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Comprueba si e^(-2x) y xe^(-2x) son soluciones 

The given differential equation is:

\[
x y'' - y' - 2(2x - 1)y = 0
\]

We need to check if the functions \( e^{-2x} \) and \( xe^{-2x} \) are solutions to this equation. Let's proceed step by step for each function.

### 1. Verifying \( y_1(x) = e^{-2x} \):

- First, compute the derivatives of \( y_1(x) \):

\[
y_1'(x) = \frac{d}{dx} \left( e^{-2x} \right) = -2e^{-2x}
\]

\[
y_1''(x) = \frac{d}{dx} \left( -2e^{-2x} \right) = 4e^{-2x}
\]

- Now, substitute \( y_1(x) \), \( y_1'(x) \), and \( y_1''(x) \) into the differential equation \( x y'' - y' - 2(2x - 1)y = 0 \):

\[
x(4e^{-2x}) - (-2e^{-2x}) - 2(2x - 1)(e^{-2x}) = 0
\]

Simplifying each term:

\[
4xe^{-2x} + 2e^{-2x} - 2(2x - 1)e^{-2x} = 0
\]

Now expand the third term:

\[
4xe^{-2x} + 2e^{-2x} - 4xe^{-2x} + 2e^{-2x} = 0
\]

Combining like terms:

\[
(4xe^{-2x} - 4xe^{-2x}) + (2e^{-2x} + 2e^{-2x}) = 0
\]

\[
0 + 4e^{-2x} = 0
\]

This simplifies to:

\[
4e^{-2x} = 0
\]

which is not true for any \( x \), so \( e^{-2x} \) is **not** a solution.

### 2. Verifying \( y_2(x) = xe^{-2x} \):

- First, compute the derivatives of \( y_2(x) \):

\[
y_2'(x) = \frac{d}{dx} \left( xe^{-2x} \right) = e^{-2x} - 2xe^{-2x} = (1 - 2x)e^{-2x}
\]

\[
y_2''(x) = \frac{d}{dx} \left( (1 - 2x)e^{-2x} \right) = (-2)e^{-2x} + 4xe^{-2x} = (-2 + 4x)e^{-2x}
\]

- Now, substitute \( y_2(x) \), \( y_2'(x) \), and \( y_2''(x) \) into the differential equation \( x y'' - y' - 2(2x - 1)y = 0 \):

\[
x((-2 + 4x)e^{-2x}) - ((1 - 2x)e^{-2x}) - 2(2x - 1)(xe^{-2x}) = 0
\]

Simplify each term:

\[
x(-2 + 4x)e^{-2x} - (1 - 2x)e^{-2x} - 2(2x - 1)xe^{-2x} = 0
\]

Expanding each term:

\[
(-2x + 4x^2)e^{-2x} - (1 - 2x)e^{-2x} - 2(2x^2 - x)e^{-2x} = 0
\]

Now combine like terms:

\[
(-2x + 4x^2)e^{-2x} - (1 - 2x)e^{-2x} - (4x^2 - 2x)e^{-2x} = 0
\]

Simplify further:

\[
( -2x + 4x^2 - 1 + 2x - 4x^2 + 2x )e^{-2x} = 0
\]

\[
(-1 + 2x)e^{-2x} = 0
\]

This simplifies to:

\[
(-1 + 2x)e^{-2x} = 0
\]

Since \( e^{-2x} \neq 0 \), we have:

\[
-1 + 2x = 0
\]

Solving for \( x \):

\[
x = \frac{1}{2}
\]

Thus, \( y_2(x) = xe^{-2x} \) satisfies the differential equation only when \( x = \frac{1}{2} \), so it is **not a general solution**.

### Conclusion:
- \( e^{-2x} \) is **not a solution**.
- \( xe^{-2x} \) satisfies the equation **only for \( x = \frac{1}{2} \)**.

Would you like to explore more details about the general solution or any particular aspect of this analysis?

Here are 5 related questions to consider:
1. How do you find the general solution to second-order linear differential equations?
2. Can \( e^{-2x} \) be part of a homogeneous solution for this equation?
3. What conditions must a function satisfy to be a solution to a differential equation?
4. How would the solution change if the differential equation included a non-homogeneous term?
5. What methods can be used to solve second-order differential equations with variable coefficients?

**Tip**: For differential equations, always check the boundary or initial conditions if provided, as they help define specific solutions.

Comprueba si e^(-2x) y xe^(-2x) son soluciones para la ecuación diferencial x y'' - y' - 2(2x - 1)y = 0.

Check if the functions e^(-2x) and xe^(-2x) are solutions to the second-order differential equation x y'' - y' - 2(2x - 1)y = 0. This problem involves solving second-order linear differential equations.

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