Difference and Differential Equations with Applications | UGC NET Economics – Notes

Introduction

While static optimisation and comparative statics analyse economic equilibria at a single point in time, dynamic analysis is concerned with the path of adjustment of economic variables over time, and with whether such variables converge to, diverge from, or oscillate around an equilibrium value. The two principal mathematical tools used for this purpose in economics are difference equations, which describe the behaviour of variables that change at discrete time intervals (such as year-to-year or period-to-period), and differential equations, which describe the behaviour of variables that change continuously over time. Both tools are extensively used in macroeconomic dynamics, growth theory, business cycle theory, and the analysis of market stability.

Difference Equations

Basic Concepts

A difference equation expresses a relationship between the values of a variable at different discrete time periods, most commonly relating the value of a variable in the current period to its value in one or more previous periods. The order of a difference equation is determined by the difference between the largest and smallest time subscripts appearing in the equation. A first-order difference equation relates \(y_t\) to \(y_{t-1}\), and takes the general linear form:

$$y_t = ay_{t-1} + b$$

where (a) and (b) are constants. The solution to a difference equation is an explicit expression for \(y_t\) purely as a function of time (t) and the initial condition, that is, the value of the variable at time \(t=0\), denoted \(y_0\).

Solving First-Order Linear Difference Equations

The general solution of the first-order linear difference equation \(y_t = ay_{t-1} + b\) consists of two parts: the particular solution (also called the equilibrium or intertemporal equilibrium value, denoted \(y^*\)), and the complementary function (also called the homogeneous solution), which captures the deviation from equilibrium over time.

The particular solution is found by setting \(
y_t=y_{t-1}=y^{*}
\) (the condition for a stationary equilibrium), which gives \(
y^{*}=ay^{*}+b
\), and solving for \(y^*\) yields:

$$y^* = \frac{b}{1-a}, \quad \text{provided } a \neq 1$$

The complementary function takes the form \(A \cdot a^t\), where (A) is an arbitrary constant determined using the initial condition. The general solution is therefore:

\[
y_t=Aa^t+\frac{b}{1-a}
\]

Using the initial condition \(y_0\), we obtain \(A = y_0 – y^*\), so that the complete solution becomes:

\[
y_t=(y_0-y^{*})a^t+y^{*}
\]