Step 1: Understanding the Concept:
This is a first-order, higher-degree differential equation. It is a quadratic equation in \(p = \frac{dy}{dx}\). We can solve for \(p\) by factoring the quadratic expression, which will give us two simpler first-order differential equations to solve. The general solution will be the product of the solutions to these two equations.
Step 2: Key Formula or Approach:
1. Treat the equation as a quadratic in \(p\) and factor it.
2. Set each factor to zero to get two differential equations.
3. Solve each differential equation separately.
4. Combine the solutions into a single general solution.
Step 3: Detailed Explanation:
The given differential equation is:
\[ p^2 + p\left(\frac{x}{y} - \frac{y}{x}\right) - 1 = 0 \]
This is a quadratic equation in \(p\). Let's try to factor it.
We can rewrite the equation as:
\[ p^2 + \frac{x}{y}p - \frac{y}{x}p - 1 = 0 \]
Group the terms:
\[ \left(p^2 + \frac{x}{y}p\right) - \left(\frac{y}{x}p + 1\right) = 0 \]
Factor out common terms from each group:
\[ p\left(p + \frac{x}{y}\right) - \frac{y}{x}\left(p + \frac{x}{y}\right) = 0 \]
Now, factor out the common binomial term \((p + \frac{x}{y})\):
\[ \left(p - \frac{y}{x}\right)\left(p + \frac{x}{y}\right) = 0 \]
This gives us two separate first-order differential equations:
Case 1: \(p - \frac{y}{x} = 0\)
\[ p = \frac{dy}{dx} = \frac{y}{x} \]
This is a separable equation.
\[ \frac{dy}{y} = \frac{dx}{x} \]
Integrating both sides:
\[ \int \frac{dy}{y} = \int \frac{dx}{x} \]
\[ \ln|y| = \ln|x| + \ln|c_1| \implies y = c_1x \]
This can be written as \(y - c_1x = 0\).
Case 2: \(p + \frac{x}{y} = 0\)
\[ p = \frac{dy}{dx} = -\frac{x}{y} \]
This is also a separable equation.
\[ y \, dy = -x \, dx \]
Integrating both sides:
\[ \int y \, dy = \int -x \, dx \]
\[ \frac{y^2}{2} = -\frac{x^2}{2} + c_2 \]
\[ x^2 + y^2 = 2c_2 \]
Let \(C = 2c_2\). The solution is \(x^2 + y^2 - C = 0\).
The combined general solution is the product of the two families of solutions:
\[ (y - cx)(x^2 + y^2 - c) = 0 \]
Note on Discrepancy: The derived solution does not match any of the options. Let's re-examine the question and options. It's possible there is a typo in the question. Let's analyze the provided correct answer: (C) \((xy - c)(x^2 - y^2 - c) = 0\).
This implies two families of solutions:
1. \(xy = c \implies y = c/x \implies p = \frac{dy}{dx} = -c/x^2 = -(xy)/x^2 = -y/x\). This gives \(p + y/x = 0\).
2. \(x^2 - y^2 = c \implies 2x - 2y \frac{dy}{dx} = 0 \implies p = \frac{dy}{dx} = x/y\). This gives \(p - x/y = 0\).
The differential equations corresponding to the correct answer are \(p = -y/x\) and \(p = x/y\).
The combined differential equation would be:
\[ (p + y/x)(p - x/y) = 0 \]
\[ p^2 + p\left(\frac{y}{x} - \frac{x}{y}\right) - 1 = 0 \]
This is slightly different from the given equation \(p^2 + p(\frac{x}{y} - \frac{y}{x}) - 1 = 0\). The term in the parenthesis is flipped. Assuming the question intended to be the one that yields the marked answer, we proceed with the DE \(p^2 - p(\frac{x}{y} - \frac{y}{x}) - 1 = 0\). However, working with the given equation as written, the factorization is \((p + x/y)(p - y/x) = 0\), which gives \(p=-x/y\) and \(p=y/x\), leading to solutions \(x^2+y^2=c\) and \(y=cx\). The combined solution is \((x^2+y^2-c)(y-cx)=0\). Given the provided key, it is highly likely the question had a typo.
We will assume the question should have been \(p^2 + p(\frac{y}{x} - \frac{x}{y}) - 1 = 0\).
Step 4: Final Answer:
Assuming a typo in the question and that it should lead to the provided answer key, the solution is \((xy - c)(x^2 - y^2 - c) = 0\). This corresponds to solving \(p = -y/x\) and \(p = x/y\), which come from the factored form \((p + y/x)(p - x/y) = 0\).