Solving driven differential equations

In summary: Yes, this would take into account the sumation and then you would use the inverse laplace transform to solve the differential equation.Yes, this would take into account the sumation and then you would use the inverse laplace transform to solve the differential equation.
  • #1
danj303
15
0

Homework Statement



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?



Homework Equations


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  • #2
danj303 said:

Homework Statement



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?



Homework Equations


on attachment

What have you tried?
 
  • #3
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 I am not sure.
 
  • #4
danj303 said:
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 I am not sure.

Have you learned the method of using Laplace transforms to solve ODEs? If so, try that...
 
  • #5
Ok that's 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
danj303 said:
Ok that's 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

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
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 I am not sure about the right hand side with the sum from 0 to K of the dirac delta function.
 
  • #8
danj303 said:
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 I am not sure about the right hand side with the sum from 0 to K of the dirac delta function.

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:

[tex]\mathcal{L}\left[g_1(t)+g_2(t)\right]=\mathcal{L}\left[g_1(t)\right]+\mathcal{L}\left[g_2(t)\right][/tex]
 
  • #9
Im not sure about that part. Can you elaborate.
 
  • #10
danj303 said:
Im not sure about that part. Can you elaborate.

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
Im not sure how to take the laplace transform of δ(t-kd/v)
 
  • #12
danj303 said:
Im not sure how to take the laplace transform of δ(t-kd/v)

It's probably in your textbook, so you might try opening that up and reading part of it :wink:...or, you can use the definition of Laplace transform and the definition of delta function and calculate it directly...
 
  • #13
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
No, it doesn't take into account the summation. The Laplace transform of the righthand side is

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

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
I find

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

The only thing I am not sure about is fitting the summation back in. How do I do this??
 
  • #16
danj303 said:
I find

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

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

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

Not only is the Laplace transform linear, so is its inverse. That means if [tex]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][/tex]
 
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FAQ: Solving driven differential equations

What are driven differential equations?

Driven differential equations are mathematical equations that describe the behavior of a system in which the dynamics are influenced by an external force or input. These equations are used to model a wide range of phenomena in fields such as physics, engineering, and economics.

How do you solve driven differential equations?

The most common method for solving driven differential equations is by using numerical methods, such as Euler's method or the Runge-Kutta method. These methods involve approximating the solution at discrete points in time and using iterative calculations to find the values of the solution at each point.

Can driven differential equations have multiple solutions?

Yes, driven differential equations can have multiple solutions. This is known as the general solution, which includes all possible solutions to the equation. However, in most cases, a specific solution is sought, which is a particular solution that satisfies specific initial conditions or constraints.

What is the significance of driven differential equations in scientific research?

Driven differential equations are essential tools in scientific research as they allow for the mathematical modeling and prediction of complex systems. They are used in a wide range of fields, from physics and engineering to biology and economics, to understand and analyze the behavior of systems under external influences.

Are there any real-world applications of driven differential equations?

Yes, driven differential equations have numerous real-world applications. For example, they are used in the design of electrical circuits, the modeling of population growth, and the prediction of weather patterns. They are also crucial in understanding and controlling complex systems such as traffic flow and chemical reactions.

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