Final answer:
The general solution to 4y'' + 4y' + y = 0 is y = Ae^((-1/2 + i/2)x) + Be^((-1/2 - i/2)x), where A and B are arbitrary constants. A particular solution to y'' - 2y' + y = 5e^(2x) is y = 5e^(2x).
Step-by-step explanation:
To find the general solution to the differential equation 4y'' + 4y' + y = 0, we can assume a solution of the form y = e^(rx). Plugging this into the equation gives us the characteristic equation 4r^2 + 4r + 1 = 0. Solving for r, we find two complex conjugate roots: r = -1/2 + i/2 and r = -1/2 - i/2. Therefore, the general solution is y = Ae^((-1/2 + i/2)x) + Be^((-1/2 - i/2)x), where A and B are arbitrary constants.
To find a particular solution to the equation y'' - 2y' + y = 5e^(2x), we can try a particular solution of the form y = Ae^(2x). Plugging this into the equation gives 4Ae^(2x) - 4Ae^(2x) + Ae^(2x) = 5e^(2x). Solving for A, we find A = 5. Therefore, a particular solution is y = 5e^(2x).