Final answer:
The solution to the differential equation is y(t) = (-2 + 10t)e^{-2t}, and y(t) reaches a maximum at t = 0.5 seconds after verifying that the second derivative is negative at this point.
Step-by-step explanation:
To solve the second-order linear homogeneous differential equation y''+4y'+4y=0 with initial conditions y(0)=-2 and y'(0)=8, we first find the general solution of the equation. Notice that the characteristic equation for this differential equation is r^2 + 4r + 4 = 0, which factors to (r+2)^2 = 0. This gives us a repeated root of r = -2.
Therefore, the general solution to the differential equation is y(t) = (C1 + C2t)e^{-2t}, where C1 and C2 are constants that we will determine using the initial conditions. By plugging in y(0) = -2 and y'(0) = 8, we find that C1 = -2 and C2 = 10.
The particular solution with our initial conditions is then y(t) = (-2 + 10t)e^{-2t}. To find when y(t) reaches a maximum, we need to set the first derivative equal to zero and solve for t. The first derivative of our solution is y'(t) = 10e^{-2t} - (20t)e^{-2t}. Setting y'(t) = 0 and solving for t gives us t = 0.5.
At t = 0.5, we must verify that y''(0.5) < 0 to confirm that we indeed have a maximum. After calculating, we find that the second derivative is negative, and thus y(t) reaches its maximum at t = 0.5 seconds.