# Homework Help: Solving driven differential equations

1. Sep 20, 2010

### danj303

1. The problem statement, all variables and given/known data

The suspension system of a car is designed so that it is a damped system described by

z'' + 2z' + z = f(t)

where z is the vertical displacement of the car from its rest position. The car is driven over a (smooth!) road which has "catseye" embedded in the road surface every d metres. Assume that each time the car passes over a catseye, the suspension system receives a unit impulse. Thus, if the car is driven with velocity v m/s over K + 1 catseyes then

f(t) = (see attachment)

(a) Find z(t) when z(0) = z'(0) = 0.

(b) Plot the solutions for K = 10 and K = 50 with v = 10 What is the eff ect of a larger K? Fix K = 50. Let v = 10 and plot the solutions for d = 5 and d = 10 on the same axes. What is the e ffect of the distance between the cateyes? Let d = 10 and plot the solutions for v = 10 and v = 20. What is the e ffect of speed?

2. Relevant equations
on attachment

#### Attached Files:

• ###### equation.jpg
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2. Sep 20, 2010

### gabbagabbahey

What have you tried?

3. Sep 20, 2010

### danj303

Ive got no idea where to start. Thats the problem. I would assume f(t) needs to be expressed in terms of heaviside step functions and then the differential equation solved but im not sure.

4. Sep 20, 2010

### gabbagabbahey

Have you learned the method of using Laplace transforms to solve ODEs? If so, try that...

5. Sep 20, 2010

### danj303

Ok thats all well and good but can you explain a little bit further. We have covered laplace transforms but only recently and I am trying to work thorugh some problems to understand how it works. Thanks

6. Sep 20, 2010

### gabbagabbahey

The first step is always to take rthe Laplace transform of both sides of the ODE....what do you get when you do that?

7. Sep 20, 2010

### danj303

The left side is easy if you set it to equal zero. The laplace transform is

s^2*Y(s) + 2s*Y(s) + Y(s) = 0

but im not sure about the right hand side with the sum from 0 to K of the dirac delta function.

8. Sep 20, 2010

### gabbagabbahey

Do you know how to take the Laplace transform of a single delta function? If so, just use the fact that the Laplace transform is linear:

$$\mathcal{L}\left[g_1(t)+g_2(t)\right]=\mathcal{L}\left[g_1(t)\right]+\mathcal{L}\left[g_2(t)\right]$$

9. Sep 20, 2010

### danj303

Im not sure about that part. Can you elaborate.

10. Sep 20, 2010

### gabbagabbahey

You're not sure how to take the LT odf a delta function? Or you aren't sure how to add a bunch together?

11. Sep 20, 2010

### danj303

Im not sure how to take the laplace transform of δ(t-kd/v)

12. Sep 20, 2010

### gabbagabbahey

It's probably in your textbook, so you might try opening that up and reading part of it ....or, you can use the definition of Laplace transform and the definition of delta function and calculate it directly....

13. Sep 21, 2010

### danj303

Alright after some time I get the laplace transform of the Right hand side as e-(kd/v*s)

and then sloving the entire equation for Y(s) I get

Y(s) = e-(kd/v*s)/s2+2s+1

But does this take into account the sumation and then how do I take the inverse laplace transform to solve the DE

14. Sep 21, 2010

### vela

Staff Emeritus
No, it doesn't take into account the summation. The Laplace transform of the righthand side is

$$L[f(t)] = L\left[\sum_{k=0}^K \delta(t-kd/v)\right] = \sum_{k=0}^K L[\delta(t-kd/v)]$$

where the last step uses the linearity of the Laplace transform. What you calculated is the inside of the last summation.

The exponential factor merely indicates a delay. To find the inverse Laplace transform, you temporarily neglect it, find the inverse transform of the rest, and then time-shift the answer. For example, suppose X(s)=e2s/(s+1). To find x(t), you first find the inverse Laplace transform of 1/(s+1), which is e-t. Now you time-shift the answer by 2, so x(t)=e-(t-2)u(t-2).

15. Sep 21, 2010

### danj303

I find

z(t) = t e-(t-kd/v) Heaviside(t-kd/v)

The only thing im not sure about is fitting the summation back in. How do I do this??

16. Sep 22, 2010

### gabbagabbahey

Close, but shouldn't your $t$ out front really be $\left(t-\frac{kd}{v}\right)$?

Not only is the Laplace transform linear, so is its inverse. That means if $$Z(s)=\sum_{k=0}^K g_k(s)[/itex] , then [tex]z(t)=\mathcal{L}^{-1}\left[\sum_{k=0}^K g_k(s)\right]=\sum_{k=0}^K \mathcal{L}^{-1}\left[g_k(s)\right]$$

Last edited: Sep 22, 2010