Problem 9 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

The solution \(y = A + C e^{kt}\) satisfies the differential equation.

Step 1: Identify the given equation

Identify and write down the differential equation: \( \frac{d y}{d t} = k(y-A) \). We are tasked to check if \( y = A + C e^{kt} \) is a solution to this equation.

Step 2: Differentiate the proposed solution

Determine the derivative of the proposed solution \( y = A + C e^{kt} \) with respect to \( t \): \[ \frac{dy}{dt} = \frac{d}{dt}(A + C e^{kt}) = C \frac{d}{dt}(e^{kt}) = C k e^{kt}. \]

Step 3: Substitute the proposed solution into the differential equation

Substitute \( y = A + C e^{kt} \) into the right side of the differential equation: \[ k(y - A) = k((A + C e^{kt}) - A) = k(C e^{kt}). \]

Step 4: Verify equality of derivatives

Compare the derivative obtained from Step 2, \( C k e^{kt} \), with the right-hand side expression from Step 3, \( k(C e^{kt}) \). Both expressions are identical, verifying the proposed solution satisfies the differential equation.

Key Concepts

Solutions to Differential Equations

Differential equations are powerful tools used to describe a variety of phenomena in engineering, physics, biology, and more. A differential equation reflects a relationship between a function and its derivatives, essentially telling us how the function changes over time or space. To solve a differential equation means to find a function that satisfies this relationship.

In our example, we start with the equation \( \frac{d y}{d t} = k(y-A) \). This equation implies how the rate of change of \( y \) relates to \( y \) itself. When given a proposed solution, like \( y = A + C e^{kt} \), we need to verify whether it satisfies the condition set by the differential equation. Finding solutions to these equations allow us to predict behavior and study how variables depend on each other.

Exponential Functions

Exponential functions play a critical role in many differential equations, often modeling dynamics such as population growth, decay processes, or interest calculations. An exponential function has the form \( e^{kt} \), where \( e \) is Euler's number and a constant base, and \( kt \) is the exponent.

Such functions are uniquely characterized by proportional rates of change, meaning the rate at which the function values increase or decrease is proportional to the current value of the function. In the provided exercise, the solution \( y = A + C e^{kt} \) reflects the presence of an exponential term \( e^{kt} \), which is pivotal in ensuring that the function can match the expected rate of change described by the differential equation.

Verification of Solutions

To verify a solution to a differential equation, we begin by taking the derivative of the proposed solution and checking if it fits back into the original differential equation. The process involves substituting the solution into the differential equation and seeing if both sides simplify to the same expression.

In our case, the proposed solution is \( y = A + C e^{kt} \). Upon differentiating, we obtain \( \frac{dy}{dt} = C k e^{kt} \). If substituting \( y = A + C e^{kt} \) into the equation \( \frac{dy}{dt} = k(y-A) \) leads to an expression that matches its right-hand side, it confirms that the proposed solution is indeed correct. This step ensures we have accurately captured the behavior dictated by the differential equation.

Derivatives

Derivatives measure how a function changes as its input changes. In the context of differential equations, derivatives represent rates of change. They are fundamental tools used to analyze how one quantity changes in relation to another.

For instance, in the equation \( \frac{d y}{d t} = k(y-A) \), \( \frac{d y}{d t} \) is the derivative of \( y \) with respect to \( t \), showing how \( y \) changes over time. To verify a solution like \( y = A + C e^{kt} \), we compute its derivative, \( C k e^{kt} \), to match it with the behavior described by the differential equation. This critical step confirms that the solution's rate of change aligns perfectly with what the equation dictates.