Thoughts on a non-linear second order problem

by Someone2841
Tags: nonlinear, order
Someone2841 is offline
Feb12-13, 05:59 PM
P: 29
[itex]h''(t)=-\frac{1}{h(t)^2}, h(0) = h_0, h'(0)=v_0[/itex]

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

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

I haven't seen this explained anywhere, but I'm fairly certain that for f = f(x), [itex]\int \frac{1}{f'}df = x + c[/itex] (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):

[itex]t(h) = \int \sqrt{\frac{h}{c_1 h+1}} dh[/itex]

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 [itex]\int \frac{1}{f'}df = x + C[/itex]:

[itex]\frac{dx}{dy} = \frac{1}{\frac{dx}{dy}}[/itex]
[itex]\int \frac{1}{\frac{dx}{dy}} dy = \int 1 dx = x + C[/itex]

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