Problem 55 - Calculus and its applications

Question

Domar's capital expansion model is $$ \frac{d I}{d t}=h k I $$ where \(I\) is the investment, \(h\) is the investment productivity (constant), \(k\) is the marginal productivity to the consumer (constant), and \(t\) is the time. a) Use separation of variables to solve the differential equation. b) Rewrite the solution in terms of the condition \(I_{0}=I(0)\)

Step-by-Step Solution

Answer

The solution is \( I = I_0 e^{hkt} \).

Step 1: Set Up the Problem

We need to solve the differential equation \( \frac{dI}{dt} = hkI \) by separation of variables. In this step, we identify the variables by rearranging terms to isolate \( dI \) and \( dt \):\[ \frac{dI}{hkI} = dt \].

Step 2: Integrate Both Sides

Integrate both sides to find \( I \) as a function of \( t \). The left side is integrated with respect to \( I \) and the right side with respect to \( t \):\[ \int \frac{1}{I} \, dI = \int hk \, dt \]This leads to:\[ \ln |I| = hkt + C \]where \( C \) is the constant of integration.

Step 3: Solve for I

To solve for \( I \), we exponentiate both sides to eliminate the natural logarithm:\[ |I| = e^{hkt + C} \].This can be rewritten as:\[ I = Ce^{hkt} \]where \( C \) can be positive or negative. We consider the general form \( I = C'e^{hkt} \) by setting \( C' = \pm e^C \).

Step 4: Apply Initial Condition

Use the initial condition \( I(0) = I_0 \) to find \( C' \). Substitute \( t = 0 \) into the equation obtained:\[ I_0 = C'e^{hk \, \cdot \, 0} = C' \].Therefore, \( C' = I_0 \).

Step 5: Write the Final Solution

Substitute \( C' = I_0 \) back into the equation to get:\[ I = I_0 e^{hkt} \].This is the solution to the differential equation with the initial condition applied.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that involves some function and its derivatives. In the world of mathematics and physics, it helps us describe how a quantity changes over time. In the given exercise, the differential equation provided is \( \frac{dI}{dt} = hkI \). Here, \( dI/dt \) represents the rate of change of investment \( I \) with respect to time \( t \). The right side of the equation, \( hkI \), suggests that the rate of change is proportional to the investment itself. This is a common form of a first-order linear differential equation, where the derivative of the function \( I \) is directly proportional to the function itself. Such equations often appear in models of exponential growth or decay, economics, and natural processes.

Integration

Integration is a core calculus concept used to find the function representing accumulated values, such as areas under curves or solutions to differential equations. In the process of solving our differential equation through separation of variables, integration plays a crucial role.The equation \( \frac{dI}{dt} = hkI \) can be rearranged to separate variables: \( \frac{dI}{hkI} = dt \). It signifies moving all terms involving \( I \) to one side and \( t \) to the other. The resulting integrals become:

• \( \int \frac{1}{I} \ dI \) on the left side
• \( \int hk \ dt \) on the right side

The integration of \( \frac{1}{I} \ dI \) yields \( \ln |I| + C_1 \), where \( C_1 \) is an arbitrary constant of integration. Similarly, integrating \( hk \ dt \) results in \( hkt + C_2 \). These constants help build the general solution, assisting us in pinpointing specific solutions with boundary conditions.

Initial Condition

An initial condition provides a specific point on the solution curve of a differential equation, allowing for a unique solution. It eliminates the arbitrary constant that arises during integration.In the given task, the initial condition is \( I(0) = I_0 \). This tells us that at time \( t = 0 \), the investment \( I \) equals some initial value \( I_0 \). By applying this condition to our general solution, \( I = C'e^{hkt} \), we substitute \( t = 0 \) leading to:

• \( I_0 = C'e^{hk \cdot 0} = C' \)

This demonstrates that \( C' = I_0 \), ensuring that our particular solution is correctly aligned with the initial state described by the problem.

Exponential Functions

Exponential functions are a vital element of mathematics, describing relationships where a quantity grows or decays at a constant relative rate. They appear frequently in natural processes, finance, and many areas of science.In our solution, we end up with \( I = I_0 e^{hkt} \), an exponential function. Here:

• \( I \): investment at time \( t \)
• \( I_0 \): initial investment
• \( e \): the base of natural logarithms, approximately 2.718
• \( h \): investment productivity
• \( k \): marginal productivity to the consumer
• \( t \): time variable

This equation indicates that investment grows exponentially over time, scaling by the factor \( e^{hkt} \). A hallmark of exponential growth is that the rate of change is proportional to the current value, making it widely applicable to scenarios where continuous and proportional growth or decay is involved.