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To solve the differential equation \( xy' + y - y = 0 \) using the Frobenius method, let's simplify and proceed with the steps required for the Frobenius approach.

### Given Equation
The equation appears as:
\[
xy' + y - y = 0
\]

After simplifying, it reduces to:
\[
xy' = 0
\]

This is an unusual form for applying the Frobenius method since Frobenius is typically used for equations around regular singular points. Here, we have \( xy' = 0 \), which implies that:
1. \( y' = 0 \) if \( x \neq 0 \).
2. Consequently, \( y = C \), where \( C \) is a constant.

This indicates that the solution is simply a constant function rather than requiring a series solution.

### Solution
The solution to \( xy' = 0 \) is:
\[
y(x) = C
\]

### Summary
In this case, there is no need to apply the Frobenius method as the differential equation simplifies to a form that yields a constant solution directly.

Would you like further details on this, or do you have any other questions?

### Additional Related Questions
1. How does the Frobenius method typically apply to second-order differential equations?
2. What are regular singular points, and why are they important for the Frobenius method?
3. How would we proceed if the differential equation were in a more complex form, like \( x^2 y'' + xy' + y = 0 \)?
4. Can you provide an example of a differential equation where the Frobenius method is essential?
5. What are the steps for constructing a Frobenius series solution?

### Tip
Always check if a differential equation simplifies before applying advanced methods, as this can save time and effort.

Use Frobenius method to solve the differential equations (b) xy' + y - y = 0.

This page provides a step-by-step solution for solving the differential equation xy' + y - y = 0 using the Frobenius method. The solution highlights the simplification steps and explains why a constant solution emerges in this case, ideal for students learning about differential equations and series solutions.

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