, leading to a gradual adjustment based on recent forecast errors.

TOPIC 142: PHILLIPS CURVE: FEEDBACK FROM INFLATION TO UNEMPLOYMENT

A complete two-equation system models the interaction between inflation and unemployment. The first equation is the expectations-augmented Phillips relation (풑 = 풂 − 푻 − 흰푼 + 품흅). The second is the adaptive expectations hypothesis (풅흅/풅풕 = 풋(풑 − 흅)). The direction from unemployment to inflation (푼 → 풑) is the supply-side effect. The feedback effect from inflation to unemployment (푼 ← 풑) operates through the demand side via government fiscal and monetary policy. Using a monetary policy channel, the rate of change of unemployment is inversely related to the growth rate of real money: 풅푼/풅풕 = −풌(풎 − 풑), where 풎 is the growth rate of nominal money and 풌 > 0. Here, (풎 − 풑) is the rate of growth of real money.

🔑 Definition — Feedback Effect (Monetary Policy): An inverse relationship between the rate of change of unemployment and the growth rate of real money, implying that higher real money growth reduces unemployment.

TOPIC 143: PHILLIPS CURVE: TIME PATH OF EXPECTED RATE OF INFLATION

The three-equation system (풑 = 풂 − 푻 − 흰푼 + 품흅, 풅흅/풅풕 = 풋(풑 − 흅), and 풅푼/풅풕 = −풌(풎 − 풑)) is reduced to a single second-order differential equation for the expected rate of inflation (흅). The particular solution is 흅_풑 = 풃/풂_ퟐ, which simplifies to 흅_풑 = 풎, meaning the equilibrium expected inflation equals the growth rate of nominal money. The nature of the roots for the complementary function depends on the determinant 풂_ퟏ^ퟐ − ퟒ풂_ퟐ. Since 풂_ퟏ (흰풌 − 풋(품 − 1)) and 풂_ퟐ (풋흰풌) are both positive, but the relative magnitudes and signs of the parameters mean all three root cases (distinct real, repeated real, complex) are possible. Crucially, in all cases, the real part of the roots (or the roots themselves) are negative, making the time path dynamically stable. This means 흅(풕) converges to 풎 over time.

💡 Why this matters: The key result is that the long-run equilibrium expected inflation rate is determined only by the growth rate of nominal money, not by unemployment or other real factors.

TOPIC 144: PHILLIPS CURVE: TIME PATH OF EXPECTED RATE OF INFLATION - NUMERICAL

Using specific parameter values (흰 = 3, 품 = 1, 풋 = 3/4, 풌 = 1/2), we find 풂_ퟏ = 3/2, 풂_ퟐ = 9/8, and 풃 = (9/8)풎. Since 풂_ퟏ^ퟐ (9/4) is less than ퟒ풂_ퟐ (9/2), the roots are complex. The roots are 풓_ퟏ, 풓_ퟐ = -3/4 ± (3/4)풊, giving a damping coefficient 풉 = -3/4 and a frequency 풖 = 3/4. The general solution for expected inflation is 흅(풕) = 풆^(-ퟑ풕/ퟒ) {푨_ퟓ 퐜퐨퐬(ퟑ풕/ퟒ) + 푨_ퟔ 퐬퐢퐧(ퟑ풕/ퟒ)} + 풎. This is a dynamically stable, fluctuating path converging to 풎.

📐 Formula: General solution for a second-order differential equation with complex roots 풉 ± 풖풊: 풚(풕) = 풆^(풉풕) (푪_ퟏ 퐜퐨퐬(풖풕) + 푪_ퟐ 퐬퐢퐧(풖풕)) + 풚_풑.

TOPIC 145: PHILLIPS CURVE: TIME PATH OF INFLATION - NUMERICAL

The time path of actual inflation 풑(풕) is derived from the adaptive expectations equation: 풑 = (풅흅/풅풕)(ퟏ/풋) + 흅. Using the derivative of 흅(풕) and substituting into this formula, the general solution for actual inflation is 풑(풕) = 풆^(-ퟑ풕/ퟒ) {푨_ퟔ 퐜퐨퐬(ퟑ풕/ퟒ) − 푨_ퟓ 퐬퐢퐧(ퟑ풕/ퟒ)} + 풎. Like expected inflation, this is a dynamically stable, fluctuating path that converges to the equilibrium value of 풎.

TOPIC 146: PHILLIPS CURVE: TIME PATH OF UNEMPLOYMENT – NUMERICAL

The time path of unemployment 푼(풕) is derived by rewriting the expectations-augmented Phillips relation for 푼 and substituting the solutions for 흅(풕) and 풑(풕). The general solution is 푼(풕) = (ퟏ/ퟑ)풆^(-ퟑ풕/ퟒ) [(푨_ퟓ − 푨_ퟔ)퐜퐨퐬(ퟑ풕/ퟒ) + (푨_ퟔ − 푨_ퟓ)퐬퐢퐧(ퟑ풕/ퟒ)] + ퟏ/ퟏퟖ. This is a dynamically stable, fluctuating path that converges to the equilibrium value of ퟏ/ퟏퟖ, regardless of the rate of money growth (풎).

🔑 Definition — Natural Rate of Unemployment (푼_푵푹): The equilibrium level of unemployment to which the economy tends in the long run, independent of the rate of inflation. In this example, it is ퟏ/ퟏퟖ. 💡 Why this matters: This demonstrates the long-run vertical Phillips curve; in the long run, there is no trade-off between inflation and unemployment. Inflation is determined by money growth, while unemployment returns to its natural rate.

⭐ Key Takeaways

The lecture evolves the Phillips Curve from a simple negative relationship between wage growth and unemployment to a dynamic model including inflation expectations and feedback effects. The crucial insight is that while expectations-augmented Phillips curve allows a short-run trade-off, the long-run equilibrium is characterized by the natural rate of unemployment, which is independent of the inflation rate. The time paths of all variables converge to their equilibrium values, demonstrating dynamic stability. The numerical example confirms that the equilibrium expected and actual inflation rates are equal to the rate of money growth (풎), while equilibrium unemployment is a constant natural rate (ퟏ/ퟏퟖ). The model's solution uses second-order differential equations, showing that macroeconomic dynamics can exhibit cyclical convergence.

🧠 Quick Revision Questions

1. What is the key difference between the Original and the Expectations-Augmented Phillips Relation in terms of what determines wage growth?
2. In the adaptive expectations framework, what condition causes expected inflation to be revised upward?
3. How does the feedback effect from inflation to unemployment operate through monetary policy? Formulate the relationship.
4. In the model, what does the equilibrium value of expected inflation (흅_풑) equal, and what does it imply about the long-run neutrality of money?
5. What is the natural rate of unemployment, and how does the long-run Phillips curve relate to it?

📘 Lecture 33 — Higher Order Differential Equations

📖 Overview: This lecture extends the study of second-order differential equations to higher-order ones, focusing on finding particular integrals (푦_푝) when variable terms are present. It introduces the Method of Undetermined Coefficients and its modifications, then generalizes to nth-order equations and the complementary function, culminating in Routh's Theorem for assessing dynamic stability without calculating characteristic roots.

🗂️ Topics Covered

The lecture covers five main topics: differential equations with variable terms, a modification to the method when parts of the LHS are missing, finding particular integrals for higher-order equations, determining complementary functions for higher-order equations, and using Routh's Theorem to check convergence and stability of the time path without solving for roots.

📝 Lecture Summary

Topic 147: Differential Equation with Variable Term

This section introduces second-order differential equations with constant coefficients but a variable term f(t) instead of a constant b. The solution requires the Method of Undetermined Coefficients. This method works only if the variable term f(t) and its successive derivatives contain a finite and distinct set of expression types.

🔑 Definition — Method of Undetermined Coefficients: A technique for finding the particular integral (푦_푝) of a differential equation by assuming a trial solution with unknown coefficients, then solving for those coefficients by equating terms of the same type on both sides of the equation.

📐 Key Concept: For a variable term f(t), the particular integral (푦_푝) changes, but the complementary function (푦_푐) remains unchanged, as it depends only on the roots of the characteristic equation r² + a₁r + a₂ = 0.

📌 Example: Find the Particular Integral of: y''(t) + 5y'(t) + 3y(t) = 6t² − t − 1. Here, a₁=5, a₂=3, and f(t)=6t²−t−1. Since the RHS is quadratic in t, we assume a trial general quadratic function: y(t) = B₁t² + B₂t + B₃. Then y'(t) = 2B₁t + B₂, and y''(t) = 2B₁. Substituting into LHS: 2B₁ + 5(2B₁t + B₂) + 3(B₁t² + B₂t + B₃) = 6t² − t − 1. Simplifying: (3B₁)t² + (10B₁ + 3B₂)t + (2B₁ + 5B₂ + 3B₃) = 6t² + (−1)t + (−1). Comparing coefficients: 3B₁=6 gives B₁=2; 10(2)+3B₂=−1 gives B₂=−7; 2(2)+5(−7)+3B₃=−1 gives B₃=10. Therefore, the particular integral is: 푦_푝 = 2t² − 7t + 10.

💡 Why this matters: The Method of Undetermined Coefficients allows us to find a specific solution that satisfies the non-homogeneous equation.

Topic 148: Differential Equation with Variable Term – A Modification

When the LHS of a constant-coefficient variable-term differential equation is missing certain terms (e.g., the y(t) term is absent), the standard Method of Undetermined Coefficients fails because a direct polynomial trial solution leads to an illogical comparison. The solution requires a Modification: multiply the trial function by t.

📌 Example: Find the Particular Integral of: y''(t) + 5y'(t) = 6t² − t − 1. Here, a₁=5, a₂=0, and f(t)=6t²−t−1. If we use the trial y(t)=B₁t²+B₂t+B₃, substituting yields: LHS: (0)t² + (10B₁)t + (2B₁+5B₂) = 6t² − t − 1. Comparing coefficients gives a contradiction: 0=6 for the t² term. This is illogical, so the trial fails. Modification: Multiply the trial function by t: y(t) = t × (B₁t²+B₂t+B₃) = B₁t³+B₂t²+B₃t. Now y'(t)=3B₁t²+2B₂t+B₃, and y''(t)=6B₁t+2B₂. Substituting into LHS: (6B₁t+2B₂) + 5(3B₁t²+2B₂t+B₃) = 15B₁t² + (6B₁+10B₂)t + (2B₂+5B₃). Equating to RHS: 15B₁t² + (6B₁+10B₂)t + (2B₂+5B₃) = 6t² − t − 1. Comparing: 15B₁=6 gives B₁=2/5; 6(2/5)+10B₂=−1 gives B₂=−17/50; 2(−17/50)+5B₃=−1 gives B₃=−8/125. Therefore, the modified particular integral is: 푦_푝 = (2/5)t² − (17/50)t − (8/125).

💡 Why this matters: When the LHS lacks a y(t) term (a₂=0), the t² term vanishes in the LHS. Multiplying by t adds one degree to every term, restoring the missing term type.

Topic 149: Higher-Order Differential Equation: Particular Integral

This section extends the concept of particular integrals from second-order to nth-order (higher-order) differential equations. The approach is analogous to a polynomial, where the order of derivative replaces the degree of polynomial. The solution for 푦_푝 depends on the values of the coefficients aₙ, aₙ₋₁, etc.

For a general nth-order equation: a₀y⁽ⁿ⁾(t) + a₁y⁽ⁿ⁻¹⁾(t) + ... + aₙ₋₁y'(t) + aₙy = b.

• If aₙ ≠ 0, then: 푦_푝 = b / aₙ
• If aₙ = 0 but aₙ₋₁ ≠ 0, then: 푦_푝 = (b / aₙ₋₁) · t
• If aₙ = 0 and aₙ₋₁ = 0, use trial solution y = kt².

📌 Example 1: Find the Particular Solution of: y⁽⁴⁾(t) + 6y‴(t) + 14y″(t) + 16y'(t) + 8y = 24. Here a₁=6, a₂=14, aₙ₋₁=a₃=16, aₙ=a₄=8≠0. Since aₙ≠0, 푦_푝 = b/aₙ = 24/8 = 3. This is a static equilibrium (constant value).

📌 Example 2: Find the Particular Solution of: y‴(t) + y″(t) + 3y'(t) = 1. Here a₁=1, aₙ₋₁=a₂=3, aₙ=a₃=0. Since aₙ=0 but aₙ₋₁≠0, 푦_푝 = (b/aₙ₋₁)·t = (1/3)·t. This is a moving equilibrium (contains t).

📌 Example 3: Find the Particular Solution of: y‴(t) + 3y″(t) = 1/3. Here a₁=3, aₙ₋₁=a₂=0, aₙ=a₃=0. Use trial y=kt². Then y'(t)=2kt, y″(t)=2k, y‴(t)=0. Substituting: 0 + 3(2k) = 1/3 → 6k=1/3 → k=1/18. Therefore, 푦_푝 = (1/18)t². This is also a moving equilibrium.

Topic 150: Higher-Order Differential Equation: Complementary Function

For higher-order differential equations, the complementary function (푦_푐) builds upon the methods for second-order equations. The characteristic equation for an nth-order equation extends naturally: for 4th order, it is r⁴ + a₁r³ + a₂r² + a₃r + a₄ = 0.

📌 Example: Find the Complementary Function of: y⁽⁴⁾(t) + 6y‴(t) + 14y″(t) + 16y'(t) + 8y = 24. The characteristic equation is r⁴ + 6r³ + 14r² + 16r + 8 = 0. Factorizing gives: (r+2)(r−2)(r²+2r+2)=0. The roots are: r₁=−2, r₂=2 (distinct real roots), and r₃, r₄ = [−2 ± √(4−8)]/2 = −1 ± i (complex conjugate roots). The complementary function is therefore hybrid: 푦_푐 = A₁e⁻²ᵗ + A₂e²ᵗ + e⁻ᵗ(A₃cos t + A₄sin t).

💡 Why this matters: Complementary functions for higher-order equations can be a hybrid combination of distinct real roots, repeated real roots, and complex roots, all combined into one solution.

Topic 151: Convergence and the Routh Theorem

As the order of the differential equation increases, calculating roots becomes complex. Routh's Theorem provides a qualitative method to judge dynamic stability without solving for the roots. Its limitation is that it does not allow plotting the time path.

The theorem is based on the characteristic equation of the higher-order DE: a₀rⁿ + a₁rⁿ⁻¹ + ... + aₙ₋₁r + aₙ = 0. All roots will be negative (indicating dynamic stability) only if the first n of the following determinants are positive: |a₁|, |a₁ a₃ / a₀ a₂|, |a₁ a₃ a₅ / a₀ a₂ a₄ / 0 a₁ a₃|, |a₁ a₃ a₅ a₇ / a₀ a₂ a₄ a₆ / 0 a₁ a₃ a₅ / 0 a₀ a₂ a₄|, ...

Pattern: 1) Number of determinants = order of DE. 2) A V-like shape forms along the principal diagonal. 3) Elements fill with ascending order vertically from bottom to top. 4) Zeros are used where elements are below a₀. Subscripts can exceed the order of the DE.

📌 Example 1: Apply Routh Theorem to: y⁽⁴⁾(t) + 6y‴(t) + 14y″(t) + 16y'(t) + 8y = 24. Here a₀=1, a₁=6, a₂=14, a₃=16, a₄=8. Evaluate 4 determinants: 1st: |a₁| = |6| = 6 > 0. 2nd: |6 16 / 1 14| = (6)(14)−(1)(16) = 84−16 = 68 > 0. 3rd: |6 16 0 / 1 14 8 / 0 6 16| = 6(224−48) − 16(16−0) + 0 = 6(176) − 256 = 1056−256 = 800 > 0. 4th: |6 16 0 0 / 1 14 8 0 / 0 6 16 0 / 0 1 14 8| = 6(14168 + 800 + 0014 − 0160 − 8146 − 800) = 6400 > 0. Since all four determinants are positive, the time path is dynamically stable.

📌 Example 2: Apply Routh Theorem to: y‴(t) − 2y″(t) − y'(t) + 2y = 4. Here a₀=1, a₁=−2, a₂=−1, a₃=2. Evaluate 3 determinants: 1st: |a₁| = |−2| = −2 < 0. Since the first determinant is negative, the time path is dynamically unstable. (The 2nd and 3rd determinants also turn out to be non-positive, confirming this conclusion.)

💡 Why this matters: Routh's Theorem allows for a quick check of stability without the onerous task of finding roots, especially useful for higher-order systems where root calculation is complex.

⭐ Key Takeaways

The Method of Undetermined Coefficients provides a systematic way to find particular integrals for variable-term DEs, but when the LHS lacks the y(t) term, the trial function must be multiplied by t. For higher-order DEs, the particular integral follows simple rules based on which coefficients vanish (b/aₙ, then b·t/aₙ₋₁, then kt²). The complementary function for higher-order equations can be a hybrid of distinct, repeated, and complex roots. Routh's Theorem offers a powerful shortcut—if any of its determinants for the characteristic equation's coefficients is negative, the system is dynamically unstable without needing to find the actual roots.

🧠 Quick Revision Questions

1. What is the precondition for applying the Method of Undetermined Coefficients to a variable-term differential equation?
2. Why does the standard trial function fail when a₂=0, and how is this corrected?
3. For a 5th-order differential equation with a₅≠0 and b≠0, what is the particular integral?
4. How is the complementary function for a higher-order DE constructed when the roots include a pair of complex conjugates and a repeated real root?
5. According to Routh's Theorem, what must be true about the sequence of determinants for a time path to be dynamically stable?

📘 Lecture 34 — First Order Difference Equations

📖 Overview: This lecture introduces discrete time analysis, where time is treated as a sequence of distinct periods rather than a continuous flow. It explains first-order difference equations, their structure, and two solution methods: the iterative method and the general formula method. Understanding these equations is essential for modeling economic variables that change at specific intervals, such as seasonal prices or annual GDP.

🗂️ Topics Covered

The lecture begins by comparing continuous and discrete time analysis, introducing key terminology like lags and leads. It then defines the standard form of a first-order difference equation. Two solution methods are presented: the iterative (crude) method and the general (formula) method. The general method is further divided into two cases: when the coefficient a ≠ -1 and when a = -1, each with its own formula for the particular integral and the complete solution.

📝 Lecture Summary

Topic 152: Discrete Time: First Order Difference Equation

This section introduces the concept of discrete time, where the variable ( t ) takes only integer values. This contrasts with continuous time analysis, which treats time as a continuous variable.

🔑 Definition — Discrete Time: Time is treated as a non-continuous variable consisting of separate, distinct periods or intervals. Successive gaps between time periods are always equal ((\Delta t = 1)).

Key differences between continuous and discrete analysis:

• Nature of Values: Continuous allows decimals and integers; Discrete allows only integers.
• Analogy: Continuous is Point Analysis; Discrete is Period Analysis.
• Tool: Continuous uses derivatives; Discrete uses differences, lags, and leads.
• Equation Type: Continuous yields Differential Equations; Discrete yields Difference Equations.

💡 Why this matters: A period does not have to be a calendar period; it is an analytical length of time before the variable ( y ) undergoes a change. This flexibility makes discrete analysis useful for modeling many economic phenomena.

The first difference of ( y ) is defined by the difference operator: 📐 Formula: (\Delta y_t \equiv y_{t+1} - y_t)

Symbology:

• ( y_t ) = value of ( y ) in current time period ( t ).
• ( y_{t+1} ) = value of ( y ) in the next time period (a lead variable).
• ( y_{t-1} ) = value of ( y ) in the previous time period (a lag variable).

The standard form of a first-order difference equation is: 📐 Formula: ( y_{t+1} + a y_t = c ) Here, ( a ) and ( c ) are constants.

Topic 153: Solving Difference Equation Using Iterative Method

The iterative method is a crude but direct way to find a solution by calculating ( y ) for successive periods starting from an initial condition. It involves repeatedly substituting the value from the previous step into the equation.

📌 Example 1: Solve ( y_{t+1} = y_t + 1 ) by iteration, given ( y_0 = 10 ).

• For ( t=0 ): ( y_1 = y_0 + 1 = 10 + 1 )
• For ( t=1 ): ( y_2 = y_1 + 1 = (10+1) + 1 = 10 + 2 )
• For ( t=2 ): ( y_3 = y_2 + 1 = (10+2) + 1 = 10 + 3 )
• Generalizing: ( y_t = y_0 + t = 10 + t )

📌 Example 2: Solve ( y_{t+1} = a y_t ) by iteration, given ( y_0 = \alpha ).

• For ( t=0 ): ( y_1 = a y_0 )
• For ( t=1 ): ( y_2 = a y_1 = a(a y_0) = a^2 y_0 )
• For ( t=2 ): ( y_3 = a y_2 = a (a^2 y_0) = a^3 y_0 )
• Generalizing: ( y_t = a^t y_0 = a^t \alpha )

Topic 154: Solving Difference Equation Using General Method: A≠-1

This section develops a general formula to solve any first-order difference equation when ( a \neq -1 ). The general solution has two components: ( y_t = y_c + y_p )

• ( y_c ) = Complementary Function (the time-path deviation from equilibrium).
• ( y_p ) = Particular Integral (the intertemporal equilibrium value of ( y )).

Complementary Function: For the homogeneous reduced form ( y_{t+1} + a y_t = 0 ), the solution is found by assuming the form ( y_t = A b^t ). Substituting into the homogeneous equation gives ( b = -a ). 🔑 Formula: ( y_c = A (-a)^t ) The base (( -a )) in discrete time plays the same role in determining dynamic stability as the base (( e^r )) in continuous time.

Particular Integral: For ( a \neq -1 ), the simplest trial solution is a constant ( y_t = k ). Substituting into the standard form ( y_{t+1} + a y_t = c ) yields: 🔑 Formula: ( y_p = \frac{c}{1+a} )

General Solution: Combining these two parts gives: 📐 Formula: ( y_t = A (-a)^t + \frac{c}{1+a} )

Definite Solution: To find a unique time path, an initial condition is used. Setting ( t=0 ) in the general solution, we solve for ( A ): 📐 Formula: ( A = y_0 - \frac{c}{1+a} ) The definite solution is then: 📐 Formula: ( y_t = \left( y_0 - \frac{c}{1+a} \right) (-a)^t + \frac{c}{1+a} ) This formula is valid only if ( a \neq -1 ).

Topic 155: Difference Equation Using General Method: A≠-1 Numerical

This is a numerical application of the general formula from Topic 154.

📌 Example: Solve ( y_{t+1} - 5y_t = 1 ), given ( y_0 = \frac{7}{4} ).

1. Identify parameters: The standard form is ( y_{t+1} + a y_t = c ). Comparing, we get ( a = -5 ) and ( c = 1 ). Note that ( a \neq -1 ).
2. Calculate the Particular Integral: ( y_p = \frac{c}{1+a} = \frac{1}{1 + (-5)} = \frac{1}{-4} = -\frac{1}{4} ).
3. Solve for the arbitrary constant ( A ): ( A = y_0 - \frac{c}{1+a} = \frac{7}{4} - \left( -\frac{1}{4} \right) = \frac{7}{4} + \frac{1}{4} = 2 ).
4. Write the Definite Solution: ( y_t = A (-a)^t + y_p = 2 \cdot (-(-5))^t + \left( -\frac{1}{4} \right) = 2 (5)^t - \frac{1}{4} ).

The time path is ( y_t = 2(5)^t - \frac{1}{4} ), where the static equilibrium is ( y_p = -\frac{1}{4} ).

Topic 156: Solving Difference Equation Using General Method: A=-1

When ( a = -1 ), the previous formula for the Particular Integral becomes invalid because it involves division by zero (( 1 + a = 0 )). A different trial solution is required.

Complementary Function: The formula ( y_c = A (-a)^t ) remains valid. Since ( a = -1 ), we have ( -a = -(-1) = 1 ), so ( y_c = A (1)^t = A ).

Particular Integral: A constant trial solution fails. The next simplest trial solution is a linear function of time: ( y_t = k t ).

1. Assume ( y_t = k t ). Its lead form is ( y_{t+1} = k (t+1) ).
2. Substitute into the standard form: ( k(t+1) + a(k t) = c ).
3. Simplify: ( k t + k + a k t = c \implies k(t + 1 + a t) = c ).
4. Since ( a = -1 ), ( t + a t = t - t = 0 ). The equation simplifies to ( k (1) = c \implies k = c ). 🔑 Formula: ( y_p = c t )

General Solution: The complete solution for ( a = -1 ) is: 📐 Formula: ( y_t = A (-a)^t + c t ) Since ( -a = 1 ), this is ( y_t = A + c t ).

Definite Solution: Using the initial condition at ( t=0 ): ( y_0 = A + c(0) \implies A = y_0 ) 📐 Formula: ( y_t = y_0 (-a)^t + c t ), which simplifies to ( y_t = y_0 + c t ).

Topic 157: Difference Equation Using General Method: A=-1 Numerical

This is a numerical application of the formula for the ( a = -1 ) case.

📌 Example: Solve ( y_{t+1} - y_t = 2 ), given ( y_0 = 0 ).

1. Identify parameters: The standard form is ( y_{t+1} + a y_t = c ). The equation is ( y_{t+1} + (-1) y_t = 2 ), so ( a = -1 ) and ( c = 2 ).
2. Calculate the Particular Integral: ( y_p = c t = 2t ).
3. Solve for the arbitrary constant ( A ): ( A = y_0 = 0 ).
4. Write the Definite Solution: ( y_t = y_0 (-a)^t + c t = 0 \cdot (1)^t + 2t ). The time path is ( y_t = 2t ).

In this solution, the complementary function is zero (( y_c = 0 )), meaning the time path has no deviation from the moving equilibrium represented by ( y_p = 2t ).

⭐ Key Takeaways

The most critical concepts from this lecture are the distinction between continuous and discrete time, and the two solution methods for first-order difference equations. The iterative method provides insight but is not practical for economic models. The general formula method is powerful: the standard form is ( y_{t+1} + a y_t = c ), and its solution depends critically on the value of ( a ). If ( a \neq -1 ), the particular integral is a constant equilibrium ( c/(1+a) ). If ( a = -1 ), the particular integral is a moving equilibrium ( ct ). The arbitrary constant ( A ) is always determined by the initial condition ( y_0 ), and the base (( -a )) in the complementary function dictates whether the time path converges to or diverges from equilibrium.

🧠 Quick Revision Questions

1. What is the standard form of a first-order difference equation?
2. In the difference equation ( y_{t+1} + a y_t = c ), what trial solution is used for the particular integral if ( a \neq -1 )?
3. What special modification to the particular integral is required when ( a = -1 )?
4. Given the definite solution ( y_t = (y_0 - \frac{c}{1+a})(-a)^t + \frac{c}{1+a} ), what condition on the base (( -a )) ensures that the time path is dynamically stable (i.e., converges to equilibrium)?
5. Solve the difference equation ( y_{t+1} - 3y_t = 4 ) with ( y_0 = 1 ) using the general method.

📘 Lecture 35 — Dynamic Stability of First Order Difference Equations

📖 Overview: This lecture examines the dynamic stability of first-order difference equations, focusing on how the complementary function determines whether the time-path is convergent or divergent. It introduces the seven cases (regions) of the parameter ( b ) that produce non-oscillatory and oscillatory time-paths, and explains the role of the coefficient ( A ) in scaling and mirroring these paths. Understanding these concepts is essential for analyzing economic models that evolve over discrete time periods.

🗂️ Topics Covered

The lecture covers dynamic stability of first-order difference equations beginning with non-oscillatory cases (Regions I–IV) and oscillatory cases (Regions V–VII). It then examines the role of the coefficient ( A ) in the complementary function, detailing its magnitude effect (scale) and sign effect (mirror). Finally, numerical examples demonstrate how to classify time-paths as dynamically stable or unstable based on the value of ( b ).

📝 Lecture Summary

TOPIC 158: DYNAMIC STABILITY OF FIRST ORDER DIFFERENCE EQUATION: NON-OSCILLATORY CASES

The dynamic stability of difference equations depends on the complementary function ( y_c ). Here, ( y_c = A(b)^t ). Setting ( A = 1 ) gives ( y_c = (b)^t ), so the expression ( (b)^t ) determines the dynamics of ( y_c ).

The base ( b ) is related to the coefficient ( a ) in the first-order difference equation ( y_t + a y_{t-1} = c ), where ( b = -a ).

The behavior of ( (b)^t ) is tabulated for positive, zero, and negative values of ( b ). When ( b = 2 > 0 ), the time-path shows exponential growth with positive signs, lying in the 1st quadrant (smooth graph). When ( b = 0 ), there is neither growth nor decay (horizontal straight line). When ( b = -2 < 0 ), the time-path shows exponential change with alternating positive/negative signs, lying in the 1st and 4th quadrants (ups and downs).

Terminology: A time-path is non-oscillatory when ( b \not< 0 ) (i.e., ( b \geq 0 )) and oscillatory when ( b < 0 ).

First-order difference equations have an additional feature of pattern (non-oscillatory or oscillatory) beyond the convergent or divergent nature seen in differential equations.

Using a number line to visualize regions of ( b ), seven values are identified: ( b > 1, b = 1, 0 < b < 1, b = 0 ) (non-oscillatory), and ( -1 < b < 0, b = -1, b < -1 ) (oscillatory). The first four (I–IV) are non-oscillatory.

Region – I (( b > 1, |b| > 1 )): Example ( y_c = (2)^t ). Values are 1, 2, 4, 8, 16... — all positive (non-oscillatory). Dynamic Stability: Time-path is divergent and hence dynamically unstable.

Region – II (( b = 1, |b| = 1 )): Example ( y_c = (1)^t ). Values are 1, 1, 1, 1, 1... — all positive (non-oscillatory). Dynamic Stability: Time-path is neither divergent nor convergent.

Region – III (( 0 < b < 1, |b| < 1 )): Example ( y_c = (1/2)^t ). Values are 1, 1/2, 1/4, 1/8, 1/16... — all positive (non-oscillatory). Dynamic Stability: Time-path is convergent and hence dynamically stable.

🔑 Definition — Dynamic Stability: A time-path that converges to equilibrium over time. For first-order difference equations, this occurs when ( |b| < 1 ).

Region – IV (( b = 0, |b| = 0 )): Example ( y_c = (0)^t ). Values are 0, 0, 0, 0, 0... — all zero (non-oscillatory). Dynamic Stability: Time-path is neither divergent nor convergent. It is pre-converged at equilibrium; since ( y_c = 0 ), the time-path is already on equilibrium.

Summary: Cases I–IV have non-negative ( b ). A non-negative number raised to any power generates a positive number, giving a graph in the 1st quadrant only — hence non-oscillatory. Only when ( |b| < 1 ) is the time-path convergent.

Tabulation:

• Region I: ( b > 1 ), non-oscillatory, ( |b| > 1 ), non-convergent (divergent)
• Region II: ( b = 1 ), non-oscillatory, ( |b| = 1 ), non-convergent (neither)
• Region III: ( 0 < b < 1 ), non-oscillatory, ( |b| < 1 ), convergent
• Region IV: ( b = 0 ), non-oscillatory, ( |b| = 0 < 1 ), pre-converged (( y_c = 0 ), no deviation)

TOPIC 159: DYNAMIC STABILITY OF FIRST ORDER DIFFERENCE EQUATION: OSCILLATORY CASES

The remaining three cases (V–VII) are oscillatory because ( b < 0 ).

Region – V (( -1 < b < 0, |b| < 1 )): Example ( y_c = (-1/2)^t ). Values are 1, -1/2, 1/4, -1/8, 1/16... — alternating signs (oscillatory). Dynamic Stability: Time-path is convergent and hence dynamically stable. It moves above and below equilibrium with decreasing vertical distance.

Region – VI (( b = -1, |b| = 1 )): Example ( y_c = (-1)^t ). Values are 1, -1, 1, -1, 1... — alternating signs (oscillatory). Dynamic Stability: Time-path is neither divergent nor convergent. It moves above and below equilibrium with uniform vertical distance.

Region – VII (( b < -1, |b| > 1 )): Example ( y_c = (-2)^t ). Values are 1, -2, 4, -8, 16... — alternating signs (oscillatory). Dynamic Stability: Time-path is divergent and hence dynamically unstable. It moves above and below equilibrium with increasing vertical distance.

Summary: Cases V–VII have negative ( b ). A negative number raised to any power generates numbers with alternating signs, giving a graph in the 1st and 4th quadrants — hence oscillatory. Only when ( |b| < 1 ) is the time-path convergent.

Tabulation:

• Region V: ( -1 < b < 0 ), oscillatory, ( |b| < 1 ), convergent
• Region VI: ( b = -1 ), oscillatory, ( |b| = 1 ), non-convergent (neither)
• Region VII: ( b < -1 ), oscillatory, ( |b| > 1 ), non-convergent (divergent)

TOPIC 160: DYNAMIC STABILITY OF FIRST ORDER DIFFERENCE EQUATION: ROLE OF ( A )

Recall ( y_c = A(b)^t ). Among the seven cases, the convergent ones (Case III: non-oscillatory and Case V: oscillatory) are the desired cases. The coefficient ( A ) can affect ( y_c ) in two ways: magnitude and sign.

The dual effect of ( A ) on ( y_c ):

• Magnitude: ( |A| = 0 ) nullifies (yields ( y_c = 0 )); ( |A| < 1 ) minifies; ( |A| = 1 ) neutral; ( |A| > 1 ) amplifies.
• Sign: ( A > 0 ) retains the sign of ( (b)^t ); ( A < 0 ) reverses it.

1. Scale Effect: In Case III (convergent, non-oscillatory):

• Original: ( y_c = (1/2)^t ) starts from 1.
• Doubled: ( y_c = 2(1/2)^t ) starts from 2.
• Halved: ( y_c = (1/2)(1/2)^t ) starts from 0.5.

The magnitude of ( A ) (( \gtrless 1 )) scales the time-path up or down. This is called the Scale Effect of ( A ).

In Case V (convergent, oscillatory):

• Original: ( y_c = (-1/2)^t ) starts from 1.
• Doubled: ( y_c = 2(-1/2)^t ) starts from 2.
• Halved: ( y_c = (1/2)(-1/2)^t ) starts from 0.5.

Again, the magnitude of ( A ) scales the oscillatory time-path up or down.

2. Mirror Effect: In Case III (convergent, non-oscillatory):

• Original: ( y_c = (1/2)^t ) starts from +1.
• Reversed: ( y_c = -1(1/2)^t ) starts from -1.

The negative sign of ( A ) reverses the shape, creating a mirror image of the original time-path. This is called the Mirror Effect of ( A ).

In Case V (convergent, oscillatory):

• Original: ( y_c = (-1/2)^t ) starts from +1.
• Reversed: ( y_c = -1(-1/2)^t ) starts from -1.

Again, the negative sign of ( A ) creates a mirror image of the oscillatory time-path.

TOPIC 161: DYNAMIC STABILITY OF FIRST ORDER DIFFERENCE EQUATION: NUMERICALS

Example # 1: Check the dynamic stability of ( y_t = 2(-\frac{4}{5})^t + 9 ).

The formula is ( y_t = A(b)^t + y_p ), where ( y_p = \frac{c}{1+a} ).

Breaking up: ( y_c = 2(-\frac{4}{5})^t ), ( y_p = 9 ). So ( b = -\frac{4}{5} ).

Interpretation: Here, ( b = -\frac{4}{5} < 0 ) (oscillatory) and ( |b| = \frac{4}{5} < 1 ) (convergent). Therefore, the time-path is oscillatory in pattern and convergent in nature. The time-path is dynamically stable.

📌 Example: ( b = -\frac{4}{5} ). Since ( -\frac{4}{5} < 0 ), the path oscillates. Since ( |-\frac{4}{5}| = 0.8 < 1 ), the oscillations converge to equilibrium (the particular solution ( y_p = 9 )).

Note: This path differs from ( y_t = -2(\frac{4}{5})^t + 9 ), where the sign inside the parentheses is positive, giving a non-oscillatory but possibly convergent or divergent time-path.

Example # 2: Check the dynamic stability of ( y_t = 3(2)^t + 4 ).

The formula is ( y_t = A(b)^t + y_p ).

Breaking up: ( y_c = 3(2)^t ), ( y_p = 4 ). So ( b = 2 ).

Interpretation: Here, ( b = 2 \not< 0 ) (non-oscillatory) and ( |b| = 2 \not< 1 ) (non-convergent, i.e., divergent). Therefore, the time-path is non-oscillatory in pattern and non-convergent in nature. The time-path is dynamically unstable.

📌 Example: ( b = 2 ). Since ( 2 > 0 ), the path is smooth (non-oscillatory). Since ( |2| = 2 > 1 ), the path diverges away from equilibrium (( y_p = 4 )).

⭐ Key Takeaways

1. Dynamic stability of first-order difference equations depends entirely on the absolute value of ( b ) (where ( b = -a )): if ( |b| < 1 ), the time-path is convergent (dynamically stable); if ( |b| > 1 ), it is divergent (dynamically unstable); if ( |b| = 1 ), it is neither convergent nor divergent. 2. The sign of ( b ) determines the pattern: ( b \geq 0 ) yields non-oscillatory (smooth) time-paths while ( b < 0 ) yields oscillatory (alternating) time-paths. 3. The seven regions of ( b ) cover all possible time-path behaviors: non-oscillatory cases (I–IV: ( b > 1 ), ( b = 1 ), ( 0 < b < 1 ), ( b = 0 )) and oscillatory cases (V–VII: ( -1 < b < 0 ), ( b = -1 ), ( b < -1 )). 4. The coefficient ( A ) in the complementary function ( y_c = A(b)^t ) has a Scale Effect (magnitude ( |A| ) scales the path up or down) and a Mirror Effect (negative sign ( A < 0 ) reverses the path). 5. For any numerical example, always compute both ( b ) and ( |b| ) to classify the time-path as oscillatory/non-oscillatory and convergent/divergent, then state whether the system is dynamically stable (convergent) or unstable (divergent).

🧠 Quick Revision Questions

1. Under what condition on ( |b| ) is the time-path of a first-order difference equation dynamically stable (convergent)?
2. What is the difference between a non-oscillatory and an oscillatory time-path in terms of the sign of ( b )?
3. In Example #1, ( y_t = 2(-\frac{4}{5})^t + 9 ), what are the values of ( b ) and ( |b| ), and what do they imply about the time-path?
4. Explain the Scale Effect and Mirror Effect of the coefficient ( A ) in the complementary function ( y_c = A(b)^t ).
5. In Region VI (( b = -1 )), why is the time-path neither convergent nor divergent even though it oscillates?

📘 Lecture 36 — Income Dynamics of Difference Equations

📖 Overview: This lecture examines how national income evolves over time when consumption and investment decisions depend on past income levels. Using first-order difference equations, we model income dynamics under various economic scenarios, including lagged consumption, induced investment, capital market equilibrium, money market conditions, and balanced trade. Understanding these dynamics is crucial for analyzing economic stability and policy effects.

🗂️ Topics Covered

This lecture covers six topics: a lagged income determination model framework with consumption depending on last period's income, its numerical solution, income dynamics with induced investment alongside autonomous investment, income dynamics in the capital market where savings equal investment from previous period's income, income dynamics in the money market where money supply and demand depend on current and past income, and finally income dynamics for balanced foreign trade where imports and exports are functions of current and lagged income.

📝 Lecture Summary

Topic 162: Lagged Income Determination Model: Framework

Unlike the simple income determination model, we introduce lags in the consumption function, where current consumption depends on last period's income. The model is:

( C_t = C_0 + cY_{t-1} ) where ( c = MPC ) (0<c<1), and ( I_t = I_0 ) (autonomous investment).

Income identity: ( Y_t = C_t + I_t )

Substituting: ( Y_t = C_0 + cY_{t-1} + I_0 ), thus ( Y_t - cY_{t-1} = C_0 + I_0 )

This is a first-order difference equation (FODE) of the form ( Y_t + aY_{t-1} = c ), where ( a = -c ) and the constant is ( C_0 + I_0 ). Since ( a \neq -1 ) (as MPC<1), the definite time-path formula is used.

🔑 Definition — First-Order Difference Equation: An equation expressing a variable's current value as a function of its immediately preceding value.

📐 Formula: ( Y_t = (Y_0 - \frac{C_0 + I_0}{1-c})(c)^t + \frac{C_0 + I_0}{1-c} )

This is the definite time-path. Dynamic stability depends on the exponential expression ( (c)^t ). Since ( |c| > 0 ) (non-zero) and ( c < 1 ), the time-path is non-oscillatory and convergent (dynamically stable).

💡 Why this matters: The time-path converges to equilibrium because the MPC is less than 1, meaning each period's income moves closer to the steady state.

As ( t \to \infty ), the complementary function ( Y_c = (Y_0 - Y^*)(c)^t \to 0 ), so ( Y_t \to Y_p = \frac{C_0 + I_0}{1-c} ), the particular solution or equilibrium income.

Topic 163: Lagged Income Determination Model: Numerical

We solve the model numerically with specific values:

( C_t = 200 + 0.9Y_{t-1} ) (MPC = 0.9), ( I_t = 100 ), ( Y_0 = 4500 )

Substituting: ( Y_t = 200 + 0.9Y_{t-1} + 100 ) → ( Y_t - 0.9Y_{t-1} = 300 )

This is ( Y_t + (-0.9)Y_{t-1} = 300 ), so ( a = -0.9 \neq -1 ), ( c = 300 ).

📐 Formula: ( Y_t = (4500 - \frac{300}{1-0.9})(0.9)^t + \frac{300}{1-0.9} )

( Y_t = (4500 - 3000)(0.9)^t + 3000 = 1500(0.9)^t + 3000 )

Since ( |0.9| > 0 ) (non-zero) and ( 0.9 < 1 ), the time-path is non-oscillatory and convergent. As ( t \to \infty ), ( 1500(0.9)^t \to 0 ), so ( Y_t \to 3000 ) (equilibrium).

The complementary function ( Y_c = 1500(0.9)^t ) represents the deviation from equilibrium, which diminishes over time.

Topic 164: Income Dynamics with Induced Investment Using Difference Equations

We now introduce induced investment alongside autonomous investment.

( C_t = 200 + 0.4Y_{t-1} ) (MPC = 0.4), ( I_t = 100 + 0.2Y_{t-1} ) (autonomous + induced), ( Y_0 = 950 )

Substituting: ( Y_t = 200 + 0.4Y_{t-1} + 100 + 0.2Y_{t-1} ) → ( Y_t - 0.6Y_{t-1} = 300 )

So ( a = -0.6 \neq -1 ), ( c = 300 ).

📐 Formula: ( Y_t = (950 - \frac{300}{1-0.6})(0.6)^t + \frac{300}{1-0.6} = (950 - 750)(0.6)^t + 750 = 200(0.6)^t + 750 )

Since ( |0.6| > 0 ) and ( 0.6 < 1 ), the time-path is non-oscillatory and convergent to ( Y^* = 750 ).

💡 Why this matters: The induced investment reduces the coefficient of the exponential term to 200 and the equilibrium to 750, showing how investment sensitivity affects dynamics.

Topic 165: Income Dynamics & Capital Market Using Difference Equations

In the capital market, savings are a proportion of current income, while investment depends on last period's income.

Savings: ( S_t = sY_t ) where ( s = MPS ) (marginal propensity to save, 0<s<1)

Investment: ( I_t = vY_{t-1} ) where ( v < 1 ) is the proportion of last year's income invested

Equilibrium: ( S_t = I_t ) → ( sY_t = vY_{t-1} ) → ( sY_t - vY_{t-1} = 0 )

Normalizing: ( Y_t - \frac{v}{s}Y_{t-1} = 0 ) → ( Y_t + (-\frac{v}{s})Y_{t-1} = 0 )

Here, ( a = -\frac{v}{s} \neq -1 ) (provided ( v \neq s )) and ( c = 0 ).

📐 Formula: ( Y_t = Y_0(\frac{v}{s})^t )

Dynamic stability depends on ( b = \frac{v}{s} ):

• Nature: If ( |b| < 1 ) (i.e., ( v < s )) → convergent; if ( |b| > 1 ) → non-convergent
• Pattern: Since ( b > 0 ) (positive), the path is non-oscillatory

Therefore, the time-path is convergent only if ( v < s ), and it moves in a non-oscillatory pattern.

Topic 166: Income Dynamics & Money Market Using Difference Equations

On the money market, the government sets money supply as a proportion of previous period's income, while money demand depends on current income.

( M_s = sY_{t-1} ) (proportion s of last year's income) ( M_D = a + vY_t ) (transaction demand, 0 < v < s < 1, a > 0)

Equilibrium: ( M_s = M_D ) → ( sY_{t-1} = a + vY_t ) Rearranging: ( vY_t - sY_{t-1} = -a ) → ( Y_t - \frac{s}{v}Y_{t-1} = -\frac{a}{v} )

So ( a = -\frac{s}{v} \neq -1 ) (since v < s), ( c = -\frac{a}{v} ).

📐 Formula: ( Y_t = (Y_0 + \frac{a}{v-s})(\frac{s}{v})^t - \frac{a}{v-s} )

Dynamic stability depends on ( b = \frac{s}{v} ):

• Since ( v < s ), ( |b| = \frac{s}{v} > 1 ) → non-convergent
• Since ( b > 0 ) → non-oscillatory

💡 Why this matters: When money supply growth (s) exceeds money demand sensitivity (v), the system diverges, potentially leading to instability.

Topic 167: Income Dynamics & Balanced Trade Using Difference Equations

For balanced trade, imports depend on current income, while exports depend on both current and previous period's income.

( M_t = mY_t ) (0 < m < 1) ( X_t = aY_t + vY_{t-1} ) (0 < a, v < 1)

Balanced trade: ( M_t = X_t ) → ( mY_t = aY_t + vY_{t-1} ) ( (m-a)Y_t - vY_{t-1} = 0 )

Normalizing: ( Y_t - \frac{v}{m-a}Y_{t-1} = 0 ) → ( Y_t + (-\frac{v}{m-a})Y_{t-1} = 0 )

Here, ( a = -\frac{v}{m-a} \neq -1 ) (provided m-a ≠ v), ( c = 0 ).

📐 Formula: ( Y_t = Y_0(\frac{v}{m-a})^t )

Dynamic stability depends on ( b = \frac{v}{m-a} ):

• Nature: If ( m-a > v ), then ( |b| < 1 ) → convergent
• Pattern: If convergent, ( m-a > 0 ) → ( b > 0 ) → non-oscillatory

Therefore, the time-path converges if ( m-a > v ), moving in a non-oscillatory pattern.

⭐ Key Takeaways

The fundamental result is that all lagged income models reduce to first-order difference equations whose stability depends on the absolute value of the coefficient on the lagged term. When the coefficient's absolute value is less than one, the time-path converges to equilibrium; when greater than one, it diverges. All models in this lecture produce non-oscillatory patterns because the coefficients are positive. The particular solution (equilibrium income) is given by the constant term divided by one minus the coefficient on lagged income, generalizing the multiplier formula. Finally, the specific economic structure—whether lags appear in consumption, investment, savings, money supply, or trade—determines the exact coefficient values and thus the dynamic properties of national income.

🧠 Quick Revision Questions

1. What is the general solution for the lagged income determination model ( Y_t - cY_{t-1} = C_0 + I_t ), and under what condition is the time-path dynamically stable?
2. In the capital market model ( S_t = sY_t ) and ( I_t = vY_{t-1} ), what condition ensures convergence of the income time-path?
3. For the money market model with ( M_s = sY_{t-1} ) and ( M_D = a + vY_t ), why does the time-path diverge when ( s > v )?
4. In the balanced trade model, if ( m = 0.3 ), ( a = 0.1 ), and ( v = 0.15 ), will the income time-path converge? Explain.
5. Why do all the income dynamics models in this lecture produce non-oscillatory time-paths?

📘 Lecture 37 — Various Economic Applications of Difference Equations

📖 Overview: This lecture applies first-order difference equations to model dynamic economic phenomena including investment behavior, price dynamics in money markets, fish population management, compound interest calculations, and the Harrod growth model. These applications demonstrate how difference equations provide quantitative frameworks for understanding economic dynamics and equilibrium paths over time.

🗂️ Topics Covered

The lecture covers five economic applications: investment dynamics where investment depends on its previous value, price dynamics in the money market relating inflation to excess money supply, fish stock analysis with population growth and harvesting, compound interest formula derivation from difference equations, and the Harrod model of economic growth including the warranted rate of growth concept.

📝 Lecture Summary

TOPIC 168: INVESTMENT DYNAMICS USING DIFFERENCE EQUATIONS

Iₜ is the amount invested in period t: Iₜ = 0.9Iₜ₋₁ + 60. This is a first-order difference equation where investment depends on 90% of previous period's investment plus a constant term of 60. Given initial investment I₀ = £900m, the time-path is found by solving.

Comparing to standard form yₜ + a yₜ₋₁ = c: here a = -0.9 and c = 60. Since a ≠ -1, the definite solution uses formula yₜ = (y₀ − c/(1+a)) (−a)ᵗ + c/(1+a).

📐 Formula: Iₜ = (900 − 60/(1−0.9)) (0.9)ᵗ + 60/(1−0.9) → Iₜ = 300(0.9)ᵗ + 600

The time-path has Iₚ (particular integral) = 600 (the equilibrium level) and I₍ (complementary function) = 300(0.9)ᵗ (deviation from equilibrium).

🔑 Definition — Dynamic stability: Depends on the base b = 0.9 of exponential term (0.9)ᵗ.

Since |b| = |0.9| < 1: the time-path is convergent.

Since b = 0.9 > 0: the time-path is non-oscillatory.

📌 Example: As t→∞, (0.9)ᵗ → 0, so Iₜ → I = 600*. Investment converges to equilibrium of £600m.

💡 Why this matters: The coefficient 0.9 being less than 1 ensures the system moves toward equilibrium over time, representing a stable investment process.

TOPIC 169: PRICE DYNAMICS & MONEY MARKET USING DIFFERENCE EQUATIONS

Money supply Mₛ is government-set. Money demand Lₜ depends on price level Pₜ and real output Y: Lₜ = aPₜY (a > 0). Inflation (price change) Pₜ₊₁ − Pₜ depends on excess money supply: Pₜ₊₁ − Pₜ = j(Mₛ − Lₜ) (j > 0).

Substituting demand into the inflation equation: Pₜ₊₁ − Pₜ = j(Mₛ − aPₜY) → Pₜ₊₁ − (1 − ajY)Pₜ = jMₛ

Comparing to yₜ₊₁ + a yₜ = c: here a = −(1 − ajY), c = jMₛ, y₀ = P₀.

📐 Formula: Pₜ = (P₀ − Mₛ/(aY)) (1 − ajY)ᵗ + Mₛ/(aY)

The exponential expression is (1 − ajY)ᵗ.

🔑 Definition — Intertemporal equilibrium price: *P = Mₛ/(aY)**, which depends directly on money supply and inversely on the product of the demand parameter a and output Y.

Stability depends on |b| = |1 − ajY|:

• If ajY < 1: |1 − ajY| < 1 → time-path is convergent
• Since b = (1 − ajY) and with ajY < 1: (1 − ajY) > 0 → non-oscillatory

📌 Example: With parameters such that ajY < 1, the price level converges to equilibrium Mₛ/(aY) without oscillations.

TOPIC 170: FISH STOCK ANALYSIS USING DIFFERENCE EQUATION

A fish population increases by 5% each generation, with a catch of 8000 removed from each generation. Let Fₜ represent population size for generation t.

(a) The difference equation: Fₜ = 1.05Fₜ₋₁ − 8000

Introducing one generation lag: Fₜ₊₁ = 1.05Fₜ − 8000 → Fₜ₊₁ − 1.05Fₜ = −8000

Comparing to yₜ₊₁ + a yₜ = c: here a = −1.05, c = −8000.

Since a ≠ −1, using solution formula with F₀ = 1000000:

📐 Formula: Fₜ = (1000000 − (−8000)/(1−1.05)) (−(−1.05))ᵗ + (−8000)/(1−1.05)

Simplifying: Fₜ = (1000000 − 160000)(1.05)ᵗ + 160000

Fₜ = (−60000)(1.05)ᵗ + 160000

Since |b| = |1.05| > 1: time-path is non-convergent (explosive).

Since b = 1.05 > 0: time-path is non-oscillatory.

📌 Example: As t→∞, (1.05)ᵗ → ∞, so Fₜ → −∞ (the population first declines then becomes negative, indicating extinction at some finite time).

💡 Why this matters: The 5% growth rate is insufficient to sustain the population against the constant catch of 8000 per generation, demonstrating unsustainable harvesting.

TOPIC 171: INITIAL AMOUNT DEPOSITED IN COMPOUND INTEREST

For compound interest annually: Pₜ₊₁ = Pₜ + i×Pₜ where i is interest rate.

📐 Formula: Pₜ₊₁ = (1+i)Pₜ → Pₜ₊₁ − (1+i)Pₜ = 0

Comparing to yₜ₊₁ + a yₜ = c: here a = −(1+i), c = 0.

Since a ≠ −1 (given i > 0), using solution formula:

📐 Formula: Pₜ = P₀(1+i)ᵗ

Stability: |b| = |1+i| > 1 (since i > 0) → time-path is non-convergent (growing).

Pattern: b = (1+i) > 0 → time-path is non-oscillatory.

📌 Example: If P₀ = 1000 and i = 0.05 (5%), then P₁ = 1000(1.05) = 1050, P₂ = 1000(1.05)² = 1102.5, etc.

💡 Why this matters: This derivation shows the standard compound interest formula emerges naturally from a first-order difference equation, confirming the underlying mathematical structure.

TOPIC 172: HARROD MODEL OF GROWTH

The Harrod model has two key assumptions:

• Sₜ = sYₜ: saving is constant proportion s of income (MPS = APS)
• Iₜ = a(Yₜ − Yₜ₋₁): investment follows acceleration principle where a is capital-output ratio (COR)

In equilibrium: Iₜ = Sₜ → a(Yₜ − Yₜ₋₁) = sYₜ

aYₜ − aYₜ₋₁ = sYₜ → (s − a)Yₜ + aYₜ₋₁ = 0

Dividing by (s − a): Yₜ + (a/(s−a))Yₜ₋₁ = 0

Introducing lag to match standard form: Yₜ₊₁ + (a/(s−a))Yₜ = 0

Here a = a/(s−a), c = 0.

📐 Formula: Yₜ = Y₀ (a/(a−s))ᵗ

Note the sign change: a/(s−a) = −a/(a−s), so the base becomes a/(a−s).

Since a > 1 (capital-output ratio) and 0 < s < 1 (marginal propensity to save), we have a > s.

|b| = |a/(a−s)|: Since denominator (a−s) < a, we have a/(a−s) > 1 → non-convergent.

Since b = a/(a−s) > 0: non-oscillatory.

📌 Example: If a = 3, s = 0.2, then Yₜ = Y₀ (3/(3−0.2))ᵗ = Y₀(3/2.8)ᵗ = Y₀(1.0714)ᵗ, showing growth at about 7.14% per period.

TOPIC 173: WARRANTED RATE OF GROWTH

From the Harrod model solution: Yₜ = Y₀ (a/(a−s))ᵗ

The warranted rate of growth is the growth rate that maintains equilibrium between saving and investment each year.

Growth rate: G = (Y₁ − Y₀)/Y₀

Substituting Y₁ = (a/(a−s))Y₀:

G = ((a/(a−s))Y₀ − Y₀)/Y₀ = a/(a−s) − 1

G = (a − (a−s))/(a−s) = s/(a−s)

🔑 Definition — Warranted growth rate (G₍): G₍ = s/(a−s), the rate of growth that, if achieved, keeps the economy on an equilibrium path where planned saving equals planned investment.

📌 Example: With a = 3, s = 0.2: G₍ = 0.2/(3−0.2) = 0.2/2.8 = 0.0714 or 7.14% per period.

💡 Why this matters: The warranted rate depends on the saving rate s and capital-output ratio a. Any deviation from this path leads to cumulative divergence (instability) in the Harrod model.

⭐ Key Takeaways

All five applications follow the same solution framework for first-order difference equations: identify coefficients a and c, check that a ≠ −1, compute the particular integral (equilibrium) and complementary function (deviation), and analyze dynamic stability using |b| and sign of b. The investment and price dynamics models converge to equilibrium when |b|<1, while the fish stock, compound interest, and Harrod growth models are non-convergent when |b|>1. The money market equilibrium price depends on money supply and output, while the Harrod warranted growth rate G₍=s/(a−s) represents the knife-edge growth path necessary for continuous equilibrium.

🧠 Quick Revision Questions

1. In the investment dynamics problem, what does the value b=0.9 indicate about the time-path's nature and pattern?

2. In the price dynamics model, what condition must hold for the price level to converge to its intertemporal equilibrium?

3. For the fish stock problem with F₀=1,000,000, growth rate 5%, and catch 8,000 per generation, what happens to the population as t→∞?

4. Derive the compound interest formula Pₜ=P₀(1+i)ᵗ using the standard first-order difference equation solution method.

5. In the Harrod model, derive the warranted rate of growth G₍=s/(a−s) and explain why the time-path is non-convergent given a > s.

📘 Lecture 38 — Cobweb Model and Difference Equations

📖 Overview: This lecture explains how first-order difference equations model price fluctuations in markets with production lags, such as agriculture. It introduces the cobweb model framework, derives the time-path of price, analyzes convergence and divergence conditions based on relative slopes of demand and supply, and contrasts adaptive expectations in differential versus difference equations.

🗂️ Topics Covered

The lecture covers the framework of the cobweb model using demand and supply functions with time lags, deriving the first-order difference equation for price and its definite solution. It then examines the time-path diagrammatically, analyzing convergent versus divergent oscillatory patterns. A rare case where slopes are equal is considered, followed by a numerical example. Finally, the lecture introduces adaptive expectations and compares their formulation in differential versus difference equations.

📝 Lecture Summary

TOPIC 174: COBWEB MODEL: FRAMEWORK

Using first-order difference equations, we can understand price fluctuations of products with lags between production and sale, such as agricultural products where planting and harvest occur with a substantial gap.

The Cobweb model is based on demand and supply forces. The demand side has no lags; a buyer is concerned with the current price of agricultural products. Mathematically: Qd(t) = D(Pt), where current demand depends on current price.

The supply side involves a lag; it takes a season (time period) for a crop to get ready. Supply decisions are based on past prices. Mathematically: Qs(t+1) = S(Pt) — future supply depends on current price. Since Qs(t+1) cannot be equated Qd(t) due to different time periods, we convert Qs(t+1) into Qs(t) by introducing a lag (-1): Qs(t) = S(Pt-1).

🔑 Definition — Unlagged Demand Function: Qd(t) = D(Pt) 🔑 Definition — Lagged Supply Function: Qs(t) = S(Pt-1) 📐 Formula — Unlagged Demand Equation: Qd(t) = a - α(Pt) 📐 Formula — Lagged Supply Equation: Qs(t) = -β + δ(Pt-1)

For equilibrium in the agricultural market: Qd = Qs. Setting Qd(t) = Qs(t): a - α(Pt) = -β + δ(Pt-1) Rearranging: α(Pt) + δ(Pt-1) = a + β Normalizing by dividing by α: Pt + (δ/α)(Pt-1) = (a + β)/α

🔑 Definition — First Order Difference Equation Standard Form: y(t+1) - ay(t) = b

To match the standard form, we introduce a time lead: Pt+1 + (δ/α)(Pt) = (a + β)/α, where a = δ/α, c = (a + β)/α, with yt+1 = Pt+1, yt = Pt, and a ≠ -1.

The definite solution formula for a first-order difference equation with a ≠ -1 is: yt = (y0 - c/(1+a))(-a)^t + c/(1+a)

Applying this to the price path: Pt = {P0 - [(a+β)/α] / [1 + (δ/α)]} (-δ/α)^t + [(a+β)/α] / [1 + (δ/α)] Simplifying: Pt = {P0 - (a+β)/(α+δ)}(-δ/α)^t + (a+β)/(α+δ)

🔑 Definition — Stationary Equilibrium Price (P)**: P = (a+β)/(α+δ)

This gives the final time-path solution: Pt = (P0 - P)(-δ/α)^t + P**

The term A = (P0 - P)* plays two roles:

• Scale Effect: The absolute value |P0 - P*| affects the magnitude of the complementary function Pc = A(-δ/α)^t. If |A|=0, Pc=0; if |A|<1, Pc is minified; if |A|=1, neutral; if |A|>1, amplified.
• Mirror Effect: The sign of (P0 - P*) determines the sign of Pc. If (P0 - P*) > 0, Pc > 0 (retain); if (P0 - P*) < 0, Pc < 0 (reverse).

💡 Why this matters: The cobweb model shows how past prices influence current supply, creating a dynamic price path that may converge to or diverge from equilibrium.

TOPIC 175: COBWEB MODEL: DIAGRAMMATICALLY

The time-path Pt = (P0 - P*)(-δ/α)^t + P* contains the exponential expression (-δ/α)^t which determines the nature and pattern of the time path.

🔑 Definition — Nature: Whether the time-path converges or diverges, determined by |b| = | -δ/α | ⋚ 1. 🔑 Definition — Pattern: Whether the time-path oscillates or not, determined by b = (-δ/α) ⋚ 0.

Since both δ and α are positive (δ, α > 0), the base b = -δ/α is always negative, meaning the pattern is always oscillatory.

The nature depends on relative slopes:

• |b| < 1 (δ < α): Convergent — slope of supply is less than slope of demand. This produces damped oscillations where the price path converges toward equilibrium while oscillating above and below it.
• |b| > 1 (δ > α): Non-convergent (Divergent) — slope of supply is greater than slope of demand. This produces explosive oscillations where the price path diverges away from equilibrium while oscillating.

The left panels show the converging/diverging dynamics in the market with supply and demand curves. The right panels plot the time-path of price (Pt) oscillating around equilibrium price (pe).

📌 Example: Convergent case shows the price oscillating with decreasing amplitude toward equilibrium. Divergent case shows increasing amplitude away from equilibrium.

Conclusion: The nature of the time-path is primarily dependent on the relative slope of demand and supply curves (δ ⋚ α).

TOPIC 176: COBWEB MODEL: RARE CASE

The rare case occurs when δ = α, meaning the slopes of supply and demand are equal.

When δ = α, the base b = -δ/α = -1.

• |b| = |-1| = 1 (neither convergent nor divergent): Non-convergent
• b = -1 < 0: Oscillatory

This produces uniform oscillations where the price uniformly oscillates above and below the equilibrium price by a constant amplitude.

The left panel shows equi-distant dynamics in the market (δ = α). The right panel plots the time-path of price (Pt) uniformly oscillating around equilibrium price (pe).

Conclusion: The existence of the uniform oscillation case is a rare event as the equality of slopes is a matter of chance (δ = α). Any slight inequality can lead to convergence or divergence cases.

TOPIC 177: COBWEB MODEL: NUMERICAL

Given: Qdt = 18 - 3Pt and Qst = -3 + 4Pt-1, find the time-path of price.

Solution: Here, a = 18, α = 3, β = 3, δ = 4.

For equilibrium: Qd(t) = Qs(t) 18 - 3Pt = -3 + 4Pt-1 -3Pt - 4Pt-1 = -21 3Pt + 4Pt-1 = 21 Normalizing by dividing by 3: Pt + (4/3)Pt-1 = 7

Introducing time lead: Pt+1 + (4/3)Pt = 7 Comparing with standard form: a = δ/α = 4/3, c = 7

Since a ≠ -1, using the definite solution formula: Pt = {P0 - c/(1+a)}(-a)^t + c/(1+a) Pt = {P0 - 7 / [1 + (4/3)]}(-4/3)^t + 7 / [1 + (4/3)] Pt = {P0 - 7 / (7/3)}(-4/3)^t + 7 / (7/3) Pt = {P0 - 3}(-4/3)^t + 3

📌 Example: The time-path is Pt = (P0 - 3)(-4/3)^t + 3

Analysis:

• |b| = | -4/3 | = 1.33 > 1: Non-convergent (Divergent)
• b = (-4/3) < 0: Oscillatory

Conclusion: The time-path is divergent in nature and oscillatory in pattern — dynamically unstable.

TOPIC 178: DIFFERENCE EQUATIONS AND ADAPTIVE EXPECTATIONS

Consider a market model with adaptive expectations:

• Qdt = Qst
• Qdt = -a + αPt
• Qst = β - δP*t

Where P*t is the expected price in the current time period t.

🔑 Definition — Adaptive Expectations: Pt = Pt-1 + η(Pt-1 - P*t-1) where 0 < η ≤ 1. The expectation in the current period equals the previous expectation plus a fraction η of the discrepancy between actual and expected price in the previous period.

Effect of Adaptive Price Expectations:

• If (Pt-1 > P*t-1) — price in previous period was greater than expected: under-expectation. Expectations are revised upward.
• If (Pt-1 < P*t-1) — price in previous period was less than expected: over-expectation. Expectations are revised downward.

📌 Example: If Pt-1 = 20, Pt-1 = 10, η = 0.5, then Pt = 20 + 0.5(20-10) = 25 (revised upward). If Pt-1 = 20, Pt-1 = 30, η = 0.5, then Pt = 20 + 0.5(20-30) = 15 (revised downward).

Role of η = 1: When η = 1, the expectation equation becomes: Pt = Pt-1 + 1(Pt-1 - Pt-1) Pt = Pt-1

This means the seller expects the price this year to be the same as last year — the situation observed in the cobweb model.

🔑 Definition — Adaptive Expectations in Differential vs. Difference Equations:

Observation: Both differential and difference equations have the same structure when including adaptive expectations. The adjustment coefficient j in differentials corresponds to η in difference equations, and the discrepancy term (p-π) in differentials corresponds to (Pt-1 - P*t-1) in difference equations.

⭐ Key Takeaways

The cobweb model uses first-order difference equations to model price fluctuations in markets with production lags. The time-path solution is Pt = (P0 - P*)(-δ/α)^t + P*, where convergence occurs when δ < α (supply slope flatter than demand) producing damped oscillations, and divergence when δ > α (supply slope steeper) producing explosive oscillations. The rare case δ = α produces uniform oscillations. In a numerical example with Qd = 18-3Pt and Qs = -3+4Pt-1, the time-path Pt = (P0-3)(-4/3)^t+3 shows divergent oscillations. Finally, adaptive expectations (Pt = Pt-1 + η(Pt-1 - P*t-1)) generalize the cobweb model and have the same structure in both differential and difference equations, with η=1 reducing to the simple cobweb case where current expectations equal last period's actual price.

🧠 Quick Revision Questions

1. Derive the cobweb model equilibrium condition from Qd(t) = a - αPt and Qs(t) = -β + δPt-1, and show how it becomes a first-order difference equation in Pt.

2. Given the solution Pt = (P0 - P*)(-δ/α)^t + P*, explain how the relative values of δ and α determine whether the time-path is convergent, divergent, or uniformly oscillatory.

3. In the numerical example with Qd = 18 - 3Pt and Qs = -3 + 4Pt-1, compute the equilibrium price P* and determine whether the time-path is convergent or divergent.

4. Explain the difference between the scale effect and mirror effect of the coefficient A = (P0 - P*) on the time-path of price.

5. Write the adaptive expectations equation and explain how setting η = 1 simplifies it to the cobweb model's assumption about price expectations.

📘 Lecture 39 — A Market Model with Inventory Using Difference Equations

📖 Overview: This lecture develops a dynamic market model that incorporates inventory, moving beyond the simple Cobweb model which assumes instantaneous market clearing. It introduces a price adjustment mechanism driven by inventory changes, derives the resulting first-order difference equation for the price path, and classifies the possible time paths based on parameter values. This model is crucial for understanding real-world market dynamics where goods can be stored.

🗂️ Topics Covered

The lecture begins by establishing the framework for a market model with inventory, outlining its three key premises. It then derives the solution as a first-order difference equation and obtains the definite time path for price. The core of the lecture is a detailed classification of seven possible time paths based on the value of the characteristic root, with a focus on dynamic stability. Finally, the types of time paths are illustrated graphically and with a numerical example.

📝 Lecture Summary

TOPIC 179: A MARKET MODEL WITH INVENTORY: FRAMEWORK

The Cobweb model assumes a perishable or non-storable product, leading to market clearance each period. However, many real-world products can be stored, necessitating a model that includes inventory.

The model is built on three premises:

Premise I: The current quantity demanded (Q_dt) and quantity supplied (Q_st) are unlagged linear functions of the current price (P_t).

• Q_dt = a - αP_t (a, α > 0)
• Q_st = -β + δP_t (β, δ > 0)

Premise II: Price adjustment is not driven by market clearance but by sellers considering the inventory situation. An accumulated inventory from the previous period leads sellers to drop the current price to move the merchandise. Conversely, a decumulated inventory raises the current price. The price adjustment equation is:

• P_t+1 = P_t - σ(Q_st - Q_dt) (σ > 0)

Premise III: The price adjustment from one period to the next is inversely proportional to the observed change in inventory (stock).

Mathematizing the premises:

• σ is the stock-induced-price-adjustment coefficient.
• The term (Q_st - Q_dt) represents the inventory (stock), while (Q_dt - Q_st) is excess demand. 💡 Why this matters: This framework directly links price changes to inventory levels, providing a more realistic model of price determination than the Cobweb model.

TOPIC 180: A MARKET MODEL WITH INVENTORY: SOLUTION

To find the time path of price P_t, we substitute the demand and supply functions into the price-adjustment function:

• P_t+1 = P_t - σ[(-β + δP_t) - (a - αP_t)]
• Simplifying: P_t+1 - [1 - σ(α + δ)]P_t = σ(a + β)

This is a First Order Difference Equation representing the market model with inventory. Comparing it to the standard form y_t+1 + ay_t = c:

• a = -[1 - σ(α + δ)]
• c = σ(a + β)
• y_t+1 = P_t+1
• y_t = P_t

Using the formula for the definite solution of a first-order difference equation (assuming a ≠ -1):

• P_t = (P₀ - P) [1 - σ(α + δ)]^t + P**

Here, P* = (a + β)/(α + δ) is the stationary equilibrium or intertemporal equilibrium price. The component (P₀ - P)* represents the deviation from equilibrium at time 0. The term b = [1 - σ(α + δ)] is the characteristic root that determines the dynamic behavior of the price path.

🔑 Definition — *Stationary Equilibrium (P)**: The long-run equilibrium price towards which the market price converges if the time path is convergent. It is the price where the difference equation stabilizes.

📐 Formula: P_t = (P₀ - P)b^t + P** → The time path of price is a function of the initial deviation from equilibrium, the characteristic root 'b' raised to the power of 't', and the stationary equilibrium price.

TOPIC 181: A MARKET MODEL WITH INVENTORY: TYPES OF TIME PATH

The time path's nature and pattern are determined by the characteristic root b = [1 - σ(α + δ)]. Seven theoretical regions exist, though two are ruled out by the parametric restriction σ > 0. The dynamic stability depends on the absolute value and sign of 'b'.

Conclusion: Out of the seven theoretical cases, only five (III to VII) are feasible given the constraint σ > 0. The stability of the market model with inventory is entirely governed by the magnitude of the stock-induced-price-adjustment coefficient (σ) relative to the sum of the slopes of the demand and supply curves (α + δ).

TOPIC 182: A MARKET MODEL WITH INVENTORY: GRAPHICALLY

The five feasible regions (III to VII) can be plotted on a graph with σ (the stock-induced-price-adjustment coefficient) on the vertical axis and (α + δ) (the sum of demand and supply curve slopes) on the horizontal axis. Only the first quadrant is relevant as both variables are positive.

The graph maps the conditions from the table above into distinct regions. The lines σ = 1/(α + δ) and σ = 2/(α + δ) are rectangular hyperbolas that serve as boundaries between the different stability regimes. As σ increases relative to (α + δ), the time path moves from convergent non-oscillatory (Region III) to divergent oscillatory (Region VII). 💡 Why this matters: This graph provides a powerful visual tool for determining the dynamic stability of a market with inventory for any given set of parameter values (α, δ, σ).

TOPIC 183: A MARKET MODEL WITH INVENTORY: NUMERICAL

Example #1: Given the demand, supply, and price-adjustment functions:

• Q_dt = 21 - 2P_t (a=21, α=2)
• Q_st = -3 + 6P_t (β=3, δ=6)
• P_t+1 = P_t - 0.3(Q_st - Q_dt) (σ=0.3)

Find the time path P_t and determine its dynamic stability.

Solution:

1. Find the difference equation:

   • P_t+1 = P_t - 0.3[(-3 + 6P_t) - (21 - 2P_t)]
   • Simplifying: P_t+1 + 1.4P_t = 7.2

2. Solve for the definite time path:

   • Using the formula P_t = (P₀ - P)(-1.4)^t + P**, where P = (a + β)/(α + δ) = (21 + 3)/(2 + 6) = 3*.
   • P_t = (P₀ - 3)(-1.4)^t + 3 (This is the definite time path)

3. Determine dynamic stability using 'b':

   • Here, b = 1 - σ(α + δ) = 1 - 0.3(2 + 6) = 1 - 2.4 = -1.4
   • Check the absolute value: |b| = |-1.4| = 1.4 > 1
   • Check the sign: b = -1.4 < 0
   • Conclusion: The time path is Divergent (Unstable) and Oscillatory.
   • This places the model in Region VII.

4. Verify dynamic stability using parameters σ, α, δ:

   • The condition for Region VII (Divergent & Oscillatory) is: σ > 2/(α + δ)
   • Here, σ = 0.3 and 2/(α + δ) = 2/(2 + 6) = 0.25
   • Since 0.3 > 0.25, we reach the same conclusion: Divergent & Oscillatory.

⭐ Key Takeaways

The lecture presents a more realistic market model where price adjusts based on inventory levels, unlike the instant-clearing Cobweb model. The core dynamic equation, P_t+1 - [1 - σ(α + δ)]P_t = σ(a + β), and its solution, P_t = (P₀ - P)b^t + P**, are essential for understanding price evolution. The characteristic root b = [1 - σ(α + δ)] is the sole determinant of the time path's stability and oscillatory nature, leading to five feasible regions. Dynamic stability requires |b| < 1, which translates to a small σ relative to (α + δ), specifically σ < 2/(α + δ) for oscillatory convergence. The numerical example demonstrates how to derive the time path and classify its stability using both 'b' and the parameters directly.

🧠 Quick Revision Questions

1. What are the three key premises of the market model with inventory, and how do they differ from the premises of the Cobweb model?
2. Derive the first-order difference equation for price in this model, starting from the functions Q_dt = a - αP_t, Q_st = -β + δP_t, and P_t+1 = P_t - σ(Q_st - Q_dt). Identify the characteristic root 'b'.
3. State the conditions on σ, α, and δ for the time path to be: (a) convergent and oscillatory, and (b) divergent and oscillatory.
4. In the numerical example, what is the intertemporal equilibrium price P*, and why is the time path classified as divergent and oscillatory?
5. Explain how the graphical representation with σ and (α + δ) helps in determining the dynamic stability of this market model.

📘 Lecture 40 — SECOND ORDER DIFFERENCE EQUATIONS

📖 Overview: This lecture introduces second-order difference equations, focusing on how to find the particular integral (yp) and the complementary function (yc) for these equations. It covers the standard form, homogenous and non-homogenous cases, and the three subcases for finding particular solutions. The lecture also explains how to find the complementary function for distinct real roots, repeated real roots, and complex roots, which are essential for solving economic models involving intertemporal equilibrium.

🗂️ Topics Covered

This lecture covers the standard form of second-order difference equations and the identification of homogenous and non-homogenous cases based on the constant term. It details the three subcases for finding the particular integral (yp) based on the values of coefficients, including the caveats for each subcase. The lecture then explains the complementary function (yc), adapting concepts from differential equations and solving through the characteristic roots, covering distinct real roots, repeated real roots, and complex roots with numerical examples.

📝 Lecture Summary

TOPIC 184: SECOND-ORDER DIFFERENCE EQUATIONS: PARTICULAR INTEGRAL

Similar to second-order differential equations, difference equations can also acquire a second order of difference (recurrence). The second difference of yt, denoted as ∆²yt, is defined as ∆(∆yt). Expanding this, ∆²yt = yt+2 - 2yt+1 + yt.

The standard form of a linear, non-homogeneous second-order difference equation with constant coefficients (a1, a2) and constant term (c) is: yt+2 + a1yt+1 + a2yt = c.

The order is 2nd (due to the 2nd difference), the degree is linear/first degree, and the coefficients are constant. The course deals only with linear cases and does not include cross-terms.

Based on the value of c, there are two main cases: Homogenous Case (c = 0, yp = 0) and Non-Homogenous Case (c ≠ 0, yp ≠ 0). Both cases are further divided into three sub-cases based on the values of a1 and a2.

🔑 Definition — Particular Integral (yp): Any particular solution (containing no arbitrary constants) of the complete equation. yp provides the equilibrium value of yt in an intertemporal sense. Formula: The general solution is yt = yc + yp, where yc is the deviation from equilibrium and yp is the equilibrium value.

Subcase – I: Since yp can be any solution, we first try the simplest, y = k (a constant). Substituting yt = yt+1 = yt+2 = k into the general form yt+2 + a1yt+1 + a2yt = c gives: k + a1k + a2k = c → k(1 + a1 + a2) = c. 📐 Formula: yp = c / (1 + a1 + a2), provided that a1 + a2 ≠ -1.

Subcase – II: If a1 + a2 = -1, the previous formula gives an undefined answer. We then try a non-constant solution, y = kt. Substituting yt = kt, yt+1 = k(t+1), and yt+2 = k(t+2) into the equation and simplifying, using the condition a1 + a2 = -1, we get k[a1 + 2] = c. 📐 Formula: yp = (c / (a1 + 2)) * t, provided that a1 ≠ -2.

Subcase – III: If a1 = -2 (and already a1 + a2 = -1), both previous subcase formulas fail. We then try y = kt². Substituting yt = kt², yt+1 = k(t+1)², and yt+2 = k(t+2)² into the equation and simplifying, using a1 + a2 = -1 and a1 = -2, we get k[2] = c. 📐 Formula: yp = (c / 2) * t². 💡 Why this matters: This hierarchy of "tries" (constant, then t, then t²) ensures we always find a valid particular integral for any non-homogenous second-order difference equation with constant coefficients.

TOPIC 185: SECOND-ORDER DIFFERENCE EQUATIONS: YP NUMERICALS

Non-Homogenous Subcase – I: c ≠ 0, a1 + a2 ≠ -1. 📌 Example: Solve yt+2 - 3yt+1 + 4yt = 12. Here, a1 = -3, a2 = 4, c = 12. Check: a1 + a2 = 1 ≠ -1. Using formula yp = c / (1 + a1 + a2) = 12 / (1 + (-3) + 4) = 12 / 2 = 6.

Non-Homogenous Subcase – II: c ≠ 0, a1 + a2 = -1, a1 ≠ -2. 📌 Example: Solve yt+2 + yt+1 - 2yt = 12. Here, a1 = 1, a2 = -2, c = 12. Check: a1 + a2 = -1, a1 = 1 ≠ -2. Using formula yp = (c / (a1+2)) * t = (12 / (1+2)) * t = 4t.

Non-Homogenous Subcase – III: c ≠ 0, a1 + a2 = -1, a1 = -2. 📌 Example: Solve yt+2 - 2yt+1 + yt = 12. Here, a1 = -2, a2 = 1, c = 12. Check: a1 + a2 = -1, a1 = -2. Using formula yp = (c/2) * t² = (12/2) * t² = 6t².

The summary tables show that for the Homogenous Case (c = 0), the particular integral is yp = 0 for all subcases. For the Non-Homogenous Case (c ≠ 0), the yp values are as calculated above.

TOPIC 186: SECOND-ORDER DIFFERENCE EQUATIONS: YC DISTINCT REAL ROOTS

The Complementary Function (yc) is the general solution (containing arbitrary constants) of the reduced equation. yc provides the deviation of the time path y(t) from its equilibrium value at every point in time. The solution is y(t) = yc + yp.

The form of yc for difference equations is adapted from differential equations, replacing natural exponentials with exponential functions based on roots (b). For second-order equations with distinct real roots (b1 ≠ b2): 📐 Formula: yc = A1(b1)^t + A2(b2)^t.

Nature of Roots: The discriminant (a1² - 4a2) determines the nature. If a1² - 4a2 > 0, roots are distinct real roots. The roots (b1, b2) are found using the quadratic formula. 📐 Formula: b1, b2 = (-a1 ± √(a1² - 4a2)) / 2.

📌 Example: Solve yt+2 + yt+1 - 2yt = 12, given y0 = 4, y1 = 5. From Topic 185, yp = 4t. Calculate roots: a1 = 1, a2 = -2 → a1² - 4a2 = 1 - 4(-2) = 9 > 0. Roots: b1, b2 = (-1 ± √9)/2 = 1 and -2. yc = A1(1)^t + A2(-2)^t. General solution: y(t) = A1 + A2(-2)^t + 4t. Definite Solution: Use initial conditions. For t=0: y0 = 4 = A1 + A2(1) + 0 → A1 + A2 = 4. For t=1: y1 = 5 = A1 + A2(-2) + 4 → A1 - 2A2 = 1. Solving: A2 = 1, A1 = 3. Definite solution: y(t) = 3 + (-2)^t + 4t.

TOPIC 187: SECOND-ORDER DIFFERENCE EQUATIONS: YC REPEATED REAL ROOTS

For second-order equations with repeated real roots (b1 = b2 = b): 📐 Formula: yc = A3(b)^t + A4 * t * (b)^t.

Nature of Roots: If a1² - 4a2 = 0, roots are repeated real roots.

📌 Example: Solve yt+2 + 6yt+1 + 9yt = 4. Here, a1 = 6, a2 = 9, c = 4. Check: a1 + a2 = 15 ≠ -1, so Subcase I applies. yp = c / (1 + a1 + a2) = 4 / (1 + 6 + 9) = 4/16 = 1/4. Calculate roots: a1² - 4a2 = 36 - 36 = 0. Root: b = -a1/2 = -6/2 = -3. yc = A3(-3)^t + A4 * t * (-3)^t. General solution: y(t) = A3(-3)^t + A4 * t * (-3)^t + 1/4. 💡 Why this matters: The inclusion of the "t" factor in the repeated roots case changes the dynamics of the time path, leading to a different convergence behavior compared to distinct roots.

TOPIC 188: SECOND-ORDER DIFFERENCE EQUATIONS: YC COMPLEX ROOTS

For second-order equations with complex roots (h ± ui): 📐 Formula: yc = R^t (A5 cos θt + A6 sin θt).

Nature of Roots: If a1² - 4a2 < 0, roots are complex roots. Here, h = -a1/2 and u = √(4a2 - a1²)/2 are defined. The modulus R (or amplitude) is calculated as R = √(h² + u²), and the angle θ is found via sin θ = u/R or cos θ = h/R.

📌 Example: Solve yt+2 + 0*yt+1 + (1/4)yt = 5. (Note: "self generated" in text indicates a1=0) Here, a1 = 0, a2 = 1/4, c = 5. Check: a1 + a2 = 1/4 ≠ -1, so Subcase I applies. yp = c / (1 + a1 + a2) = 5 / (1 + 0 + 1/4) = 5 / (5/4) = 4. Calculate roots: a1² - 4a2 = 0 - 4(1/4) = -1 < 0. h = -a1/2 = 0, u = √(4a2 - a1²)/2 = √(1 - 0)/2 = 1/2. R = √(h² + u²) = √(0 + (1/2)²) = 1/2. θ: sin θ = u/R = (1/2) / (1/2) = 1 → θ = 90° (or π/2 radians). yc = (1/2)^t (A5 cos(π/2 * t) + A6 sin(π/2 * t)). General solution: yt = (1/2)^t {A5 cos(π/2 * t) + A6 sin(π/2 * t)} + 4.

⭐ Key Takeaways

The core of solving second-order difference equations is decomposing the solution into a particular integral (yp) for the equilibrium and a complementary function (yc) for deviations from equilibrium. The formula for yp depends on three subcases, determined by the coefficients a1 and a2: if a1 + a2 ≠ -1, use the constant formula; if not, and a1 ≠ -2, use the t formula; otherwise, use the t² formula. The form of yc is determined by the nature of the roots from the characteristic equation, leading to distinct real, repeated real, or complex root solutions, which dictate the dynamic behavior of the time path. Finally, initial conditions are used to find the definite solution by solving for the arbitrary constants in the general solution.

🧠 Quick Revision Questions

1. What is the standard form of a second-order linear difference equation with constant coefficients?
2. What are the three subcases for finding the particular integral (yp), and what conditions define each subcase?
3. How do you determine the nature of the roots (distinct real, repeated real, complex) for a second-order difference equation?
4. What is the formula for the complementary function (yc) in the case of repeated real roots?
5. In the complex roots case, how are the modulus R and the angle θ defined and used in the yc formula?

📘 Lecture 41 — Dynamic Stability: Distinct Real Roots

📖 Overview: This lecture explores the concept of dynamic stability in difference equations, focusing on the conditions under which a system converges to equilibrium over time. It covers three distinct cases: distinct real roots, repeated real roots, and complex roots, each with its own stability criteria. Understanding these stability conditions is crucial for predicting the long-run behavior of economic models.

🗂️ Topics Covered

The lecture covers dynamic stability for distinct real roots, repeated real roots, and complex roots, with numerical examples for each case. It explains how the absolute values of the roots determine whether the complementary function (deviation from equilibrium) converges to zero, leading to stability, or diverges, leading to instability. The "hair splitting" approach is used to analyze the behavior of the exponential terms.

📝 Lecture Summary

TOPIC 189: DYNAMIC STABILITY: DISTINCT REAL ROOTS

For the distinct real roots case, the general solution of a difference equation is given by:

(y_t = A_1(b_1)^t + A_2(b_2)^t + y_p)

where the complementary function, (y_c = A_1(b_1)^t + A_2(b_2)^t), represents the deviation from the equilibrium (y_p). (A_1) and (A_2) are constants determined by initial conditions, and (t) is the independent variable spanning ((0, \infty)). The dynamic stability of the system depends entirely on the roots (b_1) and (b_2).

🔑 Definition — Dynamic Stability: A system is dynamically stable if, over time, the complementary function ((y_c)) converges to zero, meaning the deviation from equilibrium fades away. This occurs when both roots ((b_1) and (b_2)) are negative, causing the exponential terms to decay.

The "hair splitting" analysis shows that if both (b_1) and (b_2) are negative, the exponential terms (A_1(b_1)^t) and (A_2(b_2)^t) will fade away due to exponential decay. Consequently, their sum, (y_c), also fades away. A caveat is that both (b_1) and (b_2) must be negative; otherwise, convergence may not occur.

📐 Formula: Dynamic stability condition for distinct real roots: (|b_1| < 1) and (|b_2| < 1).

📌 Example: The lecture provides numerical examples to illustrate the concept.

• Example 1: (y(t) = 3(1)^t + (2)^t + 4t). Here, (|b_1| = 1 \not< 1) and (|b_2| = 2 \not< 1). The first term remains constant, and the second term increases; the deviation increases, leading to Divergence and Dynamic Instability.
• Example 2: (y(t) = 3(0.5)^t + (-2)^t + 4t). Here, (|b_1| = 0.5 < 1) but (|b_2| = 2 \not< 1). The first term decreases, but the second term increases; the deviation increases, leading to Divergence and Dynamic Instability.
• Example 3: (y(t) = 3(-0.8)^t + (0.1)^t + 4t). Here, (|b_1| = 0.8 < 1) and (|b_2| = 0.1 < 1). Both terms decrease; the deviation decreases, leading to Convergence and Dynamic Stability.

This confirms that for dynamic stability, the absolute value of both roots must be less than 1: ((|b_1| \text{ and } |b_2|) < 1).

TOPIC 190: DYNAMIC STABILITY: REPEATED REAL ROOTS

For the repeated real roots case, the solution is:

(y_t = A_3(b)^t + A_4 t (b)^t + y_p)

The complementary function is (y_c = A_3(b)^t + A_4 t (b)^t). The deviation from equilibrium depends on the single repeated root (b). (A_3) and (A_4) are constants.

🔑 Definition — Repeated Real Roots Case: A condition in difference equations where the characteristic equation has a single real root with multiplicity, requiring a modified solution form that includes a term multiplied by (t).

The "hair splitting" analysis shows that if (b) is negative, the exponential terms (A_3(b)^t) and (A_4 t(b)^t) will fade away. Although the second term has an additional (t) multiplier, it does not reverse the dynamics of the exponential decay. The sum of the two terms will also converge to zero.

📐 Formula: Dynamic stability condition for repeated real roots: (|b| < 1).

📌 Example: Numerical examples are provided.

• Example 1: (y = 2(3)^t + t(3)^t + \frac{1}{4}t). Here, (|b| = 3 \not< 1). Both exponential terms increase, the deviation increases, and the system is Dynamically Unstable (Divergence).
• Example 2: (y = 2(0.3)^t + t(0.3)^t + \frac{1}{4}t). Here, (|b| = 0.3 < 1). Both exponential terms decrease, the deviation decreases, and the system is Dynamically Stable (Convergence).

This confirms that for dynamic stability, the absolute value of the repeated root (b) must be less than 1: (|b| < 1).

TOPIC 191: DYNAMIC STABILITY: COMPLEX ROOTS

For the complex roots case, the solution involves trigonometric functions:

(y_t = R^t (A_5 \cos \theta t + A_6 \sin \theta t) + y_p)

The complementary function is (y_c = R^t (A_5 \cos \theta t + A_6 \sin \theta t)). The deviation from equilibrium depends on the modulus (R) and the angle (\theta) of the complex roots.

🔑 Definition — Complex Roots Case: A condition in difference equations where the characteristic equation has a pair of complex conjugate roots, leading to a solution that involves sinusoidal functions and a multiplicative factor (R^t).

The "hair splitting" analysis shows that the multiplicand ((A_5 \cos \theta t + A_6 \sin \theta t)) produces a "stepped fluctuation" with a wave-like graph of uniform amplitude. Convergence depends on the multiplier (R^t). If (R < 1), then (R^t \rightarrow 0) as (t \rightarrow \infty), damping the fluctuations and leading to convergence.

📐 Formula: Dynamic stability condition for complex roots: (R < 1).

📌 Example: The lecture provides three numerical examples.

• Example 1: (y_t = (2)^t { \cos(1)t + 3 \sin(1)t } + 4). Here, the modulus (R = 2 \not< 1). The multiplier (R^t) increases, creating an Explosive Fluctuation and Divergence; the system is Dynamically Unstable.
• Example 2: (y_t = (1)^t { \cos(1)t + 3 \sin(1)t } + 4). Here, the modulus (R = 1). The multiplier (R^t) is constant, creating a Uniform Fluctuation that is neither divergent nor convergent; the system is Neither Dynamically Stable nor Unstable.
• Example 3: (y_t = (\frac{1}{2})^t { \cos(1)t + 3 \sin(1)t } + 4). Here, the modulus (R = 0.5 < 1). The multiplier (R^t) decreases, creating a Damped Fluctuation and Convergence; the system is Dynamically Stable.

This confirms that for dynamic stability in the complex root case, the condition (R < 1) must hold.

⭐ Key Takeaways

A student must remember that dynamic stability for all three cases depends on the absolute value (or modulus for complex roots) of the characteristic roots. For distinct real roots, both (|b_1|) and (|b_2|) must be less than 1 for stability. For repeated real roots, (|b|) must be less than 1. For complex roots, the modulus (R) must be less than 1. The complementary function (y_c), which represents the deviation from equilibrium, must converge to zero as time (t) approaches infinity. If any root's absolute value or modulus is greater than or equal to 1, the system will exhibit divergence or non-convergent fluctuations, making it dynamically unstable. The numerical examples confirm that the "exponential decay" of terms like ((b)^t) is the core mechanism driving convergence when (|b|<1).

🧠 Quick Revision Questions

1. What is the general condition for dynamic stability when a difference equation has two distinct real roots?
2. In the repeated real roots case, why does the additional 't' factor in the term (A_4 t (b)^t) not prevent dynamic stability if (|b|<1)?
3. For a system with complex roots, what does it mean if the modulus (R) is exactly 1?
4. In the distinct real roots numerical example, why does the system (y(t) = 3(0.5)^t + (-2)^t + 4t) diverge despite (|0.5|<1)?
5. What is the key difference in the solution form between the distinct real roots case and the repeated real roots case?

📘 Lecture 42 — Samuelson Multiplier-Acceleration Interaction Model

📖 Overview: This lecture introduces Samuelson's Multiplier-Acceleration Interaction Model, a seminal application of second-order difference equations in economics. It explores how the interaction of the Keynesian multiplier and the acceleration principle can generate endogenous cyclical fluctuations in national income, without relying on external shocks.

🗂️ Topics Covered

The lecture is structured into three main topics: the framework of the Samuelson Multiplier-Acceleration Interaction Model, its solution using second-order difference equations, and numerical examples to determine dynamic stability. The framework defines consumption, induced investment, and exogenous government expenditure. The solution derives the particular integral and complementary function, analyzing cases for distinct real, repeated real, and complex roots. Numerical examples apply the characteristic root formula to assess time-path convergence or divergence based on parameter values.

📝 Lecture Summary

TOPIC 192: SAMUELSON MULTIPLIER-ACCELERATION INTERACTION MODEL: FRAMEWORK

The Samuelson Multiplier-Acceleration Interaction Model, attributed to Professor Paul A. Samuelson, is an application of second-order difference equations in economics. It seeks to explore the dynamic process of income determination when the acceleration principle operates alongside the Keynesian multiplier. The model shows that the mere interaction of the multiplier and accelerator can generate cyclical fluctuations endogenously.

The model presents a self-reinforcing cycle. An initial increase in autonomous investment (∆Iₒ) works through the multiplier to increase income (∆Y). The increase in income, through investors' expectations regarding consumption (MPC), triggers the accelerator, leading to an increase in induced investment (∆Iᵢ). This induced investment again works through the multiplier to further increase income (∆Y), continuing the process in a boom or recovery. The process works identically in reverse for a recession or depression, with decreases cascading through the system.

National income (Yₜ) has three component expenditure streams: consumption (Cₜ), investment (Iₜ), and government expenditure (Gₜ).

Consumption (Cₜ): Consumption is a function not of current income (Yₜ) but of the income of the prior period (Yₜ₋₁). For simplicity, it is assumed to be strictly proportional to Yₜ₋₁:

🔑 Definition — Cₜ: Cₜ = β Yₜ₋₁ where (0 < β < 1) and β is actually the MPC (Marginal Propensity to Consume), with no intercept term.

Investment (Iₜ): Induced investment (Iᵢ) is a function of the prevailing trend of consumer spending (Cₜ). An increase in income (∆Y) works through the multiplier, and then through the MPC, it triggers the acceleration mechanism in the model. It is assumed that investment is induced based on consumption patterns.

The formula for induced investment is:

📐 Formula — Induced Investment: Iₜ = a (Cₜ − Cₜ₋₁) = a (∆Cₜ₋₁) Plain-English Meaning: Investment in period t is equal to the acceleration coefficient (a) multiplied by the change in consumption from the previous period. Economic Significance of a: The parameter a is the Acceleration Coefficient. It is considered a fixed ratio, where a = Iₜ / (Cₜ − Cₜ₋₁) = Iₜ / ∆Cₜ, and a > 0.

Government Expenditure (Gₜ): The third component is government expenditure (Gₜ), which is considered exogenous (Gₒ).

Therefore, the framework of the Samuelson Multiplier-Acceleration Interaction Model is composed of the following system of equations:

1. Cₜ = β Yₜ₋₁
2. Iₜ = a (Cₜ − Cₜ₋₁)
3. Gₜ = G₀

💡 Why this matters: This framework lays the foundation for modeling how consumption and investment decisions based on past behavior can create complex, cyclical dynamics in the entire economy.

TOPIC 193: SAMUELSON MULTIPLIER-ACCELERATION INTERACTION MODEL: SOLUTION

Borrowing the framework, the solution involves combining all three components in the expenditure approach equation: Yₜ = Cₜ + Iₜ + Gₜ.

Substituting the consumption and investment equations into the national income identity: Yₜ = βYₜ₋₁ + a(Cₜ − Cₜ₋₁) + Gₒ

By introducing a lag, we get Cₜ = βYₜ₋₁ and Cₜ₋₁ = βYₜ₋₂. Substituting these into the income equation yields: Yₜ = βYₜ₋₁ + a(βYₜ₋₁ − βYₜ₋₂) + Gₒ Yₜ = βYₜ₋₁ + aβYₜ₋₁ − aβYₜ₋₂ + Gₒ Yₜ = (β + aβ)Yₜ₋₁ − aβYₜ₋₂ + Gₒ Yₜ − (β + aβ)Yₜ₋₁ + aβYₜ₋₂ = Gₒ

Introducing a lead of 2 on both sides transforms this into a standard second-order difference equation: Yₜ₊₂ − (β + aβ)Yₜ₊₁ + aβYₜ = Gₒ

This is comparable with the standard second-order difference equation: yₜ₊₂ + a₁yₜ₊₁ + a₂yₜ = c. Here: y = Y, a₁ = −(β + aβ), a₂ = aβ, and c = Gₒ.

Particular Integral (Yₚ): The particular integral depends on a₁ and a₂. Since the sum a₁ + a₂ = −(β + aβ) + aβ = −β ≠ −1, we can use the standard formula:

📐 Formula — Particular Integral: yₚ = c / (1 + a₁ + a₂) Substituting the values: Yₚ = Gₒ / (1 − (β + aβ) + aβ) Yₚ = Gₒ / (1 − β − aβ + aβ) Yₚ = Gₒ / (1 − β) This gives the intertemporal equilibrium of national income (Y*).

Complementary Solution (Yc): The complementary solution depends on the discriminant (a₁² ⋛ 4a₂).

Case – I: Distinct Real Roots (a₁² > 4a₂) Condition: {−(β + aβ)}² > 4(aβ) → (β + aβ)² > 4aβ → β > 4aβ / (β + aβ)²

Case – II: Repeated Real Roots (a₁² = 4a₂) Condition: {−(β + aβ)}² = 4(aβ) → (β + aβ)² = 4aβ → β = 4aβ / (β + aβ)²

Case – III: Complex Roots (a₁² < 4a₂) Condition: {−(β + aβ)}² < 4(aβ) → (β + aβ)² < 4aβ → β < 4aβ / (β + aβ)²

💡 Why this matters: These conditions, expressed in terms of the acceleration coefficient (a) and the MPC (β), determine the qualitative nature of the time path (e.g., monotonic, oscillatory, or cyclical) without needing to compute the full solution.

TOPIC 194: SAMUELSON MULTIPLIER-ACCELERATION INTERACTION MODEL: NUMERICALS

Under the Samuelson Multiplier-Acceleration Interaction Model, the characteristic equation is similar to the standard form of a second-order difference equation.

The reduced form is: yₜ₊₂ + a₁yₜ₊₁ + a₂yₜ = 0 Using trial solutions yₜ = Abᵗ, we get the characteristic equation: b² + a₁b + a₂ = 0

The characteristic roots are:

📐 Formula — Characteristic Roots: b₁, b₂ = (−a₁ ± √(a₁² − 4a₂)) / 2

Recalling a₁ and a₂ from the Samuelson model, with a₁ = −(β + aβ) and a₂ = aβ: b₁, b₂ = (−{−(β + aβ)} ± √({−(β + aβ)}² − 4(aβ))) / 2

📌 Formula — Samuelson Model-Specific Characteristic Roots: b₁, b₂ = (β(1 + a) ± √(β²(1 + a)² − 4aβ)) / 2

Numerical # 1: For the Samuelson Multiplier-Acceleration Interaction Model, if a = 3.5 and β = 0.8, find the characteristic roots and judge the dynamic stability of the time-path.

b₁, b₂ = (0.8(1 + 3.5) ± √(0.8²(1 + 3.5)² − 4(3.5)(0.8))) / 2 b₁, b₂ = (3.6 ± √(1.76)) / 2 b₁, b₂ = (3.6 ± 1.3266) / 2 b₁ = (3.6 + 1.3266) / 2 = 2.46 b₂ = (3.6 − 1.3266) / 2 = 1.14 Interpretation: Both |b₁| > 1 and |b₂| > 1, therefore, the interaction time-path shall be divergent and hence dynamically unstable.

Numerical # 2: For the Samuelson Multiplier-Acceleration Interaction Model, if a = 0.2 and β = 0.9, find the characteristic roots and judge the dynamic stability of the time-path.

b₁, b₂ = (0.9(1 + 0.2) ± √(0.9²(1 + 0.2)² − 4(0.2)(0.9))) / 2 b₁, b₂ = (1.08 ± √(0.4464)) / 2 b₁, b₂ = (1.08 ± 0.6681) / 2 b₁ = (1.08 + 0.6681) / 2 = 0.87 b₂ = (1.08 − 0.6681) / 2 = 0.21 Interpretation: Both |b₁| < 1 and |b₂| < 1, therefore, the interaction time-path shall be convergent and hence dynamically stable.

⭐ Key Takeaways

The Samuelson Multiplier-Acceleration Interaction Model demonstrates that the dynamic interaction between the multiplier (driven by MPC) and the accelerator (driven by the investment-consumption relationship) can generate endogenous business cycles. The model is formalized as a second-order difference equation, Yₜ₊₂ − (β + aβ)Yₜ₊₁ + aβYₜ = Gₒ, with the intertemporal equilibrium income given by Y* = Gₒ/(1−β). The dynamic stability of the time path is determined by the characteristic roots; if the absolute values of both roots are less than one, the path converges to equilibrium (stable), and if greater than one, it diverges (unstable). The numerical examples confirm that higher values of the accelerator (a) and MPC (β) tend to produce divergence, while lower values lead to convergence.

🧠 Quick Revision Questions

1. What are the three main equations that form the framework of the Samuelson Multiplier-Acceleration Interaction Model?
2. How is the induced investment function (Iₜ) expressed in terms of consumption (Cₜ) and the acceleration coefficient (a)?
3. What is the particular integral (Yₚ) in this model, and what does it represent economically?
4. What condition on the absolute values of the characteristic roots (b₁ and b₂) indicates a dynamically stable time path?
5. In Numerical #1 (a=3.5, β=0.8), what were the computed characteristic roots, and what was the conclusion about the time-path's stability?

📘 Lecture 43 — SECOND ORDER DIFFERENCE EQUATION WITH VARIABLE TERM

📖 Overview: This lecture extends second-order difference equations from constant-term to variable-term cases, introducing the Method of Undetermined Coefficients. It covers both exponential and power function variable terms, then extends analysis to higher-order difference equations and introduces Schur Theorem for testing dynamic stability without solving characteristic equations.

🗂️ Topics Covered

The lecture begins with second-order difference equations with variable terms, specifically exponential functions (cₘᵗ) and power functions (ctⁿ). It then extends to higher-order linear difference equations (n>2), covering standard forms, complementary functions with repeated and distinct real roots, and hybrid solutions. Finally, it introduces dynamic stability analysis through Schur Theorem determinants and provides numerical examples for convergence testing.

📝 Lecture Summary

SECOND-ORDER DIFFERENCE EQUATION WITH VARIABLE TERM

When the right-hand side of a second-order difference equation contains a variable term {f(t)} instead of a constant c, we use the Method of Undetermined Coefficients. This method requires that the variable term and its successive differences together contain only distinct types of expressions.

🔑 Definition — Method of Undetermined Coefficients: A technique for finding the particular integral of a difference equation with a variable term by assuming a trial solution of the same functional form as the variable term, then determining coefficients by substitution.

For exponential terms like 7ᵗ:

• yₜ = 7ᵗ
• Δyₜ = 6(7ᵗ)
• Δ²yₜ = 36(7ᵗ)

TOPIC 196: SECOND-ORDER DIFFERENCE EQUATION WITH VARIABLE TERM (CTN)

When the variable term is a power function (ctⁿ) rather than an exponential, the method adapts accordingly. For quadratic terms t²:

• yₜ = t²
• Δyₜ = 2t + 1
• Δ²yₜ = 2

📐 Pattern recognition: For exponential terms, assume yₜ = B(mᵗ); for power functions of degree n, assume a polynomial of degree n.

📌 Example 1 (Exponential): Find particular integral of yₜ₊₂ + yₜ₊₁ − 3yₜ = 7ᵗ

• Assume yₜ = B(7ᵗ)
• Then yₜ₊₁ = B(7ᵗ⁺¹) and yₜ₊₂ = B(7ᵗ⁺²)
• Substituting: B(7ᵗ⁺²) + B(7ᵗ⁺¹) − 3B(7ᵗ) = 7ᵗ
• B(49 + 7 − 3)(7ᵗ) = 7ᵗ
• 58B(7ᵗ) = 7ᵗ → B = 1/58
• Therefore, yₚ = (1/58)(7ᵗ)

📌 Example 2 (Quadratic): Find particular integral of yₜ₊₂ + 5yₜ₊₁ + 2yₜ = t²

• Assume yₜ = B₁t² + B₂t + B₃
• Generate yₜ₊₁ and yₜ₊₂, substitute, and compare coefficients of t², t, and constants
• Result: B₁ = 1/8, B₂ = −7/32, B₃ = 13/256
• Therefore, yₚ = (1/8)t² − (7/32)t + 13/256

💡 Why this matters: The complementary function (yₜ) remains unchanged as it depends on b² + a₁b + a₂ = 0; only the particular integral changes with the variable term.

TOPIC 197: HIGHER-ORDER LINEAR DIFFERENCE EQUATION

A linear, non-homogeneous higher-order difference equation with n constant coefficients has the standard form: a₀yₜ₊ₙ + a₁yₜ₊₍ₙ₋₁₎ + a₂yₜ₊₍ₙ₋₂₎ + ... + aₙ₋₁yₜ₊₁ + aₙyₜ = c

Key properties:

• Order: nᵗʰ (due to nᵗʰ difference)
• Degree: Linear/First degree
• An nᵗʰ order implies n number of characteristic roots
• A hybrid complementary solution can exist combining repeated real roots and complex roots

📌 Example (Third-order): Find general solution of yₜ₊₃ − (7/8)yₜ₊₂ + (1/8)yₜ₊₁ + (1/32)yₜ = 9

• Test Case I: a₁ + a₂ + a₃ = −7/8 + 1/8 + 1/32 = −23/32 ≠ −1
• Particular integral: yₚ = c/(1 + a₁ + a₂ + a₃) = 9/(1 − 23/32) = 32
• Characteristic equation: b³ − (7/8)b² + (1/8)b + 1/32 = 0
• Roots: b₁ = 1/2, b₂ = 1/2 (repeated), b₃ = −1/8
• Complementary function: yₜ = A₁(1/2)ᵗ + A₂t(1/2)ᵗ + A₃(−1/8)ᵗ
• General solution: yₜ = A₁(1/2)ᵗ + A₂t(1/2)ᵗ + A₃(−1/8)ᵗ + 32

🔑 Definition — Dynamic Stability: For convergence, all characteristic roots must be less than 1 in absolute value. Here |b₁| = |b₂| = 1/2 < 1 and |b₃| = 1/8 < 1, confirming convergence.

TOPIC 198: CONVERGENCE AND THE SCHUR THEOREM

The Schur Theorem provides a qualitative method for testing dynamic stability without solving for all n roots. Determinants are developed using coefficients in the difference equation.

🔑 Definition — Schur Theorem: For an nᵗʰ order difference equation to be dynamically stable (convergent), all determinants ∆₁, ∆₂, ..., ∆ₙ must be positive.

For second-order difference equation: ∆₁ and ∆₂ must both be > 0 for convergence.

📐 Determinant formulas for second-order:

• ∆₁ = |a₀ a₂; a₂ a₀|
• ∆₂ = |a₀ 0 a₂ a₁; a₁ a₀ 0 a₂; a₂ 0 a₀ a₁; a₁ a₂ 0 a₀|

📌 Example 1 (Divergent): yₜ₊₂ + 3yₜ₊₁ + 2yₜ = 12

• a₀ = 1, a₁ = 3, a₂ = 2
• ∆₁ = |1 2; 2 1| = 1 − 4 = −3 < 0
• ∆₂ = 0
• Since ∆₁ < 0 and ∆₂ = 0, the system is DIVERGENT
• Verification: roots are −1 and −2, both with absolute value ≥ 1

📌 Example 2 (Convergent): yₜ₊₂ + (1/6)yₜ₊₁ − (1/6)yₜ = 2

• a₀ = 1, a₁ = 1/6, a₂ = −1/6
• ∆₁ = |1 −1/6; −1/6 1| = 1 − 1/36 = 35/36 > 0
• ∆₂ = 49/54 > 0
• Since both determinants are positive, the system is CONVERGENT

⚠️ Limitations: (1) Dynamic stability is investigated but characteristic roots remain unknown; (2) For n > 2, determinants become huge, contrary to expectation of ease.

⭐ Key Takeaways

The Method of Undetermined Coefficients is the essential technique for solving second-order difference equations with variable terms—exponential terms require assuming B(mᵗ) while power functions require polynomials of matching degree. For higher-order equations (n > 2), the solution structure extends from second-order: particular integrals follow the same case formulas, complementary functions involve n characteristic roots, and hybrid solutions combine repeated real roots with distinct or complex roots. The Schur Theorem offers a powerful qualitative shortcut for testing dynamic stability without solving for all n roots—if all determinants are positive, the time path converges—though this method becomes impractical for very high orders due to determinant size.

🧠 Quick Revision Questions

1. What is the precondition for applying the Method of Undetermined Coefficients to a variable-term difference equation?

2. For the equation yₜ₊₂ + 2yₜ₊₁ − 5yₜ = 3ᵗ, what trial solution form should be assumed for the particular integral?

3. In Schur Theorem testing for a second-order difference equation, what must be true about ∆₁ and ∆₂ for the system to be convergent?

4. How does the complementary function (yₜ) for a third-order difference equation differ when there are two repeated real roots and one distinct real root?

5. For the third-order equation yₜ₊₃ − (7/8)yₜ₊₂ + (1/8)yₜ₊₁ + (1/32)yₜ = 9, what is the value of the particular integral?

📘 Lecture 44 — LINEAR AND NON-LINEAR PROGRAMMING

📖 Overview: This lecture introduces linear and nonlinear programming techniques for optimization in economics. It covers the three-step process of mathematizing, graphing, and selecting optimal solutions for both profit maximization and cost minimization problems, and extends to nonlinear problems using Kuhn-Tucker conditions.

🗂️ Topics Covered

The lecture covers four main topics: linear programming for profit maximization using a glassware making example, linear programming for cost minimization using an insurance company employment problem, nonlinear programming for utility maximization using Kuhn-Tucker conditions, and nonlinear programming for cost minimization. It demonstrates the full process from problem formulation to solution selection, including the need for integer programming when dealing with whole units like workers.

📝 Lecture Summary

TOPIC 199: LINEAR PROGRAMMING: GLASSWARE MAKING

Linear programming addresses real-life problems of allocating finite resources such as labor, machine time, or raw materials to maximize profits or minimize costs. The process involves mathematizing given information into Decision Variables, an Objective Function, and (multiple) Constraints. It only deals with linear objective and constraint functions and is generally done in three steps.

Numerical Example: A glassware firm produces bowls (x) and plates (y). Profit per bowl is $150, per plate is $100. Resources are limited: glassblowing time (70 hours/week, bowls take 2 hours, plates take 1 hour), annealing time (130 hours/week, bowls take 4 hours, plates take 1 hour), and silica sand (45 kg, each product uses 1 kg). Production cannot be negative.

🔑 Definition — Decision Variables: Quantities to be determined in the optimization problem (x = number of bowls, y = number of plates). 📐 Formula: Objective Function → π = 150x + 100y (to be maximized) 📐 Formula: Constraints: 2x + y ≤ 70 (glassblowing); 4x + y ≤ 130 (annealing); x + y ≤ 45 (silica sand); x ≥ 0; y ≥ 0 📌 Example: After plotting constraints and finding the feasible region, corner points are evaluated: (0,0) profit=$0; (0,45) profit=$4,500; (25,20) profit=$5,750; (30,10) profit=$5,500; (32.5,0) profit=$4,875. The optimal solution is (x, y) = (25, 20)** producing maximum profit of $5,750.

TOPIC 200: LINEAR PROGRAMMING: EMPLOYMENT COST MINIMIZATION

Numerical Example: An insurance company employs full-time staff (x) who work 40 hours/week for $800/week, and part-time staff (y) who work 20 hours/week for $320/week. Part-time staff must not exceed one-third of full-time staff. Total work required is at least 900 worker-hours per week.

🔑 Definition — Objective Function: C = 800x + 320y (to be minimized) 📐 Formula: Constraints: 40x + 20y ≥ 900 (worker hours); y ≤ (1/3)x (proportion); x ≥ 0; y ≥ 0 📌 Example: Corner points evaluated: (0,0) cost=$0; (19.3, 6.4) cost=$17,485.7; (22.5,0) cost=$18,000. The optimal solution is (x, y) = (19.3, 6.4)** with minimum cost of $17,485.7.

Integer Programming: Since workers cannot be fractions, whole numbers are tested. (20,5) gives cost=$17,600, which is the minimum integer solution, though higher than the fractional optimum.

TOPIC 201: NONLINEAR PROGRAMMING: UTILITY MAXIMIZATION

For nonlinear objective and constraint functions, Nonlinear Programming is used. It uses Kuhn-Tucker conditions (1951), which are similar to first-order conditions.

🔑 Definition — Kuhn-Tucker Conditions for Maximization: For j variables: ∂Z/∂xj ≤ 0, xj ≥ 0, xj(∂Z/∂xj) = 0 (complementary slackness). For i multipliers: ∂Z/∂λi ≥ 0, λi ≥ 0, λi(∂Z/∂λi) = 0.

Numerical Example: Maximize U = xy subject to x + y ≤ 100, x ≤ 40, and x,y ≥ 0.

🔑 Definition — Lagrangian Function: Z(x,y,λ₁,λ₂) = xy + λ₁(100 - x - y) + λ₂(40 - x) 📌 Example: Trial solution Case I with λ₁,λ₂ > 0 gives x=40, y=60. Using complementary slackness: since x>0, ∂Z/∂x=0 gives y-λ₁-λ₂=0; since y>0, ∂Z/∂y=0 gives x-λ₁=0 → λ₁=40, λ₂=20. Verification: U = 40×60 = 2400; x+y=100≤100; x=40≤40. Solution: (x, y) = (40, 60)** with maximum utility 2,400.

TOPIC 202: NONLINEAR PROGRAMMING: COST MINIMIZATION

🔑 Definition — Kuhn-Tucker Conditions for Minimization: For j variables: ∂Z/∂xj ≥ 0, xj ≥ 0, xj(∂Z/∂xj) = 0. For i multipliers: ∂Z/∂λi ≤ 0, λi ≥ 0, λi(∂Z/∂λi) = 0.

Numerical Example: Minimize C = (x₁-4)² + (x₂-4)² subject to 2x₁+3x₂ ≥ 6, -3x₁-2x₂ ≥ -12, and x₁,x₂ ≥ 0.

📐 Formula: Lagrangian Function: Z = (x₁-4)² + (x₂-4)² + λ₁(6-2x₁-3x₂) + λ₂(-12+3x₁+2x₂) 📌 Example: Trial Solution Case I (λ₁,λ₂>0) gives x₂=-6/5, violating nonnegativity. Trial Solution Case II (x₁,x₂>0) with λ₁=0 gives x₁=28/13, x₂=36/13, λ₂=16/13. Verification: C = 68/13; 2(28/13)+3(36/13)=12.6≥6; -3(28/13)-2(36/13)=-12≥-12. Solution: (x₁, x₂) = (28/13, 36/13)** with minimum cost 68/13.

💡 Why this matters: Kuhn-Tucker conditions provide the necessary conditions for optimality in constrained nonlinear optimization problems, extending the method of Lagrange multipliers to handle inequality constraints through complementary slackness.

⭐ Key Takeaways

Linear programming solves allocation problems with linear objective and constraints through three steps: mathematizing, graphing, and selecting corner points of the feasible region. When fractional solutions don't make sense for whole units, integer programming tests nearby whole numbers. For nonlinear problems, Kuhn-Tucker conditions replace standard first-order conditions and require testing trial cases for variables and multipliers using complementary slackness. The key difference between maximization and minimization Kuhn-Tucker conditions is the direction of inequalities for partial derivatives of the Lagrangian function.

🧠 Quick Revision Questions

1. What are the three steps in solving a linear programming problem?
2. In the glassware maximization problem, why was (25,20) chosen over (30,10) even though both satisfy constraints?
3. What is the purpose of integer programming, and how does it relate to linear programming?
4. How do Kuhn-Tucker conditions for maximization differ from those for minimization in terms of inequality direction?
5. In the utility maximization example, what does complementary slackness mean when x>0?