Final answer:
To solve the initial value problem y'' + 8y' + 16y = 0, y(0) = 2, y'(0) = 1, we can use the characteristic equation method. The solution to the initial value problem is y(x) = (2 - 7x)e^(-4x).
Step-by-step explanation:
To solve the initial value problem y'' + 8y' + 16y = 0, y(0) = 2, y'(0) = 1, we can use the characteristic equation of a linear homogeneous second-order differential equation. The characteristic equation for the given differential equation is r^2 + 8r + 16 = 0. Solving this quadratic equation gives us two repeated roots, -4. Therefore, the solution to the initial value problem is y(x) = (A + Bx)e^(-4x), where A and B are constants determined by the initial conditions.
To find the values of A and B, we can substitute the initial conditions into the solution. Plugging in x = 0, y(0) = 2 gives us 2 = (A + 0)e^(0), which simplifies to A = 2. Plugging in x = 0, y'(0) = 1 gives us 1 = (0 + B)e^(0) - 4(2e^(0)), which simplifies to B + 8 = 1. Therefore, B = -7.
Substituting the values of A = 2 and B = -7 back into the solution, we get the final solution to the initial value problem as y(x) = (2 - 7x)e^(-4x).