Final answer:
The general solution to the homogeneous differential equation is y(t) = C1e^-3t + C2te^-3t. The functions normalized with f1(0)=1 and f2(0)=0 are f1(t) = 1 and f2(t) = te^-3t.
Step-by-step explanation:
The given homogeneous second-order linear differential equation is dt^2 / d^2y + 6 dt/dy + 9y = 0. To find the general solution, we assume a solution of the form y = ert. Substituting into the differential equation, we get the characteristic equation r^2 + 6r + 9 = 0, which can be factored as (r + 3)2 = 0. The repeated root is r = -3.
For repeated roots, the general solution can be expressed as y(t) = C1e-^3t + C2te^-3t, where f1(t) = e^-3t and f2(t) = te^-3t. Normalizing f1(t) and f2(t) gives f1(t) = e^-3t/e0 = 1 when t = 0 and f2(t) = t * e^-3t/e0 = 0 when t = 0.