Use the general solution of x'' + 2 x = 0 given in Problem 7 to

Use the general solution of x'' + ω²x =­ 0 given in Problem 7 to show that a solution satisfying the initial conditions x(t₀) ­ x₀, x'(t₀) ­= x₁ is the solution given in Problem 7 shifted by an amount t₀:

x(t) = x₀ cos ω(t - t₀) + x₁/ω sin ω(t - t₀).