Question

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

Answer 1

General solution:
`x = C cos sqrt(3)t + D sin sqrt(3)t`
where `C` and `D` are constants.

Explanation:

`x''(t) + 3x = 0` is a linear homogeneous second order ordinary differential equation.

Suppose we try the solution:
`x = e^(pt)`
Then:
`x'' = p^2e^(pt)`
`(p^2 + 3)e^(pt) = 0`
`p = +-sqrt(3)i`

The linear combination of the individual solutions is also a solution. Hence the general solution is:
`x = Ae^(isqrt(3)t) + Be^(-isqrt(3)t)`
where `A` and `B` are constants.

Since `e^(itheta) = 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`.