1. **State the problem:** Solve the differential equation $$y'' + 2y' + y = 0$$ with initial conditions $$y(0) = 2$$ and $$y'(0) = 10$$. 2. **Identify the type of equation:** This is a second-order linear homogeneous differential equation with constant coefficients. 3. **Write the characteristic equation:** Replace $$y''$$ by $$r^2$$, $$y'$$ by $$r$$, and $$y$$ by 1 to get: $$r^2 + 2r + 1 = 0$$ 4. **Solve the characteristic equation:** $$r^2 + 2r + 1 = (r + 1)^2 = 0$$ So, $$r = -1$$ is a repeated root. 5. **Write the general solution:** For repeated roots $$r$$, the solution is: $$y = (C_1 + C_2 x)e^{rx} = (C_1 + C_2 x)e^{-x}$$ 6. **Apply initial conditions:** - At $$x=0$$, $$y(0) = C_1 = 2$$ - Derivative: $$y' = C_2 e^{-x} - (C_1 + C_2 x)e^{-x} = (C_2 - C_1 - C_2 x)e^{-x}$$ At $$x=0$$: $$y'(0) = (C_2 - C_1) = 10$$ Substitute $$C_1 = 2$$: $$C_2 - 2 = 10 \Rightarrow C_2 = 12$$ 7. **Final solution:** $$y = (2 + 12x)e^{-x}$$ This function satisfies the differential equation and initial conditions.