Final answer:
To solve the initial value problem y'' + 6y' + 9y = 0 with y(0) = 2, we use the characteristic equation method and find a repeated root. The general solution includes exponentials and constants determined by initial conditions. Given the provided initial condition, the particular solution is y(t) = 2*e^{-3t}.
Step-by-step explanation:
We are asked to solve the initial value problem y'' + 6y' + 9y = 0 with the initial condition y(0) = 2. This is a second order linear homogeneous differential equation with constant coefficients. To solve it, we can use characteristic equations.
First, we set up the characteristic equation by replacing y with r and its derivatives with powers of r: r^2 + 6r + 9 = 0. This factors as (r + 3)^2 = 0, which has a repeated root of -3.
Because we have a repeated root, our general solution will be in the form y(t) = (C1 + C2*t)*e^{-3t}, where C1 and C2 are constants that we'll determine using our initial conditions.
Applying the initial condition y(0) = 2, we find that C1 = 2 because (2 + C2*0)*e^{0} = 2. Since there is no first derivative condition provided, we cannot solve for C2. However, the particular solution that satisfies the given initial condition is y(t) = 2*e^{-3t}.