Key Concepts and Formulas

• Homogeneous Differential Equation: A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(\lambda x, \lambda y) = f(x,y)$ for all $\lambda \neq 0$. Alternatively, $f(x,y)$ can be expressed as a function of $\frac{y}{x}$.
• Solution Method for Homogeneous Equations: Substitute $y = vx$, which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$. This transforms the original equation into a separable differential equation in terms of $v$ and $x$.
• Standard Equation of a Circle: The equation $(x-h)^2 + (y-k)^2 = r^2$ represents a circle with center $(h,k)$ and radius $r$.

Step-by-Step Solution

1. Rearrange the Differential Equation into $\frac{dy}{dx}$ form: We are given the differential equation $(x^2 - y^2)dx + 2xy dy = 0$. Our goal is to isolate $\frac{dy}{dx}$ on one side. Subtract $(x^2-y^2)dx$ from both sides: $2xy dy = -(x^2 - y^2) dx$ $2xy dy = (y^2 - x^2) dx$ Divide both sides by $2xy \, dx$, assuming $x \neq 0$ and $y \neq 0$: $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ Now, we have the differential equation in the form $\frac{dy}{dx} = f(x,y)$, where $f(x,y) = \frac{y^2 - x^2}{2xy}$.

2. Identify as a Homogeneous Differential Equation: To verify homogeneity, let's check if $f(\lambda x, \lambda y) = f(x,y)$: $f(\lambda x, \lambda y) = \frac{(\lambda y)^2 - (\lambda x)^2}{2(\lambda x)(\lambda y)} = \frac{\lambda^2 y^2 - \lambda^2 x^2}{2\lambda^2 xy} = \frac{\lambda^2 (y^2 - x^2)}{\lambda^2 (2xy)} = \frac{y^2 - x^2}{2xy} = f(x,y)$ Since $f(\lambda x, \lambda y) = f(x,y)$, the differential equation is homogeneous.

3. Apply the Substitution $y = vx$: We substitute $y = vx$. Then, differentiating with respect to $x$ using the product rule, we get: $\frac{dy}{dx} = v + x \frac{dv}{dx}$

4. Substitute $y=vx$ and $\frac{dy}{dx}$ into the Differential Equation: Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$, we have: $v + x \frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)} = \frac{v^2 x^2 - x^2}{2vx^2} = \frac{x^2 (v^2 - 1)}{2vx^2} = \frac{v^2 - 1}{2v}$

5. Separate Variables: Now, we separate the variables $v$ and $x$: $x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-v^2 - 1}{2v} = -\frac{v^2 + 1}{2v}$ So, we have: $x \frac{dv}{dx} = -\frac{v^2 + 1}{2v}$ Separating variables, we get: $\frac{2v}{v^2 + 1} dv = -\frac{1}{x} dx$

6. Integrate Both Sides: Integrate both sides: $\int \frac{2v}{v^2 + 1} dv = \int -\frac{1}{x} dx$ The left-hand side integral is $\ln(v^2 + 1)$ and the right-hand side integral is $-\ln|x| + C$. Therefore, $\ln(v^2 + 1) = -\ln|x| + C$ Let $C = \ln|c|$. Then, $\ln(v^2 + 1) = -\ln|x| + \ln|c| = \ln\left|\frac{c}{x}\right|$ Taking the exponential of both sides, we get: $v^2 + 1 = \frac{c}{x}$

7. Substitute back $v = y/x$: Substitute $v = \frac{y}{x}$ back into the equation: $\left(\frac{y}{x}\right)^2 + 1 = \frac{c}{x}$ $\frac{y^2}{x^2} + 1 = \frac{c}{x}$ Multiplying by $x^2$, we get: $y^2 + x^2 = cx$ $x^2 + y^2 - cx = 0$

8. Apply the Initial Condition: The curve passes through $(1, 1)$. Substituting $x = 1$ and $y = 1$, we get: $(1)^2 + (1)^2 - c(1) = 0$ $1 + 1 - c = 0$ $c = 2$

9. Write the Particular Solution and Identify the Curve: Substituting $c = 2$ into $x^2 + y^2 - cx = 0$, we get: $x^2 + y^2 - 2x = 0$ Completing the square for the $x$ terms: $(x^2 - 2x + 1) + y^2 = 1$ $(x - 1)^2 + y^2 = 1$ This is the equation of a circle with center $(1, 0)$ and radius $1$.

Common Mistakes & Tips:

• Forgetting the Product Rule: Remember to use the product rule when differentiating $y = vx$ to find $\frac{dy}{dx}$.
• Sign Errors: Be careful with signs when rearranging and simplifying equations.
• Completing the Square: Practice completing the square to identify the equation of the curve accurately.

Summary: We identified the differential equation as homogeneous, applied the substitution $y=vx$, separated variables, integrated both sides, substituted back to the original variables, and used the initial condition to find the particular solution. By completing the square, we found the equation of the curve to be $(x-1)^2 + y^2 = 1$, which represents a circle of radius 1.

The final answer is \boxed{\text{a circle of radius one.}}, which corresponds to option (A).