Looking for a solution to this differential equation

[Someone2841]

I have been interested for a while now in the differential equation x''(t) = -1/x(t)² where x(0) = h and x'(0) = 0. I've only made limited progress on this, and recently have been looking into a more specific case where x(0) = -1.

When x(0) = h, the solution for x'(t) is x'(t) = sqrt(1/x(t) - 1/h), which can be easily verified via implicit differentiation.

When h = -1, x'(t) =


I would like to find the explicit form (in terms of elementary or special functions) of x(t). So far I have hit a dead end at:

t = integral{1/sqrt[(x(t)+1)/x(t)]}dx(t), which after integrating would give at best an implicit solution.

Any ideas/suggestions?

 

[Someone2841]

I've recently found a way to reach an implicit solution for problems such as this. First, $\displaystyle x''(t) = \frac{-1}{x(t)^2}$ can be rewritten in terms of the inverse function $\displaystyle t(x)$ like such: $\displaystyle -\frac{t''(x)}{t'(x)^3} = -\frac{1}{x(t)^2}$. Rewritting like this shows that this can be solved as a separable equation:

$\displaystyle -\frac{t''(x)}{t'(x)^2}\cdot \frac{dx}{dt} = -\frac{1}{x(t)^2}$

After integrating:

$\displaystyle \frac{1}{2t'(x)^2} = \frac{1}{x(t)} + c$

From here you can solve for $\displaystyle t'(x)$ and integrate both sides.

$\displaystyle t(x) = \pm \int{\sqrt{\frac{x(t)}{cx(t)+2}}\, dx(t)}$

Which is, of course, a pain to integrate and the solution doesn't lend itself (at least well) to an explicit solution of $\displaystyle x(t)$.


Any thoughts on this?

 

[Deleted member 4993]

I've recently found a way to reach an implicit solution for problems such as this. First, $\displaystyle x''(t) = \frac{-1}{x(t)^2}$ can be rewritten in terms of the inverse function $\displaystyle t(x)$ like such: $\displaystyle -\frac{t''(x)}{t'(x)^3} = -\frac{1}{x(t)^2}$. Rewritting like this shows that this can be solved as a separable equation:

$\displaystyle -\frac{t''(x)}{t'(x)^2}\cdot \frac{dx}{dt} = -\frac{1}{x(t)^2}$

After integrating:

$\displaystyle \frac{1}{2t'(x)^2} = \frac{1}{x(t)} + c$

From here you can solve for $\displaystyle t'(x)$ and integrate both sides.

$\displaystyle t(x) = \pm \int{\sqrt{\frac{x(t)}{cx(t)+2}}\, dx(t)}$

Which is, of course, a pain to integrate and the solution doesn't lend itself (at least well) to an explicit solution of $\displaystyle x(t)$.


Any thoughts on this?

$\displaystyle \sqrt{\dfrac{x}{cx+a}} \ = \ \dfrac{1}{\sqrt{c}} \ * \ \sqrt{1 \ - \ \dfrac{a}{cx+a}}$

Now substitute,

$\displaystyle \dfrac{a}{cx+a} \ = cos^2(\theta)$

and continue.....

Instead of all that you can also use Wolfram-alpha