# Differential Equations

## First Order Equations

Differential equations are widely used to describe a variety of dynamic processes in Economic Sciences, Business and Marketing. Below we consider how an advertising campaign can be simulated using differential equations.

Imagine that a company has developed a new product or service. The marketing strategy of the company involves aggressive advertising. To construct the simple model we introduce two variables:

• The quantity $$q\left( t \right)$$ represents advertising activity that is described by spending rate, for example, by the amount of dollars (euros, pounds) the company spends for advertising per week;
• The quantity $$A\left( t \right)$$ represents awareness of the target group of potential consumers about the new goods or services.

Thus, we will consider the market niche as a black box (Figure $$1$$). The advertising activity $$q\left( t \right)$$ is the input variable, and awareness of consumers $$A\left( t \right)$$ is the output variable that measures response of the system to the advertising campaign.

A simple model of such type was proposed in $$1962$$ and is called the advertising model of Nerlove and Arrow (the N-A model). This model relates advertising activity $$q\left( t \right)$$ and awareness of consumers $$A\left( t \right)$$ and is given by the differential equation:

$\frac{{dA}}{{dt}} = bq\left( t \right) – kA,$

where $$b$$ is a constant describing advertising effectiveness, $$k$$ is a constant corresponding to decay (or forgetting) rate.

The given equation contains two terms in the right side. The first term $$bq\left( t \right)$$ provides the linear growth of awareness of consumers as a result of advertising. The second term $$-kA$$ reflects the opposite process, i.e. forgetting about the product. In the first approximation, we can assume that the forgetting rate is proportional to the current level of awareness $$A.$$

This equation is a Linear Differential Equation of First Order. It’s convenient to rewrite it in the standard form:

$\frac{{dA}}{{dt}} + kA = bq\left( t \right).$

The integrating factor is the exponential function:

$u\left( t \right) = {e^{\int {kdt} }} = {e^{kt}}.$

Therefore the general solution of the given differential equation is given by

${A\left( t \right) }={ \frac{{b\int {{e^{kt}}q\left( t \right)dt} + C}}{{{e^{kt}}}}.}$

As usual, the constant of integration $$C$$ can be found from the initial condition $$A\left( {{t_0}} \right) = {A_0}.$$

In the examples below we consider how awareness of customers $$A\left( t \right)$$ varies for different advertising modes.

## Solved Problems

Click or tap a problem to see the solution.

Management of a company decided to advertise a new product permanently during the year. The advertising budget is $12,000. The coefficients $$k$$ and $$b$$ are $$k = {\large\frac{1}{4}\normalsize},\;b = 25.$$ Construct and solve the differential equation describing the number of people $$A\left( t \right)$$ aware of this product. ### Example 2 Under the conditions of the previous problem $$1,$$ figure out how will the number of potential buyers vary by the end of the year if the advertising budget is spent evenly for the first $$6$$ months? ### Example 3 Investigate the dynamics of awareness $$A\left( t \right)$$ for the case of Linear Advertising Function $$q\left( t \right).$$ Use the same data as in Examples $$1,2.$$ ### Example 1. Management of a company decided to advertise a new product permanently during the year. The advertising budget is$12,000. The coefficients $$k$$ and $$b$$ are $$k = {\large\frac{1}{4}\normalsize},\;b = 25.$$ Construct and solve the differential equation describing the number of people $$A\left( t \right)$$ aware of this product.

Solution.

The equation of dynamics of $$A\left( t \right)$$ is written in the form

$\frac{{dA}}{{dt}} + kA = bq\left( t \right).$

We assume that the time $$t$$ is measured in months. By the condition of the problem, the advertising costs are constant during the year, so every month they are equal to:

${{q_0} = \frac{{12,000}}{{12}} }={ 1,000\,\frac{}{\text{month}}.}$

Substituting the known data leads to the following differential equation:

$\frac{{dA}}{{dt}} + \frac{A}{4} = 25,000.$

The integrating factor for this equation has the form:

$u\left( t \right) = {e^{\int {{\large\frac{1}{4}\normalsize}dt} }} = {e^{\large\frac{t}{4}\normalsize}}.$

The general solution is written as

${A\left( t \right) }={ \frac{{25,000\int {{e^{\large\frac{t}{4}\normalsize}}dt} + C}}{{{e^{\large\frac{t}{4}\normalsize}}}} } = {\frac{{\frac{{25,000}}{{\frac{1}{4}}}{e^{\large\frac{t}{4}\normalsize}} + C}}{{{e^{\large\frac{t}{4}\normalsize}}}} } = {100,000 + C{e^{ – \large\frac{t}{4}\normalsize}}.}$

We determine the constant of integration $$C$$ from the initial condition $$A\left( {t = 0} \right) = 0.$$ Hence, $$C = -100,000.$$ Then the particular solution is expressed by the formula

${A\left( t \right) }={ 100,000\left( {1 – {e^{ – \large\frac{t}{4}\normalsize}}} \right).}$

The graph of this function is shown in Figure $$2$$.

Thus in case of permanent advertising, the number of potential buyers aware of the product grows non-linearly approaching the maximum value

${A_{\max }} = \frac{{b{q_0}}}{k} = 100000.$

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Problem 1
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Problems 2-3