9. SECOND-ORDER LINEAR EQUATIONS
77
Then, by superposition, anything of the form
(##) x(t) = c1 cos ωt + c2 sin ωt
is also a solution of x00 + ω 2x = 0.
But do these solutions satisfy the IC of (∗ ∗ ∗)?
For (##) to be a solution of (∗ ∗ ∗), we need, noting
x0(t) = −c1ω sin ωt + c2ω cos ωt,
x(t0) = c1 cos ωt0 + c2 sin ωt0 = x0 and
x0(t0) = −c1ω sin ωt0 + c2ω cos ωt0 = x1
would have to have a solution for c1 and c2. In fact,
x1
c1 = x0 cos ωt0 − sin ωt0 and
ω
x1
c2 = x0 sin ωt0 + cos ωt0.
ω
Thus
x(t) = c1 cos ωt + c2 sin ωt
with c1 and c2 as above is a solution of (∗ ∗ ∗). By Lemma 9.2, it is the only
solution. We have proven
Theorem (9.4 — Simple Harmonic Oscillator and Solutions).
For any real numbers x1, x2, ω, where ω 6= 0, there exists a unique solution
of
(∗ ∗ ∗) x00 + ω 2x = 0, x(t0) = x0, x0(t0) = x1.
This solution can be expressed in the form
x(t) = c1 cos ωt + c2 sin ωt,
where the constants c1 and c2 are uniquely determined by x0 and x1.