91Ó°ÊÓ

Differential equations form the foundation of many mathematical models in physics, engineering, and economics. They are equations that involve a function and its derivatives. In general, differential equations help describe how a particular quantity changes over time. For the given exercise, we start by considering the family of curves defined by the equation \( y = mx \). The most crucial step here is to derive its differential equation, which tells us about the rate of change of \( y \) concerning \( x \).To achieve this, we differentiate \( y = mx \) with respect to \( x \), giving us \( \frac{dy}{dx} = m \). Knowing that \( m = \frac{y}{x} \), we substitute back to get \( \frac{dy}{dx} = \frac{y}{x} \). This differential equation is essential to find the orthogonal trajectories of our given family of curves.

In mathematics, a family of curves is a set of curves that can be described by an equation containing a variable parameter. This parameter can change, giving us different curves within the same family. In the problem we are considering, the family of curves is represented by the equation \( y = mx \). Here, \( m \) serves as the parameter, which, when varied, yields different straight lines that all pass through the origin.

• For \( m = 1 \), the line is \( y = x \).
• For \( m = -2 \), the line is \( y = -2x \).
• For \( m = 0 \), the line is simply \( y = 0 \).

Such a family of lines will be infinite, showing all possible slopes for lines through the origin. Understanding this concept of a parameter controlling the specific curve is critical when analyzing and visualizing the family's behavior.

The principle of orthogonality is a vital concept in geometry. Two curves are orthogonal if their tangents at points of intersection are perpendicular. Mathematically, this involves ensuring that the product of their slopes at the intersection is -1.To determine the orthogonal trajectories of our original family \( y = mx \), we take its differential equation \( \frac{dy}{dx} = \frac{y}{x} \) and apply the orthogonality condition. We need a differential equation whose solution gives us curves that intersect the given family at right angles. By setting the product of the original slope \( \frac{y}{x} \) and the slope of the orthogonal trajectory as -1, we find \( \frac{dy}{dx} = -\frac{x}{y} \).This new differential equation guides us to the curve solutions that are orthogonal to the initial family, thus providing the desired geometric property.

Integration is the process of finding a function given its derivative, often called "the antiderivative". It is a fundamental technique used to solve differential equations. Once we have derived the differential equation \( \frac{dy}{dx} = -\frac{x}{y} \) for the orthogonal trajectories in our problem, we integrate both sides to find the actual equation of the curves.Starting from \( \int y \, dy = - \int x \, dx \), we perform the integration of both sides:

• Left side: Integrate \( y \, dy \) to get \( \frac{y^2}{2} \).
• Right side: Integrate \(- x \, dx \) to get \(-\frac{x^2}{2} + C \).

Combining both integrals yields \( \frac{y^2}{2} = -\frac{x^2}{2} + C \). Simplifying, we arrive at \( y^2 + x^2 = C \), the equation of a circle centered at the origin. Understanding integration is key to transitioning from a differential to a geometric equation, revealing the curve's shape and position.