Register to reply

Second Order DE: Nonlinear Homogeneous

by irighti
Tags: homogeneous, nonlinear, order
Share this thread:
Oct5-07, 12:15 PM
P: 1
I am sure most of you are familiar with the equation: m(x)''+c(x)'+k(x) = 0. Then, we create an auxillary equation that looks like this: mr^2+cr+k = 0. And, then we find the roots of this auxillary equation, calling them r1 and r2. And, if the roots are r1,r2>0 we consider the system to be overdamped and we develop the following general form of the equation to be: x=A*e^(r1*t)+B*e^(r2*t) etc...

I have an inverted pendulum with damping and understand the equation to be in the form of a second order nonlinear (homogenous) differential equation.

The second order nonlinear homegenous differential equation is: m(x)''+c(x)'-k*sin(x)=0.
I have tried everything I know and haven't had much luck... I mean how do you find the roots to something like this? I am sure there is a method to this madness. I would think there is some type of variation of parameters or substitution involved but do not know how to apply substitution etc... to this application.

I would appreciate any help or advice that might lead me in the right direction...

Thanks in advance, John
Phys.Org News Partner Science news on
Sapphire talk enlivens guesswork over iPhone 6
Geneticists offer clues to better rice, tomato crops
UConn makes 3-D copies of antique instrument parts
Oct5-07, 12:23 PM
Sci Advisor
HW Helper
PF Gold
P: 12,016
There are no generally effective methods to gain the exact solution(s) of non-linear equations, differential or algebraic!

The "method to this madness" in order to get approximate solutions is to make a guess, giving you a wrong answer, then make a second guess, giving you (hopefully!) a less wrong answer, and so on ad infinitum.

Such systematized guess-working techniques are very much in use, and to improve those techniques is one of the main branches of applied mathematics areas like computational fluid dynamics and suchlike fields.
Oct6-07, 12:37 PM
Sci Advisor
PF Gold
P: 39,339
Here, for example, since sin(x)= x- (1/6)x3+ ..., a first approximation would be to replace sin(x) by x: mx"+ cx'- kx= 0.

A slightly more sophisticated method is "quadrature": Let u= x' so that x"= u'. By the chain rule, u'= (du/dx)(dx/dt)= u u' so the equation becomes u u'+ cu- kx= 0. That is now a first order equation for u as a function of x. The problem typically is that even after you have found u, integrating x'= u, to find x as a function of t, in closed form may be impossible (without the damping, this gives elliptic integrals).

Register to reply

Related Discussions
Homogeneous second order linear ODE Calculus & Beyond Homework 4
Higher Order homogeneous ODE Calculus & Beyond Homework 0
Reduction of order (2nd order linear ODE homogeneous ODE) Calculus & Beyond Homework 10
Homogeneous nonlinear function Calculus & Beyond Homework 36
Higher Order Homogeneous ODE (IVP) Calculus & Beyond Homework 3