Problem 25 - Applied Calculus

Question

Some people write the solution of the initial value problem $$\frac{d y}{d t}=k(y-A) \quad y=y_{0} \text { at } t=0$$ in the form $$\frac{y-A}{y_{0}-A}=e^{k t}$$ Show that this formula gives the correct solution for \(y\) assuming \(y_{0} \neq A\).

Step-by-Step Solution

Answer

The formula \( \frac{y-A}{y_0-A} = e^{kt} \) is correct for the given differential equation.

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{dy}{dt} = k(y - A) \). Here, \( k \) and \( A \) are constants, and \( y_0 \) is the initial value of \( y \) at \( t = 0 \). We are tasked with verifying that the solution to this equation can be expressed as \( \frac{y - A}{y_0 - A} = e^{kt} \).

Step 2: Reformat the Differential Equation

Rearrange the differential equation to isolate the variables: \[\frac{1}{y-A} \frac{dy}{dt} = k\]This allows us to integrate both sides with respect to their respective variables.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( t \):\[\int \frac{1}{y-A} \, dy = \int k \, dt\]The left side integrates to \( \ln|y-A| \) and the right side to \( kt + C \), where \( C \) is the constant of integration.

Step 4: Solve for the Constant of Integration

Apply the initial condition \( y = y_0 \) at \( t = 0 \):\[\ln|y_0 - A| = k \cdot 0 + C\]This simplifies to \( \ln|y_0 - A| = C \).

Step 5: Substitute the Constant Back

Substitute \( C = \ln|y_0 - A| \) back into the integrated equation:\[\ln|y-A| = kt + \ln|y_0 - A|\]

Step 6: Exponentiate to Solve for \( y \)

Exponentiate both sides to eliminate the natural logarithm:\[e^{\ln|y-A|} = e^{kt + \ln|y_0 - A|}\]This simplifies to:\[|y-A| = e^{kt} \cdot |y_0 - A|\]

Step 7: Arrive at the Solution

Assuming \( y_0 eq A \) ensures \( |y_0 - A| = y_0 - A \), and similarly for \( |y - A| = y - A \) (assuming \( y > A \) or manually adjust if otherwise):\[\frac{y-A}{y_0-A} = e^{kt}\]This matches the given form and verifies it solves the original differential equation.

Key Concepts

Initial Value Problem

Let's dive into the concept of an "Initial Value Problem" as it often arises in differential equations. An initial value problem involves finding a function that satisfies a given differential equation and also meets a specified initial condition. In our example, the differential equation is \( \frac{dy}{dt} = k(y - A) \) with the initial condition \( y = y_0 \) at \( t = 0 \).

The initial condition is crucial because it specifies the exact point on the solution curve that the function must pass through. For instance:

• The differential equation tells us the relationship between the rate of change of a variable, \( y \), and another quantity depending on \( y \),
• The initial condition indicates the starting value of \( y \) at a particular time, usually \( t = 0 \).

By combining these two elements, you can find a unique solution function \( y(t) \) that precisely fits the scenario described in the problem.

Integration Techniques

When solving differential equations, it's common to employ integration techniques. In this exercise, we need to solve an equation by integrating both sides. We started with:\[\frac{1}{y-A} \frac{dy}{dt} = k\]This form allows us to separate the variables and integrate:\[\int \frac{1}{y-A} \, dy = \int k \, dt\]By recognizing the integration forms, we get:

• The left side integrates to \( \ln|y-A| \) because the integral of \( \frac{1}{u} \, du \) is \( \ln|u| \).
• The right side is a straightforward integration, yielding \( kt + C \), with \( C \) being the constant of integration.

Integration unveils so much about the functions themselves, translating rates of change into actual positional or state values. By carefully handling integration, we derive accurate equations governing the behavior of systems over time.

Exponential Functions

Exponential functions play a significant role in the solution of differential equations like the one covered here. Once we integrated both sides, we ended up with:\[\ln|y-A| = kt + \ln|y_0 - A|\]By exponentiating both sides, we remove the logarithms. This step is essential as it transforms our equation into a form that allows us to isolate the variable \( y \):\[|y-A| = e^{kt} \cdot |y_0 - A|\]From this expression, we conclude:\[\frac{y-A}{y_0-A} = e^{kt}\]

Exponential functions, defined as functions where the variable appears in the exponent, are very powerful in modeling scenarios like growth and decay. In our context, the solution involves an exponential function because the relationship between the rate of change and the function itself leads to exponential behavior. This reflects how changes in the system multiply over time.

Understanding how exponential functions relate in differential equations helps not only in solving problems but in appreciating the underlying dynamics they describe in various applications.