To separate the variables for the differential equation $(1 - x) y' = y^2$, follow these steps:

1. Express the derivative $y'$ as $\frac{dy}{dx}$:

   $(1 - x) \frac{dy}{dx} = y^2$

2. Separate the variables $y$ and $x$ on different sides of the equation:

   $\frac{dy}{y^2} = \frac{dx}{1 - x}$

3. Integrate both sides:

   • For the left side:

     $\int \frac{dy}{y^2} = \int y^{-2} \, dy = \int -y^{-1} \, dy = \frac{1}{y} + C_1$

   • For the right side:

     $\int \frac{dx}{1 - x} = -\ln|1 - x| + C_2$

   After integrating, you get:

   $\frac{1}{y} = -\ln|1 - x| + C$

   where $C = C_2 - C_1$ is the constant of integration.

4. Simplify the solution:

   $y = \frac{1}{C - \ln|1 - x|}$

This is the general solution after separating the variables and integrating.

Would you like further details or have any questions?

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Here are some related questions:

1. How can you solve for $C$ if given an initial condition?
2. What happens if $y = 0$ in this context?
3. Can this equation be solved using a different method?
4. How does the solution behave as $x$ approaches 1?
5. What if $y$ is negative? How does the solution change?
6. Could you find a particular solution for a given $x$ and $y$ value?
7. How can you verify the solution by differentiating it?
8. What does the solution imply about the relationship between $y$ and $x$?

Tip: When integrating, always remember to include the constant of integration, as it plays a crucial role in finding the general solution.