Problem 15 - Calculus

Question

Show that \(y=A+C e^{k t}\) is a solution to the equation $$\frac{d y}{d t}=k(y-A).$$

Step-by-Step Solution

Answer

The function is a solution as it satisfies the differential equation.

Step 1: Differentiate the Given Solution

To determine if the given function is a solution, we need to differentiate it with respect to time \(t\). The function is defined as \(y = A + C e^{k t}\). Differentiating term-by-term gives: \(\frac{dy}{dt} = 0 + C k e^{k t}\).

Step 2: Substitute into the Differential Equation

Now we substitute \(y = A + C e^{k t}\) and \(\frac{dy}{dt} = Ck e^{kt}\) into the differential equation \(\frac{dy}{dt} = k(y - A)\). Substituting \(y - A\) gives \(C e^{k t}\), so the equation becomes \(C k e^{k t} = k(C e^{k t})\).

Step 3: Simplify and Verify

Both sides of the equation \(C k e^{k t} = k(C e^{k t})\) are identical, indicating the equality holds for all \(t\). Therefore, the original function \(y = A + C e^{k t}\) is indeed a solution to the differential equation \(\frac{d y}{d t} = k(y - A)\).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In the context of our differential equation exercise, the exponential term is written as \( e^{kt} \). Here, \( e \) is the base of natural logarithms, approximately equal to 2.718, and \( kt \) is the exponent. This form is characteristic of exponential functions, which are widely used in modeling growth or decay processes.

In an exponential function, the rate of change is proportional to its current value. For example, in finance, population biology, and physics, exponential growth or decay describes processes in which the quantity doubles or halves over regular intervals. The key point in this exercise is the exponential term \( C e^{kt} \) determining how the solution \( y \) evolves over time based on changing \( t \).

• Key Properties: Constant proportional growth or decay
• Rapid escalation over time if \( k > 0 \) (growth)
• Rapid reduction over time if \( k < 0 \) (decay)

Understanding these characteristics helps in analyzing how the solution behaves under the influence of the exponential term.

Solution Verification

Solution verification involves confirming a proposed solution satisfies a given differential equation. In our exercise, we check if \( y = A + C e^{kt} \) actually solves \( \frac{d y}{d t} = k(y - A) \). This requires substituting both \( y \) and its derivative \( \frac{dy}{dt} \) back into the differential equation.

The verification steps can be simplified as:

• Differentiate the proposed solution: Identify \( \frac{dy}{dt} \)
• Substitute \( y \) and \( \frac{dy}{dt} \) into the equation
• Check the equality to ensure both sides match

For our given equation, upon substitution, we reach \( C k e^{k t} = k(C e^{kt}) \). Simplifying shows both sides equal, confirming the original proposed function as a valid solution.

This method of solution verification is fundamental in ensuring our understanding of differential equations, offering insight into whether our algebraic manipulations and theoretical solutions hold under real conditions.

First-order Differential Equation

A first-order differential equation involves functions and their first derivatives. The primary characteristic is that it contains no terms with derivatives higher than the first degree. The general form for a first-order differential equation is \( \frac{dy}{dt} = f(t,y) \). In our specific problem, the equation \( \frac{dy}{dt} = k(y - A) \) illustrates the standard format, comprising only the first derivative.

First-order differential equations are used to model dynamic systems where the rate of change of a variable is directly related to the variable itself, as observed in population growth, chemical reactions, and heat transfer. These equations can frequently be solved by separation of variables or integrating factors, though in our case, the solution was given, and we've verified it.

• Model simple dynamic systems
• Analytical solutions by integration methods
• Common in real-world applications

Recognizing the fundamental nature of these equations allows us to tackle more complex systems by building on the foundational principles they represent.