Verified answer

If x = acos[√(μ)t]

dx/dt = -a√(μ)sin[√(μ)t]

d^2x/dt^2 = -aμcos[√(μ)t], so,

d^2x /dt^2 + μx = aμcos[√(μ)t] - aμcos[√(μ)t] = 0