Problem 1 - Applied Calculus

Question

Find particular solutions. $$\frac{d y}{d t}=0.5(y-200), \quad y=50\( when \)t=0$$

Step-by-Step Solution

Answer

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

Step 1: Write the Differential Equation

The given differential equation is \( \frac{dy}{dt} = 0.5(y - 200) \). We need to find a solution that satisfies this equation and the initial condition \( y(0) = 50 \).

Step 2: Separate Variables

Rearrange the terms to separate variables: \( \frac{1}{y - 200} \, dy = 0.5 \, dt \).

Step 3: Integrate Both Sides

Integrate both sides: \( \int \frac{1}{y - 200} \, dy = \int 0.5 \, dt \). The left side becomes \( \ln |y - 200| \), and the right side is \( 0.5t + C \), where \( C \) is a constant.

Step 4: Solve for y

Exponentiate both sides to solve for \( y \): \( |y - 200| = e^{0.5t + C} \). Simplify to \( y - 200 = Ce^{0.5t} \) where \( C' = e^C \).

Step 5: Apply Initial Condition

Substitute the initial condition \( y(0) = 50 \) into the equation: \( 50 - 200 = Ce^{0}\). This simplifies to \( -150 = C \).

Step 6: Write the Particular Solution

The particular solution to the differential equation is \( y(t) = 200 - 150e^{0.5t} \).

Key Concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. These equations are essential in mathematics, modeling the dynamics of systems in fields such as physics, engineering, and biology. They describe how a quantity changes over time or space. In our original exercise, the differential equation given is \( \frac{dy}{dt} = 0.5(y - 200) \). This particular type of differential equation is a first-order ordinary differential equation because it involves only the first derivative of the unknown function \( y(t) \).

In this context, \( y(t) \) could represent a physical quantity like temperature, population, or concentration, and \( \frac{dy}{dt} \) describes how this quantity changes with respect to time. The right-hand side of the equation, \( 0.5(y - 200) \), indicates that the rate of change of \( y \) depends linearly on how far \( y \) is from 200.

• First-order: Involves the first derivative.
• Ordinary: Contains functions of only one independent variable.
• Linear: The derivative and the function appear to the first power.

Initial Value Problems

Initial value problems (IVPs) consist of a differential equation and an initial condition. The goal is to find a particular solution that satisfies both. In the exercise, the initial condition is \( y(0) = 50 \), which is used to determine the value of the constant in the general solution. Solving IVPs is crucial because it ensures that the solution is tailored to meet specific initial conditions, reflecting the unique situation being modeled.
To solve an IVP, we generally follow these steps:

• Find the general solution of the differential equation.
• Apply the initial condition to determine unknown constants.
• Write the particular solution.

For example, in the exercise, the initial condition \( y = 50 \) at \( t = 0 \) ensures the solution maps to real-world data, providing a precise description of the system's state at an initial point in time.

Separation of Variables

Separation of variables is a powerful technique used to solve differential equations. This method works by rearranging the equation so that each variable and its derivative appear on opposite sides. In the exercise, we start with \( \frac{dy}{dt} = 0.5(y - 200) \). By rearranging, we get \( \frac{1}{y - 200} \, dy = 0.5 \, dt \), effectively separating the variables \( y \) and \( t \).
This is followed by integrating both sides to find the general solution. Integrating \( \frac{1}{y - 200} \, dy \) results in \( \ln |y - 200| \), and integrating \( 0.5 \, dt \) gives \( 0.5t + C \).

This technique is particularly advantageous for simple differential equations and provides direct insight into the relationship between the variables. Once integrated, you can solve for the unknown function, and subsequently apply any initial conditions to find the particular solution.