1. **Stating the problem:** Solve the differential equation $$y'' + 2xy' + x'y = 0$$ where $y''$ is the second derivative of $y$ with respect to $x$, $y'$ is the first derivative, and $x'$ is the derivative of $x$ with respect to itself. 2. **Clarifying the terms:** Since $x$ is the independent variable, its derivative with respect to itself is 1, so $x' = 1$. The equation becomes: $$y'' + 2xy' + y = 0$$ 3. **Recognizing the equation:** This is a second-order linear differential equation with variable coefficients. It resembles the Hermite differential equation: $$y'' - 2xy' + 2ny = 0$$ but with different signs and coefficients. 4. **Attempting a solution:** We try a power series solution or recognize it as a form of the Hermite equation with a sign change. Alternatively, use substitution or known methods for such equations. 5. **Using substitution:** Let’s try $y = e^{-x^2/2}u(x)$ to simplify the equation. 6. **Compute derivatives:** $$y' = e^{-x^2/2}u' - xe^{-x^2/2}u$$ $$y'' = e^{-x^2/2}u'' - 2xe^{-x^2/2}u' + (x^2 - 1)e^{-x^2/2}u$$ 7. **Substitute into the original equation:** $$y'' + 2xy' + y = e^{-x^2/2}[u'' - 2xu' + (x^2 - 1)u] + 2x e^{-x^2/2}[u' - xu] + e^{-x^2/2}u = 0$$ Simplify inside the bracket: $$u'' - 2xu' + (x^2 - 1)u + 2xu' - 2x^2 u + u = u'' + (x^2 - 1)u - 2x^2 u + u = u'' + (-x^2)u = 0$$ 8. **Simplify:** $$u'' - x^2 u = 0$$ 9. **Solve the simplified equation:** This is a form of the differential equation for the parabolic cylinder functions. The general solution is: $$u(x) = C_1 D_{-1/2}(x \\sqrt{2}) + C_2 D_{-1/2}(-x \\sqrt{2})$$ 10. **Write the general solution for $y$:** $$y = e^{-x^2/2} u(x) = e^{-x^2/2} \left[C_1 D_{-1/2}(x \\sqrt{2}) + C_2 D_{-1/2}(-x \\sqrt{2})\right]$$ **Final answer:** $$\boxed{y = e^{-x^2/2} \left[C_1 D_{-1/2}(x \\sqrt{2}) + C_2 D_{-1/2}(-x \\sqrt{2})\right]}$$ This is the general solution to the given differential equation.