Final answer:
The solution of the second-order linear differential equation with given initial conditions is y(t) = te^{-4t}. The characteristic equation reveals a repeated root, leading to a general solution that is refined by applying initial conditions.
Step-by-step explanation:
The question asks to find the solution of the second-order linear differential equation y'' + 8y' + 16y = 0 with the initial conditions y(0) = 0 and y'(0) = 1. This is a homogeneous differential equation with constant coefficients. To solve it, we look for solutions of the form y = ert, where r is a constant to be determined.
First, we find the characteristic equation:
r2 + 8r + 16 = 0. This factors as (r + 4)2 = 0, which means r = -4 is a repeated root. Therefore, the general solution of the differential equation is y(t) = c1e-4t + c2te-4t, where c1 and c2 are constants.
We use the initial conditions to determine c1 and c2. Since y(0) = 0, we have 0 = c1. The derivative of y(t) is y'(t) = -4c1e-4t - 4c2te-4t + c2e-4t. Applying the second initial condition yields y'(0) = 1 = c2. Therefore, the solution satisfying the initial conditions is y(t) = te-4t.