3. Solve the equation \( y^{\prime}-2 x y=x \) \[ \begin{array}{l}\text { A. } y(x)=C e^{x^{-2}}-\frac{1}{2} \\ \text { B. } y(x)=C e^{x^{2}}-\frac{1}{2} \\ \text { C. } y(x)=C e^{x^{-2}}+\frac{1}{2} \\ \text { D. } y(x)=C e^{x^{2}}+\frac{1}{2}\end{array} \]

To solve the differential equation \( y^{\prime}-2 x y=x \), we can rearrange it into a standard linear form: \[ y' - 2xy = x. \] The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int -2x \, dx} = e^{-x^2}. \] Now, we multiply the entire equation by the integrating factor: \[ e^{-x^2}y' - 2x e^{-x^2}y = x e^{-x^2}. \] The left-hand side simplifies to: \[ \frac{d}{dx}\left(e^{-x^2}y\right) = x e^{-x^2}. \] Integrating both sides, we get: \[ e^{-x^2}y = -\frac{1}{2} e^{-x^2} + C, \] where \( C \) is the constant of integration. Multiplying throughout by \( e^{x^2} \) gives: \[ y = C e^{x^2} - \frac{1}{2}. \] Thus, the correct option is option B: \[ y(x) = C e^{x^{2}} - \frac{1}{2}. \] --- Did you know that linear differential equations like this one were extensively studied during the 18th and 19th centuries by mathematicians such as Joseph-Louis Lagrange and Leonhard Euler? Their groundwork laid the foundation for many modern techniques we use today in solving such equations! In practical terms, understanding how to apply integrating factors can come in handy beyond pure mathematics. Engineers might use similar techniques when dealing with systems modeled by differential equations, such as in control systems or population dynamics, ensuring that solutions reflect real-time behaviors and trends.