91Ó°ÊÓ

An initial value problem is a specific type of differential equation that not only needs to find a function, but also requires the solution to satisfy a given condition at a particular point. In this exercise, we were given a differential equation \( \frac{dy}{dx} = \sec x \) with the initial condition \( y(2) = 3 \). This tells us

• how the function \( y \) changes with \( x \)
• one specific point through which the function passes.

This initial condition is essential for finding a unique solution among the infinite number of possible solutions from the differential equation. It is like being given a specific starting point on a journey. With the initial condition, we can determine exactly which path our function \( y(x) \) follows.

Integration is the main tool we use to solve a differential equation. By integrating both sides of the equation, we find a solution that describes how one quantity changes with another. In this case, the differential equation is \( \frac{dy}{dx} = \sec x \). To find \( y \), we need to integrate to undo the differentiation:\[ y = \int \sec x \ dx + C \]This integral helps us find a general form of \( y \). However, without further information, it's just a family of solutions. In practical terms, integration can be likened to finding the area under a curve – helping us understand the accumulated change of a function over an interval.

When you integrate a function, you always add a constant of integration, usually denoted by \( C \). This is because the integral of a function represents a whole set of antiderivatives. Each antiderivative differs by a constant value. In the equation \( y = \int \sec x \, dx + C \), the constant \( C \) allows for all possible vertical shifts of the antiderivative curve. These shifts mean many potential solutions to the differential equation.

Finding the Constant

To pin down \( C \), we use the initial condition. In this case, given \( y(2) = 3 \), we substitute \( x = 2 \), helping solve for \( C \). This gives the specific solution of the differential equation that passes through the point we know:\[ C = 3 - \ln |\sec 2 + \tan 2| \]Thus, the constant of integration personalizes the general solution, aligning with any particular initial condition.