Chapter 4.1, Problem 8E

Use the general solution of x″ + ω²x = 0 given in Problem 7 to show that a solution satisfying the initial conditions x(t₀) = x_0, x′(t₀) = x₁ is the solution given in Problem 7 shifted by an amount t₀:

$x(t)=x_0\cos \omega \left(t-t_0\right)+\frac{x_1}{\omega }\sin \omega \left(t-t_0\right).$

7. Given that x(t) = c₁ cos ωt + c₂ sin ωt is the general solution of x″ + ω²x = 0 on the interval (−∞, ∞), show that a solution satisfying the initial conditions x(0) = x₀, x′(0) = x₁ is given by

$x(t)=x_0\cos \omega t+\frac{x_1}{\omega }\sin \omega t.$