Problem 69 - University Calculus: Early Transcendentals

Question

Express the solutions of the initial value problems in terms of integrals. $$\frac{d y}{d x}=\sec x, \quad y(2)=3$$

Step-by-Step Solution

Answer

The solution is \( y = \int \sec x \, dx - \int^{2} \sec x \, dx + 3 \).

Step 1: Understanding the Problem

The given differential equation is \( \frac{dy}{dx} = \sec x \) with the initial condition \( y(2) = 3 \). Our task is to express the solution of this initial value problem as an integral.

Step 2: Separate Variables

The differential equation \( \frac{dy}{dx} = \sec x \) is ready to be integrated directly with respect to \( x \). There is no need for separation as it is already in a suitable form.

Step 3: Integrate Both Sides

To find \( y \), we integrate both sides of \( \frac{dy}{dx} = \sec x \) with respect to \( x \). The integral form will be: \[ y = \int \sec x \, dx + C \]where \( C \) is the constant of integration.

Step 4: Use the Initial Condition

Apply the initial condition \( y(2) = 3 \) to find the constant \( C \). Substitute \( x = 2 \) and \( y = 3 \) into the integral equation:\[ 3 = \int^2 \sec x \, dx + C \]Thus, \( C = 3 - \int^{2} \sec x \, dx \).

Step 5: Substitute Back into the Integral Equation

Substitute \( C \, \) into the integral to express the complete solution:\[ y = \int \sec x \, dx + \left( 3 - \int^{2} \sec x \, dx \right) \] \[ y = \int \sec x \, dx - \int^{2} \sec x \, dx + 3 \] This expresses \( y \) in terms of an integral dependent on the given condition.

Key Concepts

Integrals

Integrals are a fundamental part of calculus, specifically useful in solving differential equations. An integral can be thought of as the mathematical method for calculating the area under a curve. This area calculation helps in finding unknowns within equations based on their rates of change.
In the context of the problem, we're dealing with the integral of the function \( \sec x \). The integral of \( \sec x \) allows us to determine the antiderivative, which is necessary to solve for \( y \) in the given differential equation.
Essentially, to solve our specific problem, we calculate:

• The indefinite integral \( y = \int \sec x \, dx + C \), where \( C \) is a constant that adjusts based on initial conditions.
• This integration gives us a general solution, which can be applied to find specific function values.

Integrals provide us with the means to backtrack from a derived rate of change to the original function. By applying the process of integration, we account for constant shifts in values within functions. Such shifts are often represented by constants like \( C \), derived from initial conditions.

Initial Value Problem

Initial value problems are a type of differential equation that include conditions provided at a specific point, \( x = a \). These problems require solutions that satisfy both the differential equation and the given condition. The initial condition guides us in choosing the correct constant \( C \) after integration.
In our problem, the differential equation \( \frac{dy}{dx} = \sec x \) is solved with the initial condition \( y(2) = 3 \). Here's how we approach such problems:

• First, integrate the differential equation to get the general form \( y = \int \sec x \, dx + C \).
• Next, apply the initial condition by substituting \( x = 2 \) and \( y = 3 \) to solve for \( C \).

The initial value is key to transforming a generic integral solution into a specific one that fits the context of the problem. This is done by replacing \( C \) with the value derived from solving the initial condition.

Integral Calculus

Integral calculus is the branch of mathematics dealing with integrals and their properties. It's pivotal not only to solve differential equations but to understand the broader implications of rates of change in physical systems. The primary aim of integral calculus is to find total values, such as areas and accumulated quantities, from given rates of change (differential equations).
The exercise above exemplifies the use of integral calculus in solving an initial value problem. Integral calculus allows us to determine the original function from its derivative:

• The process involves calculating indefinite integrals to find antiderivatives.
• Next, constants of integration are determined via additional information, such as initial conditions.

By using integral calculus, particularly through the integration of \( \sec x \), we solve for \( y \) with a precise value at \( x = 2 \). This shows integral calculus' strength in defining relationships involving quantities that change over time or space. Understanding these principles enables one to find specific solutions grounded in real-world criteria.