Register to reply

Thoughts on a non-linear second order problem

by Someone2841
Tags: nonlinear, order
Share this thread:
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?
Phys.Org News Partner Science news on
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds

Register to reply

Related Discussions
2nd Order Linear ODE Problem Calculus & Beyond Homework 3
First-Order Linear Differential Problem Calculus & Beyond Homework 5
A problem about reduction of the order of a linear ODE Calculus & Beyond Homework 5
First order, non linear ode problem Calculus & Beyond Homework 15
Linear First order ODE problem Calculus & Beyond Homework 2