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.

Learn how to separate variables and integrate the differential equation (1 - x) y' = y^2 step-by-step. This problem covers differential equations, separation of variables, and integration by substitution. Suitable for advanced undergraduate students and those interested in differential equations.

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