2nd order linear non-homogeneous ODE - having trouble

[InSaNiUm]

1. Homework Statement :
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)

Homework Equations

[tex]my'' + cy' +ky = f(t)[/tex]
This is a non-homogeneous equation, so:
[tex]y(t) = y_h(t) + y_p(t)[/tex]

The Attempt at a Solution

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

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

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 [tex]f(t) = d(\frac{1}{\epsilon})[/tex] where d is the .02m bump. I chose t=0 for the impulse.

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

Now, how do I separate the coefficients? Have I gone astray somewhere? I don't know where to go from here.
Thanks in advance!

 

[HallsofIvy]

Can you explain WHY you chose y= A₀+A₁t as the particular solution. As long as t is less than [itex]\epsilon[/itex], your right hand side is a constant. Why not just y= A?