# Thoughts on a non-linear second order problem

by Someone2841
Tags: nonlinear, order
 P: 29 $h''(t)=-\frac{1}{h(t)^2}, h(0) = h_0, h'(0)=v_0$ 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?