Final answer:
To solve the given differential equation y'''+4y''-16y'-64y=0 with initial conditions, we can use the characteristic equation method. By finding the roots of the characteristic equation and applying the initial conditions, the general solution is obtained as y(t) = (-16/21)e^(-8t) + (8/21)e^(2t) + (8/21)e^(-4t).
Step-by-step explanation:
To solve the given differential equation, we can use the characteristic equation method. We first find the characteristic equation by substituting y = e^(mt) into the differential equation, which gives us the equation (m^3 + 4m^2 - 16m - 64)e^(mt) = 0. Since e^(mt) is never zero, we can simplify the equation to m^3 + 4m^2 - 16m - 64 = 0.
Using a numerical method or factoring, we find that the roots of the characteristic equation are m = -8, m = 2, and m = -4. Therefore, the general solution to the differential equation is y(t) = c1e^(-8t) + c2e^(2t) + c3e^(-4t), where c1, c2, and c3 are constants determined by the initial conditions.
Using the given initial conditions y(0) = 0, y'(0) = 26, and y''(0) = -16, we can solve for the constants. Substituting t = 0 into the general solution and its derivatives, we get the equations c1 + c2 + c3 = 0, -8c1 + 2c2 - 4c3 = 26, and 64c1 + 4c2 + 16c3 = -16. Solving these equations, we find c1 = -16/21, c2 = 8/21, and c3 = 8/21.
Therefore, the solution to the differential equation is y(t) = (-16/21)e^(-8t) + (8/21)e^(2t) + (8/21)e^(-4t).