# Homogenous ODE problem

1. Oct 20, 2012

### Baartzy89

Hi all,

I'm struggling with this question - I don't really know where to start. So far I have tried putting arbitrary values for 'a' into a quadratic auxiliary equation but using wolfram to calculate the roots gives me complex conjugates that I cant remember a thing about. Question as follows:

A body of mass 'm' kg attached to a spring moves with friction. The motion is described by the second Newton's law:

m(t).y" + a(t).y' + ky = 0

Where y is the body displacement in m, t is the time in s, a > 0 is the friction coefficient in kg/s and k is the spring constant in kg/s^2. Assuming m=1kg and k=4kg/s^2 find;

A) what is the range of values of a for which the body moves (i) with oscillations, (ii) w

2. Oct 20, 2012

### stephenkeiths

first you need to know m(t). I'm assuming its constant, some value m that doesn't change. Your a(t) what is that? I'm guessing its the "drag coefficient" this is also usually a constant. I've seen it represented as R. Then divide through by m to get y'' by itself. Then I always let 2b=R/m (you need to carry the 2 around, imo this is the cleanest way to do it). Then I let k/m=w^2 Now you have y''+2by'+w^2y=0 to find the solution you guess that y=e^rt. Then differentiate this and plug it into your ODE to find values of r that work. Then you have the general solution y=c*e^r1t+k*e^r2t. When I did this problem I found r1=-b+sqrt(b^2+w^2) and r2=b-sqrt(b^2+w^2).

You need initial conditions to continue from there

3. Oct 20, 2012

### Baartzy89

Yes, m is a constant and is the mass of the pendulum which is 1kg. 'a' is the friction coefficient which is >0. So I take it I need to find a general solution for the equation with variable coefficient 'a' prove that this general solution slows to a rest state as t approaches infinity and then plug the initial conditions in part c of the question and solve.

Is there a decent you tube video out there or tutorial which describes your method?

4. Oct 20, 2012

### Baartzy89

Damn, I just realized half of my original question was missing. Doing this on an iPhone is not easy

5. Oct 20, 2012

### stephenkeiths

I'm sure there is. This is the standard method for non-homogeneus second order ODEs with constant coefficients. You 'guess' what the solution is, and find out what you have to do to make it work. In this case you guess e^rt because the derivative of exponentials are exponentails and you can have them add up to zero if you choose the right values of r. Remember, when you take the derivative, an r comes down. Also, recall that e^rt is never zero. So divide by e^rt; you will be left with a quadratic equation in r. Solve this. It will give you two values of r that work. The most general is that answer that I gave above: y=c*e^r1t+k*e^r2t where r1 and r2 are the solutions to the quadratic equation you found. Now take the limit as t->infinity and show this is zero. What exactly are the other questions.

6. Oct 20, 2012

### Baartzy89

If you take the section I originally posted as the intro, and then the questions as follows;

A) what is the range of 'a' for which the body moves (i) with oscillations, (ii) without oscillations?

B) Find the general solution for any a<4 (Your solution should be a formula depending on the parameter a.) Prove that it follows from the solution obtained that the body slows down to a virtually rest state at large time (ie as t goes to infinity)?

C) Find the particular solution for a=4 subject to the initial conditions y(0)=0, dy/dt=1 m/s at t=0. Plot this solution and determine the largest displacement of the mass using calculus.

7. Oct 21, 2012

### Baartzy89

I've managed to work out part C, but the first two are still baffling me - mostly how to create a general solution for variable 'a'. Can anyone talk me through how to do it? The only Linear Homogenous equations I remember solving had constant coefficients.

Below is my answer for part C, please let me know if I've made a mistake somewhere.

C) when a=4
Auxiliary equation is;
p^2+4p+4=0

Which both roots equal -2, therefore general solution y(t)=(At+B)e^kt where k=-2

Initial conditions y(0)=0, y'(0)=1

Therefore B=1, -2B+A=1 hence A=3

Therefore final solution;
y(t)=(3t+1)e^-2t

Last edited: Oct 21, 2012