# 2nd order linear non-homogeneous ODE - having trouble

1. Nov 12, 2007

### InSaNiUm

1. The problem statement, all variables and given/known data:
This problem is in regard to a suspension system (mass, spring, dashpot) subjected to a 2 cm bump in the road. Given the mass and spring coefficient, we are to find:

a) The minimum damping coefficient, c, to avoid oscillation.
b) The expression for amplitude of vibration of the mass after the vehicle runs over the bump.
c) The amplitude of vibration 1 ms and 1 sec after running over the bump.

m = 270kg
k = 70,000N/m
d = 2cm (height of bump)

2. Relevant equations
$$my'' + cy' +ky = f(t)$$
This is a non-homogeneous equation, so:
$$y(t) = y_h(t) + y_p(t)$$

3. The attempt at a solution
For part a I used the homogeneous solution:
$$y'' + \frac{c}{m}y' + \frac{k}{m}y = 0$$
to find the characteristic:
$$r^2 + \frac{c}{m}r + \frac{k}{m} = 0$$
Inside it's quadratic, I set $$(\frac{c}{m})^2 - 4\frac{k}{m} = 0$$ for the critically damped case and got $$c_{min} = 8,695$$N-m/s.

Since this gives me two identical roots the soln becomes $$y_h(t) = c_1e^{-16t} + c_2te^{-16t}$$

I'm pretty sure that's right, but the part I'm stumped on is how to solve the particular solution.

I tried solving it with the forcing function being an impulsive function $$f(t) = d(\frac{1}{\epsilon})$$ where d is the .02m bump. I chose t=0 for the impulse.

$$y_p'' + \frac{c}{m}y_p' + \frac{k}{m}y_p = \frac{2}{m\epsilon}$$for $$0<t<\epsilon$$
I chose polynomials for the solution:
$$y_p = A_0 + A_1t$$
so:
$$y_p' = A_1$$ and $$y_p'' = 0$$
When I plug this back in I get:
$$\frac{c}{m}A_1 + \frac{k}{m}(A_0 + A_1t) = \frac{2}{m\epsilon}$$

Now, how do I separate the coefficients? Have I gone astray somewhere? I don't know where to go from here.
Can you explain WHY you chose y= A0+A1t as the particular solution. As long as t is less than $\epsilon$, your right hand side is a constant. Why not just y= A?