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

Question

Elijah Abram · Accepted Answer

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

# 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 #.

Charles Anctil · Answer

undefined To solve the differential equation x''(t) + x^3 = 0, we can use a method called the shooting method or numerical methods because this is a nonlinear differential equation and does not have a general analytical solution.

In the shooting method, we convert the second-order differential equation into a system of first-order differential equations by introducing a new variable. This allows us to apply numerical techniques like Euler's method, the Runge-Kutta method, or other numerical methods to approximate the solution.

Alternatively, software packages like MATLAB or Python libraries such as SciPy can be used to solve this differential equation numerically.

However, if you're interested in an analytical approach, it might not be possible for this specific equation due to its nonlinear nature.