Final answer:
To solve the given initial value problem, we can find the general solution to the differential equation y" - 4y' + 4y = 0, which is y = c1e^(2x) + c2xe^(2x). Using the initial conditions, we can find the specific solution as y = (-2 + 7/2 - 4x)e^(2x).
Step-by-step explanation:
To solve the given initial value problem: y" - 4y' + 4y = 0, we can solve the characteristic equation r^2 - 4r + 4 = 0, which factors as (r - 2)^2 = 0. This gives us a repeated root of r = 2. Therefore, the general solution to the differential equation is y = c1e^(2x) + c2xe^(2x), where c1 and c2 are constants.
Now, we can use the initial conditions y(0) = -2 and y'(0) = 7/2 to find the specific solution. Plugging in x = 0 and y = -2 into the general solution, we get -2 = c1e^(2*0) + c2*0*e^(2*0), which simplifies to -2 = c1. Similarly, plugging in x = 0 and y' = 7/2, we get 7/2 = 2c1e^(2*0) + c2(1 + 2*0)e^(2*0), which simplifies to 7/2 = 2c1 + c2.
Therefore, the solution to the initial value problem is y = -2e^(2x) + (7/2 - 4)e^(2x)x = (-2 + 7/2 - 4x)e^(2x).