Final answer:
To solve the given initial value problem y'' + 2y' - 8y = 0 with the initial conditions y(0) = 3 and y'(0) = -12, we need to find the general solution to the differential equation and then determine the values of the constants using the initial conditions. The general solution is y(t) = C1e^(-4t) + C2e^(2t), where C1 and C2 are constants. Substituting the initial conditions, we find C1 = 3 and C2 = 0, giving us the particular solution y(t) = 3e^(-4t).
Step-by-step explanation:
To solve the given initial value problem y'' + 2y' - 8y = 0 with the initial conditions y(0) = 3 and y'(0) = -12, we can assume the solution is of the form y = e^(rt), where r is a constant.
Substituting this into the equation, we get r^2e^(rt) + 2re^(rt) - 8e^(rt) = 0. Factoring out e^(rt), we get (r^2 + 2r - 8)e^(rt) = 0.
The equation r^2 + 2r - 8 = 0 can be factored as (r + 4)(r - 2) = 0. So we have two possible solutions for r: r = -4 and r = 2.
Therefore, the general solution to the differential equation is y(t) = C1e^(-4t) + C2e^(2t), where C1 and C2 are constants. To find the values of C1 and C2, we can use the initial conditions given: y(0) = 3 and y'(0) = -12.
Substituting t = 0 and y = 3 into the equation, we get 3 = C1e^(0) + C2e^(0), which simplifies to 3 = C1 + C2.
Substituting t = 0 and y' = -12 into the equation, we get -12 = -4C1e^(0) + 2C2e^(0), which simplifies to -12 = -4C1 + 2C2.
We now have a system of two equations with two unknowns:
C1 + C2 = 3
-4C1 + 2C2 = -12
Solving this system of equations, we find C1 = 3 and C2 = 0.
Therefore, the particular solution to the initial value problem is y(t) = 3e^(-4t).