Final answer:
The initial value problem x'' + 12x' +36x = 0, with initial conditions x(0) = 0 and x'(0) = 2, is solved by finding the repeated root of the characteristic equation, leading to the solution x(t) = 2te^{-6t}.
Step-by-step explanation:
To solve the initial value problem x'' + 12x' +36x = 0, with conditions x(0) = 0 and x'(0) = 2, we approach it as a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation for this differential equation is r^2 + 12r + 36 = 0, which simplifies to (r + 6)^2 = 0. This gives us a repeated root of r = -6.
For repeated roots r_1 = r_2 = -6, the general solution has the form x(t) = (C_1 + C_2t)e^{-6t}. Applying the initial conditions, we first take the derivative x'(t) to find C_2 from x'(0) = 2, leading to C_2 = 2. Since x(0) = 0, we can solve for C_1 which turns out to be C_1 = 0, because x(0) = C_1 = 0.
The final solution is therefore x(t) = 2te^{-6t}. This satisfies both the differential equation and the initial conditions given.