The equation you provided is:

$\frac{dy}{dx} + 2xy = 0$

This is a first-order linear differential equation. To solve it, we can use the method of separation of variables or find an integrating factor. I will proceed using the separation of variables method.

Steps to solve:

1. Rewrite the equation: $\frac{dy}{dx} = -2xy$

2. Separate the variables $y$ and $x$: $\frac{dy}{y} = -2x \, dx$

3. Integrate both sides: $\int \frac{1}{y} \, dy = \int -2x \, dx$

4. The integrals give: $\ln |y| = -x^2 + C$ where $C$ is the constant of integration.

5. Solve for $y$: $y = e^{-x^2 + C} = e^C e^{-x^2}$

6. Let $A = e^C$ (where $A$ is a constant), the solution is: $y = A e^{-x^2}$

Thus, the general solution to the differential equation is:

$y = A e^{-x^2}$

Would you like further clarification on any steps or additional details?

Here are some related questions to explore further:

1. What is the integrating factor method for solving differential equations?
2. How would the solution change if the initial condition $y(0) = 1$ is given?
3. Can this method of separation of variables be applied to nonlinear equations?
4. What is the significance of the constant of integration $C$ in the solution?
5. How can we verify the solution by substituting back into the original equation?

Tip: Always check if a given differential equation can be simplified or solved using a known method like separation of variables or integrating factors.