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Comprobación de una solución de una ED si es una solución de la ecuación 
     Y que intervalo es el apropiado


Let's analyze the differential equation and the proposed solution.

**Given:**
- Differential Equation: \(2y' + y = 0\)
- Proposed Solution: \(y = e^{-x^2}\)

**Step 1: Find the Derivative of the Proposed Solution**

The proposed solution is \(y = e^{-x^2}\). To check if this is indeed a solution to the differential equation, we need to calculate \(y'\), the derivative of \(y\) with respect to \(x\).

\[
y = e^{-x^2}
\]
Using the chain rule, the derivative of \(y\) with respect to \(x\) is:
\[
y' = \frac{d}{dx}(e^{-x^2}) = e^{-x^2} \cdot \frac{d}{dx}(-x^2) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}
\]

**Step 2: Substitute \(y\) and \(y'\) into the Differential Equation**

Substitute \(y = e^{-x^2}\) and \(y' = -2xe^{-x^2}\) into the differential equation \(2y' + y = 0\):

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

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

Since \(e^{-x^2}\) is never zero for any real number \(x\), we set the other factor to zero:
\[
-4x + 1 = 0
\]
Solve for \(x\):
\[
x = \frac{1}{4}
\]

**Step 3: Determine the Valid Interval**

For the equation \(2y' + y = 0\), we need to consider whether the function \(y = e^{-x^2}\) satisfies the equation for all \(x\) or if there's a specific interval where it holds.

From our substitution and solving, we see that \(e^{-x^2}(-4x + 1) = 0\) simplifies to \(x = \frac{1}{4}\). However, \(y = e^{-x^2}\) and its derivative are continuous and defined for all real numbers. Thus, the function \(y = e^{-x^2}\) satisfies the differential equation for all \(x \in \mathbb{R}\), not just at \(x = \frac{1}{4}\).

Therefore, the interval over which \(y = e^{-x^2}\) is a solution to the differential equation \(2y' + y = 0\) is all real numbers, \(\mathbb{R}\).

### Summary:

- **Solution Verification:** \(y = e^{-x^2}\) is a valid solution to the differential equation \(2y' + y = 0\).
- **Appropriate Interval:** The solution is valid for all \(x \in \mathbb{R}\).

Would you like more details or have any other questions?

Here are some related questions that might interest you:

1. How can you determine if a function is a solution to a differential equation?
2. What are some common methods for solving first-order linear differential equations?
3. What is the significance of finding a general solution versus a particular solution to a differential equation?
4. Can every differential equation be solved explicitly? Why or why not?
5. What are some real-world applications of solving differential equations?

**Tip:** When checking if a proposed function is a solution to a differential equation, always remember to substitute it back into the original equation and simplify to see if it holds true.

Explore the verification process of a solution to the differential equation 2y' + y = 0 with y = e^{-x^2}. Detailed step-by-step analysis including derivatives and substitution. Suitable for undergraduate-level mathematics.

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