Problem 67 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=2 x-7, \quad y(2)=0 $$

Step-by-Step Solution

Answer

The particular solution is \( y = x^2 - 7x + 10 \).

Step 1: Identify the Differential Equation

The given problem is a first-order linear differential equation: \( \frac{dy}{dx} = 2x - 7 \), with the initial condition \( y(2) = 0 \).

Step 2: 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.

Step 3: 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:\[ 0 = 4 - 14 + C \]\[ C = 10 \]

Step 4: Write the Particular Solution

Substitute \( C = 10 \) back into the general solution to find the particular solution:\[ y = x^2 - 7x + 10 \]

Key Concepts

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.

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 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.

When we integrate \( 2x - 7 \), we get \( x^2 - 7x + C \).
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.

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:

• Start with the equation \( 0 = (2)^2 - 7(2) + C \).
• Solve for \( C \) by simplifying to get \( 0 = 4 - 14 + C \), leading to \( C = 10 \).

Finally, substitute \( C = 10 \) back into the general solution to get the particular solution \( y = x^2 - 7x + 10 \), satisfying the initial condition.