y''+2y'-8y=0, y(0)=3, y'(0)=-12

Given differential equation is y'' + 2y' - 8y = 0 , with initial conditions y(0) = 3 and y'(0) = -12 .

Let's assume the solution of the differential equation is in the form of y = e^{rt} , where r is a constant.

Substitute y = e^{rt} into the differential equation:
\begin{aligned} y'' + 2y' - 8y &= 0 \ (r^2 + 2r - 8)e^{rt} &= 0 \end{aligned}

For the exponential term to be nonzero, we must have r^2 + 2r - 8 = 0 . Solve the quadratic equation to find r :
r^2 + 2r - 8 = 0 \Rightarrow (r + 4)(r - 2) = 0

So, r = -4 or r = 2 .

Therefore, the general solution is in the form y(t) = c_1e^{-4t} + c_2e^{2t} , where c_1 and c_2 are constants.

Apply the initial conditions y(0) = 3 and y'(0) = -12 :
\begin{cases} y(0) = 3 \Rightarrow c_1 + c_2 = 3 \ y'(0) = -12 \Rightarrow -4c_1 + 2c_2 = -12 \end{cases}

Solving the system of equations gives c_1 = -3 and c_2 = 6 .

Therefore, the particular solution is:
y(t) = -3e^{-4t} + 6e^{2t}

\textbf{Answer:} y(t) = -3e^{-4t} + 6e^{2t}