Final answer:
To solve the given boundary-value problem, we assume a solution of the form y = e^(rx) and find the characteristic equation to determine the values of r. Applying the boundary conditions, we find the particular solution y = e^(4x) - xe^(4x).
Step-by-step explanation:
To solve the given boundary-value problem y'' - 8y' + 16y = 0 with the boundary conditions y(0) = 1 and y(1) = 0, we can assume a solution of the form y = e^(rx). By substituting this into the differential equation, we find the characteristic equation r^2 - 8r + 16 = 0, which simplifies to (r - 4)^2 = 0. So, we have a repeated root of r = 4.
The general solution is then y = C1e^(4x) + C2xe^(4x), where C1 and C2 are constants. Applying the boundary conditions, we have y(0) = C1 = 1 and y(1) = C1e^(4) + C2e^4 = 0. Substituting C1 = 1, we can solve for C2 and find the particular solution y = e^(4x) - xe^(4x).