[latex]
\[
\textbf{Solve the differential equation:} \quad \frac{dy}{dx} = xy + 1
\]
\[
\textbf{Step 1: Rewrite the equation.}
\]
\[
\frac{dy}{dx} – xy = 1
\]
\[
\textbf{Step 2: Identify the type of differential equation.}
\]
This is a first-order linear differential equation in the form:
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
where \( P(x) = -x \) and \( Q(x) = 1 \).
\[
\textbf{Step 3: Find the integrating factor (I.F.).}
\]
\[
\text{I.F.} = e^{\int P(x) \, dx} = e^{\int -x \, dx} = e^{-\frac{x^2}{2}}
\]
\[
\textbf{Step 4: Multiply through by the integrating factor.}
\]
\[
e^{-\frac{x^2}{2}} \frac{dy}{dx} – e^{-\frac{x^2}{2}} xy = e^{-\frac{x^2}{2}}
\]
\[
\textbf{Step 5: Simplify the left-hand side.}
\]
\[
\frac{d}{dx} \left( e^{-\frac{x^2}{2}} y \right) = e^{-\frac{x^2}{2}}
\]
\[
\textbf{Step 6: Integrate both sides.}
\]
\[
\int \frac{d}{dx} \left( e^{-\frac{x^2}{2}} y \right) dx = \int e^{-\frac{x^2}{2}} dx
\]
\[
e^{-\frac{x^2}{2}} y = \int e^{-\frac{x^2}{2}} dx + C
\]
\[
\textbf{Step 7: Solve for \( y \).}
\]
\[
y = e^{\frac{x^2}{2}} \left( \int e^{-\frac{x^2}{2}} dx + C \right)
\]
\[
\textbf{Final solution:} \quad y = e^{\frac{x^2}{2}} \left( \int e^{-\frac{x^2}{2}} dx + C \right)
\]