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The problem asks for the solution of the differential equation:

\[
e^{x^2} y y' + x = 0 \quad \text{with} \quad y(0) = 2.
\]

### Step-by-step Explanation:

1. **Rewrite the differential equation**:
   Start by simplifying the given equation:
   \[
   e^{x^2} y y' + x = 0.
   \]
   Rearrange it:
   \[
   e^{x^2} y y' = -x.
   \]
   Now, divide both sides by \( e^{x^2} y \) to isolate \( y' \):
   \[
   y' = -\frac{x}{e^{x^2} y}.
   \]
   
2. **Separate variables**:
   The next step is to use separation of variables. Multiply both sides by \( y \) and \( dx \) to separate \( y \) and \( x \):
   \[
   y \, dy = -x e^{-x^2} \, dx.
   \]

3. **Integrate both sides**:
   Now, integrate both sides of the equation. The left-hand side is straightforward:
   \[
   \int y \, dy = \frac{1}{2} y^2.
   \]
   For the right-hand side:
   \[
   \int -x e^{-x^2} \, dx.
   \]
   Using the substitution \( u = x^2 \), with \( du = 2x \, dx \), the integral becomes:
   \[
   \frac{1}{2} \int e^{-u} \, du = \frac{1}{2} e^{-x^2} + C.
   \]
   So, after integration, we get:
   \[
   \frac{1}{2} y^2 = \frac{1}{2} e^{-x^2} + C.
   \]

4. **Solve for \( y^2 \)**:
   Multiply both sides by 2 to simplify:
   \[
   y^2 = e^{-x^2} + 2C.
   \]

5. **Use the initial condition \( y(0) = 2 \)**:
   To find the constant \( C \), use the initial condition \( y(0) = 2 \). Substitute \( x = 0 \) and \( y = 2 \) into the equation:
   \[
   2^2 = e^{-(0)^2} + 2C,
   \]
   \[
   4 = 1 + 2C.
   \]
   Solving for \( C \):
   \[
   2C = 3 \quad \Rightarrow \quad C = \frac{3}{2}.
   \]

6. **Final solution**:
   Substitute the value of \( C \) back into the equation:
   \[
   y^2 = e^{-x^2} + 3.
   \]

This is the final solution to the differential equation.

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

### Related Questions:
1. What substitution was used to integrate the right-hand side of the equation?
2. Why is separation of variables a useful technique for solving this type of differential equation?
3. How would the solution change if the initial condition was different, e.g., \( y(0) = 1 \)?
4. Can this equation be solved using any other methods, such as an integrating factor?
5. What is the importance of the initial condition in determining the constant \( C \)?

### Tip:
When integrating a function that includes an exponential and a polynomial, substitution often simplifies the process effectively.

Please explain how they got the answer for the differential equation e^(x^2) * y * y' + x = 0 with y(0) = 2.

Learn how to solve the first-order differential equation e^(x^2) * y * y' + x = 0 using the separation of variables technique and initial conditions. This example includes integration by substitution and solving for constants.

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