The given differential equation is:

$xy' + y = x^2 e^{-x^2}$

Step 1: Rewrite the equation

We can rewrite this equation in a standard linear form. Divide through by $x$ (assuming $x \neq 0$):

$y' + \frac{y}{x} = x e^{-x^2}$

This is now a first-order linear differential equation of the form:

$y' + P(x)y = Q(x)$

where $P(x) = \frac{1}{x}$ and $Q(x) = x e^{-x^2}$.

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int P(x)\, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln |x|} = |x|$

Since we typically assume $x > 0$, we take $\mu(x) = x$.

Step 3: Multiply through by the integrating factor

Multiply both sides of the equation by $x$ to obtain:

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

Notice that the left-hand side is now the derivative of $xy$:

$\frac{d}{dx}(xy) = x^2 e^{-x^2}$

Step 4: Integrate both sides

Now integrate both sides with respect to $x$:

$xy = \int x^2 e^{-x^2} \, dx$

We can use substitution to evaluate the integral. Let $u = -x^2$, so $du = -2x \, dx$. This gives:

$\int x^2 e^{-x^2} \, dx = -\frac{1}{2} \int (-2x e^{-x^2}) \, dx = -\frac{1}{2} e^{-x^2}$

Step 5: Solve for $y$

Now substitute back into the equation:

$xy = -\frac{1}{2} e^{-x^2} + C$

Finally, solve for $y$:

$y = -\frac{1}{2x} e^{-x^2} + \frac{C}{x}$

Final Solution:

The general solution to the differential equation is:

$y(x) = -\frac{1}{2x} e^{-x^2} + \frac{C}{x}$

Let me know if you would like further clarifications or details!

Related Questions:

1. How do you identify a first-order linear differential equation?
2. What is the role of an integrating factor in solving differential equations?
3. How do you determine the integrating factor for a given linear differential equation?
4. Can substitution methods always simplify integrals like $\int x^2 e^{-x^2} \, dx$?
5. What are the conditions under which we can assume $x > 0$ when solving differential equations?

Tip: Always verify whether your differential equation is separable, exact, or linear to decide the best method for solving it.