Problem 91 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Integrate the Differential Equation

We start by integrating the differential equation. The given equation is \( \frac{d y}{d x} = 2x - 7 \). To find \( y \), integrate with respect to \( x \): \( y = \int (2x - 7) \,dx = \int 2x\,dx - \int 7\,dx = x^2 - 7x + C \)Here, \( C \) is the integration constant.

Step 2: Apply the Initial Condition

We use the given initial condition \( y(2) = 0 \) to find the constant \( C \). Substitute \( x = 2 \) and \( y = 0 \) into the integrated function: \( 0 = 2^2 - 7(2) + C \)\( 0 = 4 - 14 + C \)Solve for \( C \): \( C = 10 \)

Step 3: Write the Particular Solution

Substitute the value of \( C \) back into the integrated function. The particular solution is:\( y = x^2 - 7x + 10\)

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) includes a differential equation along with specific values, known as initial conditions. These conditions define the value of the unknown function, usually at a certain point. In the problem exercise given, the differential equation \( \frac{dy}{dx} = 2x - 7 \) is accompanied by the initial condition \( y(2) = 0 \). This initial condition is essential because it allows us to determine a unique solution out of an infinite set of possible solutions. Without this condition, we would only have a general solution that includes an arbitrary constant.

• The differential equation indicates how \( y \) changes with \( x \).
• The initial condition pinpoints the specific path among all potential solutions by providing a concrete starting point.

By solving the IVP accurately, you can find not just a solution, but the particular solution that satisfies both the equation and the initial condition!

Integration Process

The integration process is a crucial step in solving differential equations. It involves finding the integral of the equation, essentially reversing the process of differentiation. In our exercise, we start with the equation \( \frac{dy}{dx} = 2x - 7 \). To find \( y \), we integrate both sides of the differential equation with respect to \( x \). The process is broken down like this:

• Separate the terms: \( y = \int (2x - 7) \, dx \) becomes \( y = \int 2x \, dx - \int 7 \, dx \).
• Integrate each term separately: The integration of \( 2x \) with respect to \( x \) is \( x^2 \), and the integration of \( 7 \) is \( 7x \).
• The resulting general solution is: \( y = x^2 - 7x + C \), where \( C \) is the constant of integration.

The integration process transforms the differential equation into a form that can include a constant. This constant is key to finding specific solutions when using initial conditions.

Particular Solution

A particular solution of a differential equation is a solution that satisfies both the equation itself and any given initial conditions. In our problem, after integrating the differential equation into \( y = x^2 - 7x + C \), we need to find \( C \) using the initial condition \( y(2) = 0 \).Follow these steps to determine the particular solution:

• Plug in the initial condition values into the equation: \( 0 = 2^2 - 7(2) + C \).
• Solve the resulting equation: This gives us \( 0 = 4 - 14 + C \), which simplifies to \( C = 10 \).
• Substitute \( C \) back into the integrated equation to form the particular solution: \( y = x^2 - 7x + 10 \).

The particular solution is distinguished from the general solution as it uniquely fits the initial condition given in the problem. This makes it the exact solution to the specific scenario outlined in the differential equation along with its initial condition.