Chapter 4: Problem 67
Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=2 x-7, \quad y(2)=0 $$
Short Answer
Expert verified
The particular solution is \( y = x^2 - 7x + 10 \).
Step by step solution
01
Identify the Differential Equation
The given problem is a first-order linear differential equation: \( \frac{dy}{dx} = 2x - 7 \), with the initial condition \( y(2) = 0 \).
02
Integrate the Differential Equation
To solve the differential equation, integrate both sides with respect to \( x \):\[ \int \frac{dy}{dx} \, dx = \int (2x - 7) \, dx \] This results in:\[ y = x^2 - 7x + C \] where \( C \) is the constant of integration.
03
Apply Initial Condition
Use the initial condition \( y(2) = 0 \) to find \( C \). Substitute \( x = 2 \) and \( y = 0 \) into the equation:\[ 0 = (2)^2 - 7(2) + C \]Simplify to:\[ 0 = 4 - 14 + C \]\[ C = 10 \]
04
Write the Particular Solution
Substitute \( C = 10 \) back into the general solution to find the particular solution:\[ y = x^2 - 7x + 10 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
A first-order linear differential equation is a type of equation that involves a function and its first derivative. In our initial value problem, the equation is \( \frac{dy}{dx} = 2x - 7 \). Here, the right side consists of a linear expression in terms of \( x \), making this a straightforward linear equation.
First-order refers to the highest derivative in the equation being the first derivative (\( \frac{dy}{dx} \)). The goal is to find the function \( y \) that satisfies this equation over a range of \( x \).
These types of differential equations frequently appear in various applications, such as physics, engineering, and economics. Recognizing a first-order linear differential equation is the first step in finding a solution, often along with any given initial conditions, just like \( y(2) = 0 \) in our problem.
First-order refers to the highest derivative in the equation being the first derivative (\( \frac{dy}{dx} \)). The goal is to find the function \( y \) that satisfies this equation over a range of \( x \).
These types of differential equations frequently appear in various applications, such as physics, engineering, and economics. Recognizing a first-order linear differential equation is the first step in finding a solution, often along with any given initial conditions, just like \( y(2) = 0 \) in our problem.
Integration
Integration is a mathematical process used to reverse differentiation. To solve our first-order differential equation, we integrate both sides with respect to \( x \).
The original expression turns into a function plus a constant (\( C \)), which represents an infinite family of solutions.
- The left side is the integration of the derivative \( \frac{dy}{dx} \), which simply returns \( y \).
- The right side is the integration of the expression \( 2x - 7 \), which means finding the antiderivative.
The original expression turns into a function plus a constant (\( C \)), which represents an infinite family of solutions.
Constant of Integration
The constant of integration, denoted as \( C \), plays a crucial role in integration. It represents an arbitrary constant that appears when integrating an indefinite integral.
Since integration reverses differentiation, and derivatives of constant functions are zero, the constant \( C \) accounts for any constant that might have been lost when differentiating.
In our problem, after integrating, we have \( y = x^2 - 7x + C \). Without determining \( C \), we have a general solution. The initial condition \( y(2) = 0 \) helps us find the specific \( C \) that fits the particular scenario.
Since integration reverses differentiation, and derivatives of constant functions are zero, the constant \( C \) accounts for any constant that might have been lost when differentiating.
In our problem, after integrating, we have \( y = x^2 - 7x + C \). Without determining \( C \), we have a general solution. The initial condition \( y(2) = 0 \) helps us find the specific \( C \) that fits the particular scenario.
Particular Solution
A particular solution is a specific solution that adheres to the initial condition provided. To find this specific solution, we take the general form \( y = x^2 - 7x + C \) and use an initial condition.
In our exercise, the initial condition is \( y(2) = 0 \), which allows us to solve for \( C \). Substituting \( x = 2 \) and \( y = 0 \) into the formula:
In our exercise, the initial condition is \( y(2) = 0 \), which allows us to solve for \( C \). Substituting \( x = 2 \) and \( y = 0 \) into the formula:
- Start with the equation \( 0 = (2)^2 - 7(2) + C \).
- Solve for \( C \) by simplifying to get \( 0 = 4 - 14 + C \), leading to \( C = 10 \).