# Thoughts on a non-linear second order problem

1. Feb 12, 2013

### Someone2841

$h''(t)=-\frac{1}{h(t)^2}, h(0) = h_0, h'(0)=v_0$

The first step is to, I think, reduce this to a fist-order problem:

$h'(t)h''(t)=-h'(t)\frac{1}{h(t)^2}$ --- Multiply both sides by h'(t)
$h'(t)^2=\frac{1}{h(t)}+c_1$ --- Integrate both sides
$1/h'(t) = \sqrt{\frac{h(t)}{c_1 h(t)+1}}$ --- Rearrange

I haven't seen this explained anywhere, but I'm fairly certain that for f = f(x), $\int \frac{1}{f'}df = x + c$ (could someone outline a proof/disproof for this? Please no ad verecundiam.) Therefore, integrating both sides with respect to h(t) gets a messy explicit formula for t(h):

$t(h) = \int \sqrt{\frac{h}{c_1 h+1}} dh$

I am looking for a non-recursive, explicit formula for h(t). Any ideas? Is there a better approach than this one?

Thanks in advance!

P.S., here's my thought process for $\int \frac{1}{f'}df = x + C$:

$\frac{dx}{dy} = \frac{1}{\frac{dx}{dy}}$
$\int \frac{1}{\frac{dx}{dy}} dy = \int 1 dx = x + C$

Even if this isn't valid, is it true?

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