91Ó°ÊÓ

A differential equation is a mathematical equation that relates some function with its derivatives. In this exercise, we are given the differential equation \( \frac{dy}{dx} = \frac{1}{x^2+1} - 1 \). This equation expresses the rate of change of the function \( y \) with respect to \( x \). Differential equations are crucial because they describe various real-world systems, such as motion, heat, and electrical circuits.
Solving a differential equation involves finding the original function \( y(x) \) that fulfills this rate of change description, and this typically requires techniques from calculus.

• The left side \( \frac{dy}{dx} \) indicates the derivative of \( y \), showing how \( y \) changes as \( x \) changes.
• The right side \( \frac{1}{x^2+1} - 1 \) gives the specific formula for this rate of change.

Understanding how to rearrange these equations and solve them is fundamental to exploring many scientific and engineering problems.

Integration is a key concept in calculus used to find a function when its derivative is known. In this problem, the task is to find \( y \) based on its derivative \( \frac{dy}{dx} \). Integration is the process of reversing differentiation.
In this exercise, the integral of \( \frac{1}{x^2+1} - 1 \) is calculated. This involves two main integrals:

• \( \int \frac{1}{x^2+1} \, dx \) which results in \( \tan^{-1}(x) \), since the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{x^2+1} \).
• \( \int -1 \, dx \) which gives \( -x \), because the reverse of the constant \(-1\) is linear \(-x\).

By summing these integrated parts, we approach the function solution. Don't forget to add \( C \), the constant of integration, at the end, reflecting that many functions can share the same derivative.

An initial condition specifies the value of the function at a particular point, enabling the identification of a specific solution from a family of solutions. The given initial condition in this problem is \( y(0) = 1 \). This constraint helps to determine the exact value of the constant \( C \) in the integrated function.
Use the initial condition by substituting \( x = 0 \) and \( y = 1 \) into the equation: \( 1 = \tan^{-1}(0) - 0 + C \).
The term \( \tan^{-1}(0) \) equals zero, simplifying the equation to \( 1 = 0 + C \). Thus, \( C = 1 \).
Applying initial conditions is essential in finding a specific solution that fits specific scenarios or physical conditions.

A particular solution is a solution to a differential equation that satisfies the initial conditions. For this exercise, the particular solution is the final equation that includes the constant \( C \) as determined by the initial condition.
After solving the integration and applying the initial condition, we find that the particular solution is:

• \( y = \tan^{-1}(x) - x + 1 \)

This equation not only satisfies the original differential equation but also meets the initial condition \( y(0) = 1 \).
The particular solution is essential as it represents a specific curve fitting the original problem setup. This solution can be invaluable in practical applications, where exact fits to models are necessary.