Key Concepts and Formulas

• Differential Equations: An equation involving derivatives of a function. A first-order differential equation involves only the first derivative.
• Variable Separable Differential Equations: A type of differential equation that can be written in the form $f(y) dy = g(x) dx$.
• Integration: The process of finding the antiderivative of a function. Remember the constant of integration, $C$.
• Equation of a Circle: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$.

Step-by-Step Solution

Step 1: Forming and Solving the Differential Equation

1. 1 Setting up the Differential Equation: We are given that the slope of the tangent to the curve $y = y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. This translates to the differential equation: $\frac{dy}{dx} = \frac{2y}{x^2}$ Reasoning: The slope of the tangent is represented by the derivative $\frac{dy}{dx}$.

2. 2 Separating Variables: We separate the variables to get all $y$ terms on one side and all $x$ terms on the other side: $\frac{dy}{y} = \frac{2}{x^2} dx$ Reasoning: This prepares the equation for integration with respect to each variable separately. We are assuming $y \neq 0$.

3. 3 Integrating Both Sides: Integrate both sides of the equation: $\int \frac{1}{y} dy = \int \frac{2}{x^2} dx$ $\log_e |y| = -\frac{2}{x} + C$ Reasoning: The integral of $\frac{1}{y}$ is $\log_e |y|$, and the integral of $\frac{2}{x^2} = 2x^{-2}$ is $2\frac{x^{-1}}{-1} = -\frac{2}{x}$. $C$ is the constant of integration.

Step 2: Finding the Initial Condition (Point on the Curve)

1. 1 Finding the Center of the Circle: The equation of the circle is given as $x^2 + y^2 – 2x – 2y = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we have $2g = -2$ and $2f = -2$, which gives $g = -1$ and $f = -1$. Thus, the center of the circle is $(-g, -f) = (1, 1)$. Reasoning: The center of the circle provides the point through which the curve passes, which is our initial condition.

Step 3: Determining the Constant of Integration

1. 1 Substituting the Initial Condition: Substitute $x = 1$ and $y = 1$ into the general solution: $\log_e |1| = -\frac{2}{1} + C$ $0 = -2 + C$ Reasoning: This step uses the initial condition to solve for the constant of integration, $C$.

2. 2 Solving for C: $C = 2$

Step 4: Obtaining the Particular Equation of the Curve

1. 1 Substituting C back: Substitute $C = 2$ back into the general solution: $\log_e |y| = -\frac{2}{x} + 2$

2. 2 Rearranging the Equation: Multiply both sides by $x$: $x \log_e |y| = -2 + 2x$ $x \log_e |y| = 2x - 2$ $x \log_e |y| = 2(x - 1)$ Reasoning: We manipulate the equation to match the given options.

Common Mistakes & Tips

• Missing the Constant of Integration: Always remember to add the constant of integration, $C$, when performing indefinite integrals.
• Absolute Value in Logarithms: When integrating $\frac{1}{y}$, remember to use $\log_e |y|$ to account for both positive and negative values of $y$.
• Algebraic Errors: Be careful with algebraic manipulations, especially when rearranging equations to match the given options.

Summary

We started by setting up and solving a variable separable differential equation. Then, we found the center of the given circle to obtain an initial condition. Using this initial condition, we solved for the constant of integration and substituted it back into the general solution to get the particular solution. Finally, we rearranged the equation to match one of the given options.

The final answer is $x \log_e |y| = 2(x - 1)$, which corresponds to option (A).

Final Answer The final answer is \boxed{x \log_e |y| = 2(x - 1)}, which corresponds to option (A).