# How do you solve x''(t)+x3=0?

Aug 13, 2015

General solution:
$x = C \cos \sqrt{3} t + D \sin \sqrt{3} t$
where $C$ and $D$ are constants.

#### Explanation:

$x ' ' \left(t\right) + 3 x = 0$ is a linear homogeneous second order ordinary differential equation.

Suppose we try the solution:
$x = {e}^{p t}$
Then:
$x ' ' = {p}^{2} {e}^{p t}$
$\left({p}^{2} + 3\right) {e}^{p t} = 0$
$p = \pm \sqrt{3} i$

The linear combination of the individual solutions is also a solution. Hence the general solution is:
$x = A {e}^{i \sqrt{3} t} + B {e}^{- i \sqrt{3} t}$
where $A$ and $B$ are constants.

Since ${e}^{i \theta} = \cos \theta + i \sin \theta$, we can re-arrange the above to
$x = C \cos \sqrt{3} t + D \sin \sqrt{3} t$
where $C = A + B$ and $D = A - B$.