Problem 14 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d y}{d t}=0.5(y-200), \quad y=50 \text { when } t=0$$

Step-by-Step Solution

Answer

The solution is \( y = -150e^{0.5t} + 200 \).

Step 1: Write the Differential Equation

Given the differential equation: \( \frac{d y}{d t} = 0.5(y - 200) \). We need to solve this equation using the separation of variables technique.

Step 2: Separate the Variables

Rearrange the equation to separate the variables, \(y\) and \(t\): \[ \frac{1}{y - 200} \, d y = 0.5 \, d t \] This separates the variables such that all \( y \) terms are on one side and \( t \) terms are on the other.

Step 3: Integrate Both Sides

Integrate both sides of the equation:\[\int \frac{1}{y - 200} \, d y = \int 0.5 \, d t\]This gives:\[\ln|y - 200| = 0.5t + C\]where \( C \) is the constant of integration.

Step 4: Solve for \( y \)

Solve the equation \( \ln|y - 200| = 0.5t + C \) for \( y \):First, exponentiate both sides to remove the natural logarithm:\[|y - 200| = e^{0.5t + C}\]Let \( K = e^{C} \) be a new constant. Hence,\[ y - 200 = Ke^{0.5t} \]Solve for \( y \):\[ y = Ke^{0.5t} + 200 \]

Step 5: Apply Initial Condition

Use the initial condition \( y(0) = 50 \) to find \( K \). Substitute \( y = 50 \) and \( t = 0 \) into the equation:\[ 50 = K \, e^{0.5 \times 0} + 200 \]\[ 50 = K + 200 \]\[ K = -150 \]

Step 6: Write the Particular Solution

Substitute \( K = -150 \) back into the general solution:\[y = -150e^{0.5t} + 200\]This is the particular solution that satisfies the given differential equation and initial condition.

Key Concepts

Differential Equations

A differential equation is a mathematical expression that involves functions and their derivatives. It shows the relationship between a function and the rates at which it changes. Differential equations serve as a powerful tool in modeling the behavior of complex systems, especially in science and engineering.
A simple example is the equation \(\frac{d y}{d t} = 0.5(y - 200)\), which describes how the variable \(y\), such as population or chemical concentration, changes over time \(t\). In this exercise, we use the method of separation of variables to untangle this relationship and find an explicit formula for \(y\).
This approach minimizes mistakes and ensures that our solutions are ordered and logical, while also making it easier to later apply any initial conditions for specific problems.

Integration

Integration is the process of finding the integral, or the antiderivative, of a function. It essentially reverses the act of differentiating. This tool is fundamental when solving differential equations through separation of variables.
In our problem, after separating variables, we need to integrate both sides of the equation:

• \( \int \frac{1}{y - 200} \, dy = \int 0.5 \, dt \)
• The result is a logarithmic function: \( \ln|y - 200| = 0.5t + C \)

This step captures the essence of integration as it transforms a differential equation into an expression that we can solve. The constant \(C\) is critically important as it represents any number added to the antiderivative, arising from the indefinite nature of integration.

Initial Conditions

Initial conditions are specific values provided at the start (typically for time \(t=0\)) that allow us to find the particular solution to a differential equation. These conditions eliminate the arbitrary constants introduced during integration.
In our exercise, the initial condition \(y = 50\) when \(t = 0\) is given. By substituting these values into our expression \( y = Ke^{0.5t} + 200 \), we solve for the constant \(K\):

• Substitute: \(50 = K + 200\)
• Solve: \(K = -150\)

This step is critical since it turns a general solution into a particular one that precisely matches the problem scenario. It ensures that our final answer is uniquely tailored to the given condition at \(t=0\), creating a complete and meaningful solution.