We are given the differential equation:

$(2xy \, dx + dy) e^{x^2} = 0$

This is a first-order differential equation, and we are also given the initial condition $y(0) = 2$.

Step 1: Simplify the equation

We can start by dividing both sides by $e^{x^2}$, since $e^{x^2} \neq 0$, to eliminate the exponential:

$2xy \, dx + dy = 0$

Step 2: Rewrite in differential form

We now have the equation in a simpler form:

$2xy \, dx + dy = 0$

This can be rewritten as:

$dy = -2xy \, dx$

Step 3: Separate variables

Now, separate the variables $y$ and $x$:

$\frac{dy}{y} = -2x \, dx$

Step 4: Integrate both sides

Integrate both sides of the equation:

$\int \frac{dy}{y} = \int -2x \, dx$

On the left-hand side, we get:

$\ln |y| = -x^2 + C$

where $C$ is the constant of integration.

Step 5: Solve for $y$

To solve for $y$, exponentiate both sides:

$y = e^{-x^2 + C} = e^C \cdot e^{-x^2}$

Let $A = e^C$, where $A$ is a constant. Thus, the solution is:

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

Step 6: Apply the initial condition

We are given $y(0) = 2$. Substituting $x = 0$ and $y = 2$ into the solution:

$2 = A e^{-(0)^2} = A$

Thus, $A = 2$.

Final Solution:

The solution to the differential equation is:

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

Would you like more details on any step, or do you have any further questions?

Here are 5 related questions to deepen your understanding:

1. How would the solution change if the initial condition was $y(0) = 1$?
2. Can you solve $y = A e^{-x^2}$ for another initial condition, such as $y(1) = 3$?
3. What happens to $y(x)$ as $x \to \infty$?
4. How does this method apply to solving nonlinear differential equations?
5. What if the exponential term was $e^{x}$ instead of $e^{x^2}$?

Tip: When separating variables in differential equations, always check if the equation is exact or can be made exact to simplify solving!