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

1 Answer
dani83
Aug 13, 2015

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 .

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