
#1
Feb1213, 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 fistorder 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 nonrecursive, 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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