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 correctly solves the initial value problem for \( y \).

Step 1: Rewrite the given equation

Start by separating variables in the differential equation: \( \frac{d y}{d t}=k(y-A) \). This can be rewritten as \( \frac{d y}{y-A}=k \cdot dt \).

Step 2: Integrate both sides

Integrate the left-hand side with respect to \( y \) and the right-hand side with respect to \( t \). The left integral becomes \( \int \frac{1}{y-A} \, dy = \ln|y-A| \), while the right integral result is \( \int k \, dt = kt + C \), where \( C \) is the integration constant.

Step 3: Exponentiate to solve for \( y \)

Exponentiate both sides of the equation \( \ln|y-A| = kt + C \) to clear the logarithm, yielding \( |y-A| = e^{kt+C} = e^C \cdot e^{kt} \). Let \( C' = e^C \), so \( y-A = \pm C' e^{kt} \).

Step 4: Solve for the constant \( C' \)

Using the initial condition \( y = y_0 \) at \( t = 0 \), substitute into the equation \( y-A = C' e^{kt} \) to get \( y_0 - A = C' \cdot e^0 = C' \). Hence, \( C' = y_0 - A \).

Step 5: Final expression for \( y \)

Finally, plug \( C' \) back into the expression, yielding \( y-A = (y_0 - A) e^{kt} \). Solving for \( y \), we have: \( y = A + (y_0 - A) e^{kt} \).

Step 6: Verify the solution form

The given expression is \( \frac{y-A}{y_0-A} = e^{kt} \). Substitute the previously derived expression: \( \frac{y-A}{y_0-A} = \frac{(y_0-A)e^{kt}}{y_0-A} = e^{kt} \). This confirms that the given formula provides the correct solution for \( y \).

Key Concepts

Initial Value Problem

An Initial Value Problem, often abbreviated as IVP, is a type of differential equation that comes with additional information called initial conditions. This additional information is crucial as it allows us to find a particular solution among a family of possible solutions.
In a typical initial value problem, you are tasked to solve a differential equation and at the same time satisfy a given condition at a specified point in time. For example, the problem we're dealing with is represented by the differential equation:
\[ \frac{dy}{dt} = k(y - A) \]- Here, the rate of change of the variable \( y \) is proportional to the difference between \( y \) and a constant \( A \).
- The initial condition is \( y = y_0 \) at \( t = 0 \). This tells us the precise value of \( y \) when time \( t \) equals zero.
The initial condition allows us to determine the constant of integration that appears when we integrate the differential equation. Without this condition, the solution would include an arbitrary constant, representing infinitely many solutions that fit the differential equation.

Separation of Variables

Separation of Variables is a method used to solve differential equations, where the main idea is to rewrite a differential equation such that the variables can be separated on different sides of the equation.
This technique simplifies the process of finding a solution by allowing us to integrate each side with respect to its own variable. In our problem, we start with:
\[ \frac{dy}{dt} = k(y - A) \]
By separating variables, we get:
\[ \frac{dy}{y - A} = k \, dt \]
This step is crucial because it allows us to integrate both sides independently. The integration gives us:

• The left side becomes \( \int \frac{1}{y-A} \, dy = \ln|y-A| \).
• The right side results in \( \int k \, dt = kt + C \), where \( C \) is a constant of integration.

By successfully integrating, we create a clearer path towards isolating \( y \) and determining the particular solution given the initial condition.

Exponential Functions

Exponential Functions form a critical part of solving many differential equations due to their properties. Once we have the result from separating variables and integrating, we arrive at an equation involving a natural logarithm:
\[ \ln|y-A| = kt + C \]
To solve for \( y \), we need to eliminate the logarithm by exponentiating both sides of the equation:
\[ |y-A| = e^{kt + C} \]
This results in:

• \( y-A = C' e^{kt} \), where \( C' = e^C \).
• The solution depends exponentially on \( t \), showcasing the rapid growth or decay typical of exponential functions.

Exponential functions allow us to find specific solutions considering initial conditions. In our case, using the condition \( y = y_0 \) when \( t = 0 \), we find that \( C' \) equals \( y_0 - A \). Thus, our solution becomes:
\[ y = A + (y_0 - A) e^{kt} \]This demonstrates how exponential functions can define changes in solutions based on time—offering predictive insights into dynamic systems.