If the slope of tangent at \((\mathrm{x}, \mathrm{y})\) to the curve passing through \((2,1)\) is \(\left[\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) /(2 \mathrm{xy})\right]\). The equation of the curve is: (A) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)=6 \mathrm{y}\) (B) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)=3 \mathrm{x}\) (C) \(x\left(x^{2}+y^{2}\right)=10\) (D) \(x\left(x^{2}-y^{2}\right)=6\)

Understanding the slope of a tangent line is pivotal when studying differential equations. The slope indicates how steep the line is at a certain point on the curve. A key aspect to remember is that the slope of a tangent line at any point on a curve is the derivative of the function at that point. In our exercise, the slope of the tangent line at any point \( (x, y) \) on the curve is given by \( \left[\frac{x^2+y^2}{2xy}\right] \). By setting this equal to the derivative \( \frac{dy}{dx} \) of the curve equation, we create a direct link between the slope at a specific point and the overall behavior of the curve. ewlineewline To ensure clearer understanding, visualize a curve on a graph. At any point on that curve, the tangent line is a straight line that just touches the curve without intersecting it and represents the instantaneous rate of change of the function at that point. The given slope formula in the exercise is what we use to find this rate of change, effectively telling us how the y-value changes with respect to the x-value at that specific point.

When we encounter a differential equation that is not readily separable, integrating factors can be a crucial technique to transform the equation into an integrable form. In this exercise, it's not explicitly mentioned, but the separation of variables technique we use is facilitated by an implied integrating factor of 1, which keeps the equation structurally sound during manipulation.ewlineewline

Understanding Integrating Factors

An integrating factor is essentially a function that we multiply by the original non-separable equation to make it separable. The chosen integrating factor will typically depend on the form of the differential equation and is derived by a method that ensures the left and right sides of the equation will integrate smoothly.ewlineewline In practice, integrating factors are more frequently used in linear differential equations, especially those of the first order. We look for an integrating factor when an equation cannot be separated as is, which was not the case in our exercise. However, it's important to recognize their potential role in solving more complex differential equations.

Separation of variables is a straightforward method for solving a particular type of differential equations. The main idea is to rearrange the equation so that each side contains variables of a single kind, meaning all the x's on one side and all the y's on the other. In our exercise, we execute this by reorganizing the given equation and isolating \(y\text{-terms}\) on one side and \(x\text{-terms}\) on the other, leading to \(\frac{2y}{y^2+1}dy = \frac{x}{x^2}dx\).ewlineewline Once separated, the next step is to integrate both sides independently. Through integration, we find an implicit solution to the differential equation. For students, remembering to integrate each side with respect to its own variable is key—integrating \(dy\) terms with respect to \(y\), and \(dx\) terms with respect to \(x\). As seen in the exercise, this method is incredibly powerful and elegant for simple types of differential equations, it's a fundamental technique to grasp early on in the study of differential equations.

The curve equation is the function that describes the shape of the curve in the coordinate space. It can often be expressed implicitly or explicitly, where an implicit equation includes both \(x\) and \(y\) without one being subject to the other, and an explicit equation defines \(y\) purely in terms of \(x\) (or vice versa). In our textbook exercise, after performing the necessary separation and integration, we attempted to derive an explicit curve equation that passes through the given point \( (2,1) \).ewlineewline The process introduced the natural logarithm function due to the integration of terms involving \(1/u\) and \(1/2v\), where \(u\) and \(v\) were the substituted variables representing \(y^2 + 1\) and \(x^2\), respectively. Solving for the constant using the given point, we found an equation describing the curve. While none of the answer choices matched our derived equation, understanding the steps to reach it reinforces the connection between differential equations, slopes, and their corresponding curves on a graph.