91Ó°ÊÓ

A demand function, usually denoted as \( q = f(p) \), describes the relationship between the quantity demanded (\(q\)) of a commodity and its price per unit (\(p\)). This function essentially tells us how much of a product consumers are willing to buy at different price points.
Understanding the demand function is crucial as it underpins many economic analyses, allowing businesses to strategize pricing and production levels. For instance, if the price of a product increases, the demand function typically predicts a lower quantity demanded, showing the inverse relationship between price and demand.
In the original exercise, the demand function is given an elasticity component, facilitating a more detailed analysis of how changes in price affect demand quantitatively.

A differential equation is a mathematical equation that relates a function to its derivatives. These equations are fundamental in describing various physical and economic phenomena.
In our exercise, the demand function's elasticity leads us to formulate a differential equation: \( f^{\prime}(p) = -\left( 1 + \frac{1}{p} \right) f(p) \).
This differential equation helps to find how the quantity demanded changes with price.
The equation tells us not just the static relationship between price and demand, but also how this relationship evolves, giving deeper insights into the economic behavior of consumers.

The integrating factor method is a technique used to solve linear first-order differential equations of the form \( y' + P(x) y = Q(x) \). This method simplifies the equation by multiplying through an integrating factor, which makes it easier to solve.
In our exercise, we start with the differential equation \( f^{\prime}(p) + \left( 1 + \frac{1}{p} \right) f(p) = 0 \). We find the integrating factor, \( \mu(p) = e^{p + \ln|p|} = e^{p} |p| \).
Multiplying through by this integrating factor allows us to consolidate terms and simplify the solution process, eventually helping us integrate and solve for \( f(p) \). The use of the integrating factor transforms the raw differential equation into something more manageable and solvable.

An initial condition provides a specific value of the function at a particular point, which is crucial for solving differential equations uniquely.
In our exercise, the initial condition is given as \( f(1) = 100 \).
This means that when the price is 1 unit, the quantity demanded is 100 units. We use this information to solve for the constant of integration necessary to determine the exact form of the demand function.
Applying this initial condition is vital as it anchors our general solution to a specific context, ensuring that the function accurately reflects the given economic scenario.
Thus, substituting this into our integrated solution helps us find the constant, leading to the final demand function: \( f(p) = \frac{100}{e^{p-1} p} \).