Problem 3 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d P}{d t}=0.02 P, \quad P(0)=20$$

Step-by-Step Solution

Answer

\( P(t) = 20e^{0.02t} \).

Step 1: Introduce the Differential Equation

We are given the differential equation \( \frac{dP}{dt} = 0.02P \). Our goal is to solve this using the method of separation of variables.

Step 2: Separate Variables

To employ the method of separation of variables, rearrange the equation to isolate variables: \( \frac{1}{P} \, dP = 0.02 \, dt \).

Step 3: Integrate Both Sides

Integrate both sides of the separated equation: \[ \int \frac{1}{P} \, dP = \int 0.02 \, dt \] This yields: \[ \ln |P| = 0.02t + C \] where \( C \) is the constant of integration.

Step 4: Solve for P

To obtain \( P \), exponentiate both sides: \( |P| = e^{0.02t + C} \).This can be rewritten as \( P = Ce^{0.02t} \).

Step 5: Apply Initial Condition

Use the initial condition \( P(0) = 20 \) to determine the constant \( C \).Substitute \( t = 0 \) and \( P = 20 \) into the equation: \( 20 = Ce^0 \), which means \( C = 20 \).

Step 6: Write the Particular Solution

Substitute the value of \( C \) back into the equation for \( P \): \( P(t) = 20e^{0.02t} \). This is the particular solution that satisfies the given initial condition.

Key Concepts

Differential Equations

Differential equations are a type of equation that involve functions and their derivatives, expressing relationships between varying quantities. In basic terms, these equations are used to determine how a particular set of variables changes over time or other dimensions. They are found in almost every scientific discipline, where they're used to model everything from population growth to the cooling of objects.
In the exercise provided, the differential equation is given as \( \frac{dP}{dt} = 0.02P \). This means the rate of change of \( P \) relative to \( t \) is proportional to the size of \( P \) itself. This kind of equation often models exponential processes, which is precisely the focus here – the process of separating variables to simplify and solve a differential equation.

Initial Conditions

Initial conditions are the foundation of solving a differential equation uniquely. They give the specific value of a function or its derivatives at a particular point. When we solve a differential equation, the solution often contains an arbitrary constant, which can take an infinite number of values. Initial conditions help determine exactly what this constant is, providing a specific solution rather than a family of potential ones.
In this problem, the initial condition provided is \( P(0) = 20 \). This tells us that when time \( t \) is zero, the quantity \( P \) is 20. By substituting this initial condition into our general solution, we are able to solve for the constant, ensuring our solution isn't just generally correct, but precisely correct for the given scenario.

Exponential Growth

Exponential growth refers to a process where the quantity increases at a rate proportional to its current size. The solution to the differential equation after applying separation of variables illustrates this concept.
In our example, once the initial condition is applied, the function becomes \( P(t) = 20e^{0.02t} \). The term \( e^{0.02t} \) represents the exponential growth of the population over time, with the base \( e \) indicating continuous growth. The rate \( 0.02 \) outlines how quickly this growth occurs. The initial condition of \( 20 \) acts as a starting point for this growth. As time increases, \( P(t) \) grows exponentially, illustrating how quickly even small growth rates can lead to significant increases over time.