Chapter 11: Problem 31
Give an example of: A differential equation with an initial condition.
Short Answer
Expert verified
Example: \( \frac{dy}{dt} = 3y \) with \( y(0) = 2 \).
Step by step solution
01
Define a Differential Equation
We start by defining a simple first-order differential equation. Let's choose the equation \( \frac{dy}{dt} = 3y \), which describes the rate of change of \(y\) with respect to \(t\). This is a first-order linear differential equation.
02
Specify the Initial Condition
An initial condition is needed to find a unique solution to the differential equation. Let's set the initial condition \( y(0) = 2 \), meaning that when \(t = 0\), \(y = 2\). This condition will help us determine the specific solution for the given problem.
03
Write the Complete Example
Combine the differential equation and the initial condition to form a complete example: \( \frac{dy}{dt} = 3y \) with \( y(0) = 2 \). This combination provides the necessary information to solve the differential equation uniquely.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Conditions
An initial condition is a vital element in solving a differential equation as it determines the specific solution out of the infinitely many possible solutions. Without an initial condition, the solution to a differential equation could vary widely. For instance, while solving the simple differential equation \( \frac{dy}{dt} = 3y \), we would encounter a general solution without an initial condition. The general solution of this equation is \( y(t) = Ce^{3t} \), where \( C \) is an arbitrary constant.
However, by applying an initial condition such as \( y(0) = 2 \), we can compute the exact value of \( C \). Substituting into the general solution, we get \( 2 = Ce^{0} = C \), making \( C = 2 \). Hence, the unique solution satisfying the initial condition is \( y(t) = 2e^{3t} \).
To summarize the importance:
However, by applying an initial condition such as \( y(0) = 2 \), we can compute the exact value of \( C \). Substituting into the general solution, we get \( 2 = Ce^{0} = C \), making \( C = 2 \). Hence, the unique solution satisfying the initial condition is \( y(t) = 2e^{3t} \).
To summarize the importance:
- Initial conditions "anchor" the solution in a specific value, eliminating ambiguity.
- They allow us to calculate the constants that arise in the integration process.
- They provide context to a mathematical model, aligning it with real-world scenarios.
First-Order Differential Equations
First-order differential equations involve the first derivative of the unknown function, typically represented as \( \frac{dy}{dx} \). These equations are significant in modeling various real-world phenomena where the rate of change of a quantity is crucial.
For example, the differential equation \( \frac{dy}{dt} = 3y \) is a linear first-order differential equation. Linear means that the function \( y \) and its derivative appear to the first power and no products of \( y \) and \( \frac{dy}{dx} \) are present.
There are several methods to solve first-order differential equations:
For example, the differential equation \( \frac{dy}{dt} = 3y \) is a linear first-order differential equation. Linear means that the function \( y \) and its derivative appear to the first power and no products of \( y \) and \( \frac{dy}{dx} \) are present.
There are several methods to solve first-order differential equations:
- Separable Variables: When both variables can be separated on opposite sides of the equation.
- Integrating Factors: Useful for linear first-order equations to make them easily integrable.
- Exact Equations: Applies when the differential equation can be expressed in a perfect differential form.
Rate of Change
The concept of "rate of change" is fundamental to differential equations as they primarily express how a quantity changes over time or with respect to another variable. Typically depicted as derivatives, rates of change provide insights into dynamic processes.
Consider the equation \( \frac{dy}{dt} = 3y \); the term \( \frac{dy}{dt} \) represents the rate of change of \( y \) with respect to time \( t \). It informs us how quickly \( y \) increases or decreases. In this equation, the rate of growth is proportional to the current amount, a common form found in population dynamics or financial growth models.
Key points about rate of change:
Consider the equation \( \frac{dy}{dt} = 3y \); the term \( \frac{dy}{dt} \) represents the rate of change of \( y \) with respect to time \( t \). It informs us how quickly \( y \) increases or decreases. In this equation, the rate of growth is proportional to the current amount, a common form found in population dynamics or financial growth models.
Key points about rate of change:
- It helps understand how rapidly a quantity is changing, which is crucial for predictions and modeling scenarios.
- Such understanding is applicable in various contexts, aiding in solving problems like acceleration in physics, or rates of reaction in chemistry.
- The relationship depicted often helps to explain exponential growth or decay models as seen in many natural processes.