Solve the differential equation y'' + 2y' + y = 0 with initial condition y(0) = 2.

The given differential equation is a second-order linear homogeneous equation with constant coefficients: $y'' + 2y' + y = 0$.
We assume a solution of the form $y = e^{rx}$. Then $y' = re^{rx}$ and $y'' = r^2 e^{rx}$.
Substituting these into the differential equation gives:

$$r^2 e^{rx} + 2re^{rx} + e^{rx} = 0$$

Divide by $e^{rx}$ (since $e^{rx} \neq 0$):

$$r^2 + 2r + 1 = 0$$

This is the characteristic equation.

The characteristic equation is a quadratic equation. It can be factored as a perfect square:

$$(r+1)^2 = 0$$

This yields a repeated real root:

$$r = -1$$

For a second-order linear homogeneous differential equation with a repeated real root $r$, the general solution is of the form:

$$y(x) = C_1 e^{rx} + C_2 x e^{rx}$$

Substitute the root $r = -1$:

$$y(x) = C_1 e^{-x} + C_2 x e^{-x}$$

Here, $C_1$ and $C_2$ are arbitrary constants.

We are given the initial condition $y(0) = 2$. Substitute $x=0$ and $y(0)=2$ into the general solution:

$$2 = C_1 e^{-(0)} + C_2 (0) e^{-(0)}$$ $$2 = C_1 (1) + C_2 (0)(1)$$ $$2 = C_1$$

So, $C_1 = 2$.

Substitute the value of $C_1$ back into the general solution. Since no condition for $y'(0)$ is provided, $C_2$ remains an arbitrary constant.

$$y(x) = 2e^{-x} + C_2 x e^{-x}$$

The solution to the differential equation with the given initial condition is:

$$\boxed{y(x) = 2e^{-x} + C_2 x e^{-x}}$$

where $C_2$ is an arbitrary constant.