Problem 71 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

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

Step 1: Recognize the Differential Equation

The given differential equation is \( \frac{dy}{dx} = 2x - 7 \). This is a first-order linear differential equation, and we will solve it by integrating both sides.

Step 2: Integrate the Differential Equation

To find \( y \), integrate the right side with respect to \( x \). So, the integral becomes: \[ y = \int (2x - 7) \, dx \] This can be split into two separate integrals: \[ y = \int 2x \, dx - \int 7 \, dx \]

Step 3: Perform the Integration

Integrate each term separately: - The integral of \( 2x \) is \( x^2 \). - The integral of \( 7 \) is \( 7x \). After integration, we have: \[ y = x^2 - 7x + C \] where \( C \) is the constant of integration.

Step 4: Apply the Initial Condition

Use the initial condition \( y(2) = 0 \) to find the value of \( C \). Substitute \( x = 2 \), and \( y = 0 \) into the equation: \[ 0 = 2^2 - 7(2) + C \] Calculate to find \( C \): \[ 0 = 4 - 14 + C \] Thus, \( C = 10 \).

Step 5: Write the Particular Solution

Substitute \( C = 10 \) back into the expression for \( y \). The particular solution to the initial value problem is:\[ y = x^2 - 7x + 10 \]

Key Concepts

Integration

Integration is a fundamental concept in calculus, often used to reverse the process of differentiation. In the context of differential equations, integration is the method we use to find a function when its derivative is known. When you are given a differential equation like \( \frac{dy}{dx} = 2x - 7 \), integration helps to determine the original function \( y \) that satisfies the equation.

Here's a simple breakdown of the integration process you might encounter:

• Identify the part of the equation that needs integration – in this exercise, it's \( 2x - 7 \).
• Break it into manageable pieces – integrate each term separately, here \( 2x \) and \( -7 \).
• Use known integration rules: \( \int 2x \, dx = x^2 \) and \( \int -7 \, dx = -7x \).
• Combine them to form the integrated function \( y = x^2 - 7x + C \).

Each term is treated individually before being recombined, and understanding this step-by-step process means even complex differential equations are solvable with time and practice.

Initial Value Problems

An initial value problem (IVP) is a differential equation accompanied by a specific value that the solution must satisfy at a particular point. This initial condition pinpoints the solution among the family of solutions provided by the general equation.

In the exercise, the initial value \( y(2) = 0 \) defines the specific solution. Here's how this works:

• Once the integration step is complete, you're left with a general solution with a constant \( C \): \( y = x^2 - 7x + C \).
• Apply the initial condition by substituting the particular \( x \) and \( y \) values given: \( y(2) = 0 \).
• Replacing \( x = 2 \) and \( y = 0 \) in \( x^2 - 7x + C \) helps you solve for \( C \).
• Calculation using the equation \( 0 = 4 - 14 + C \) yields \( C = 10 \).
• Concluding with the particular solution: \( y = x^2 - 7x + 10 \).

By confirming \( y(2) = 0 \), we ensure that our chosen solution line precisely fits through the initial value, personalizing the general solution.

Constant of Integration

When integrating a function, a constant of integration, typically denoted as \( C \), is always added to the equation. This is because integration can produce infinitely many solutions differing by a constant.

In our case, \( y = x^2 - 7x + C \) represents the general solution of the differential equation. However, without knowing \( C \), we have a wide range of potential solutions. Here's why \( C \) is critical:

• It represents the vertical shift in the graph of the function \( y = x^2 - 7x \).
• To find its true value, additional information like an initial condition is necessary.
• The specific initial condition \( y(2) = 0 \) allows us to calculate \( C \) precisely.
• Solving \( 0 = 4 - 14 + C \) gives \( C = 10 \), fixing the precise vertical position of the curve.
• Thus, the constant transforms a generic solution into a unique solution that fits the problem's criteria.

This constant is crucial in fine-tuning mathematically arrived solutions to match real-world scenarios represented by differential equations.