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Solve the differential equation?
I don't know how to solve this. Could someone please help? It's not an assignment or anything, I just haven't taken diffeq or anything, and I'm not sure how to solve something like this. Be it Laplace Transforms or whatever, just please tell me how to solve this.
x"(t) + k/x² = 0
Where x'(0) = v0
And x(0) = x0
It should be:
x"(t) + k/[x(t)]² = 0
x is a function of t, not a constant.
1 Answer
- ecapS trebliHLv 61 decade agoFavorite Answer
Of course this DE describes one-dimensional motion of a particle in an inverse square
force, where
x is the distance of the particle from the center of the earth (for example).
k > 0 corresponds to an attractive force (gravity); k < 0 to a repulsive (?) force,
between charges of the same sign.
Standard technique: multiply by x'(t).
x"(t) x'(t) + k x'(t) /x(t)^2 = 0
(1/2) (x'(t))^2 -- k / x(t) = E = constant
E can be interpreted as total energy; the 2 terms on the left are kinetic
energy & potential energy,
The kinetic energy is (1/2) m v^2. In this case, mass m = 1; not important,
dx/dt = +/- sqrt (2 (E + k/x(t))
The sign depends on whether the particle is rising or falling.
This is a separable DE:
dt = dx/ sqrt(2(E + k/x))
sqrt(2) t = Int dx/sqrt(E + k/x)
The integral can easily be reduced to k = E = 1 or k = -1, E = 1.
The substitution x = 1/u is supposed to put the integral in a do-able form.
But these days one should use Matlab or Mathematica.
