The given differential equation is:

\[
(x - 2xy) \, dx - dy = 0
\]

This can be rewritten as:

\[
(x - 2xy) \, dx = dy
\]

To solve this, we can attempt separation of variables or other methods like exact differential equations. Let's first check if the equation can be separated:

\[
\frac{dy}{dx} = x - 2xy
\]

Factor the right-hand side:

\[
\frac{dy}{dx} = x(1 - 2y)
\]

This is now separable, meaning we can rewrite it as:

\[
\frac{dy}{1 - 2y} = x \, dx
\]

Now, we integrate both sides:

\[
\int \frac{1}{1 - 2y} \, dy = \int x \, dx
\]

The left side can be solved using a substitution \( u = 1 - 2y \), which gives \( du = -2 \, dy \). Thus, the integral becomes:

\[
-\frac{1}{2} \ln|1 - 2y| = \frac{x^2}{2} + C
\]

Multiply both sides by -2:

\[
\ln|1 - 2y| = -x^2 + C_1
\]

Exponentiate both sides:

\[
|1 - 2y| = e^{-x^2 + C_1} = e^{C_1} e^{-x^2}
\]

Let \( A = e^{C_1} \), which is a constant:

\[
|1 - 2y| = A e^{-x^2}
\]

Finally, remove the absolute value and solve for \( y \):

\[
1 - 2y = \pm A e^{-x^2}
\]

\[
y = \frac{1 \mp A e^{-x^2}}{2}
\]

Thus, the solution is:

\[
y = \frac{1 \mp A e^{-x^2}}{2}
\]

Let me know if you'd like further details on any step or have additional questions!

Here are 5 questions related to this problem:
1. How would the solution change if the equation were \( (x - xy) dx - dy = 0 \)?
2. Can you explain the technique of separation of variables in more depth?
3. How can you verify if a differential equation is separable?
4. What would be the solution if we use different boundary conditions for this problem?
5. What happens if we add a constant term to the right-hand side of the original equation?

**Tip:** Always check if a differential equation is separable or exact before trying more complex methods.

Explore how to solve the differential equation (x - 2xy) dx - dy = 0 using separation of variables and integration techniques. Learn step-by-step how to simplify the equation and apply substitution to arrive at the solution. Ideal for students at the Calculus I or II level.

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