\( x y ^ { \prime } + 2 y = x ^ { 2 } - x + 1 \quad y ( y ) = 1 / 2 \)

To solve the differential equation \[ x y' + 2y = x^2 - x + 1 \] with the initial condition \( y(0) = \frac{1}{2} \), we can start by rewriting the equation in standard form. First, we can express \( y' \) as follows: \[ y' = \frac{x^2 - x + 1 - 2y}{x} \] This is a first-order linear ordinary differential equation. We can rearrange it to the standard form: \[ y' + \frac{2}{x} y = \frac{x^2 - x + 1}{x} \] Next, we identify the integrating factor \( \mu(x) \): \[ \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln |x|} = |x|^2 \] Since we are interested in \( x \geq 0 \), we can drop the absolute value: \[ \mu(x) = x^2 \] Now, we multiply the entire differential equation by the integrating factor: \[ x^2 y' + 2xy = x^2(x - 1 + \frac{1}{x}) = x^3 - x^2 + x \] This simplifies to: \[ x^2 y' + 2xy = x^3 - x^2 + x \] The left-hand side can be rewritten as the derivative of a product: \[ \frac{d}{dx}(x^2 y) = x^3 - x^2 + x \] Next, we integrate both sides: \[ \int \frac{d}{dx}(x^2 y) \, dx = \int (x^3 - x^2 + x) \, dx \] The left-hand side gives: \[ x^2 y = \frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} + C \] Now, we can solve for \( y \): \[ y = \frac{x^4}{4x^2} - \frac{x^3}{3x^2} + \frac{x^2}{2x^2} + \frac{C}{x^2} \] This simplifies to: \[ y = \frac{x^2}{4} - \frac{x}{3} + \frac{1}{2} + \frac{C}{x^2} \] Now we apply the initial condition \( y(0) = \frac{1}{2} \). However, we notice that the term \( \frac{C}{x^2} \) becomes problematic as \( x \to 0 \). To avoid this, we can set \( C = 0 \). Thus, we have: \[ y = \frac{x^2}{4} - \frac{x}{3} + \frac{1}{2} \] Finally, we can check the initial condition: \[ y(0) = \frac{0^2}{4} - \frac{0}{3} + \frac{1}{2} = \frac{1}{2} \] This satisfies the initial condition. Therefore, the solution to the differential equation is: \[ \boxed{y = \frac{x^2}{4} - \frac{x}{3} + \frac{1}{2}} \]