# Laplace of the function

1. May 19, 2004

### angel

hi how do u find the laplace of the function below:

x'(t) = kx(t) -hH(44-t)

k is the rate which is >0, x(t) is the actual population, h is the harvesting rate

2. May 19, 2004

### TALewis

Do you mean

$$x'(t)=kx(t)-hH(44-t)$$

where k and h are constants? If so, what is H(t)? Is it the Heaviside unit step function?

Edited for LaTeX errors.

3. May 19, 2004

### angel

hi,

yes H is the heaviside unit step.

4. May 19, 2004

### arildno

I suggest you use Laplace transform

5. May 19, 2004

### angel

thats the method i am using, but im not sure how to apply it to this equation.

6. May 19, 2004

### TALewis

Rewriting:

x' - kx + hH(44 - t) = 0

You probably already know how to find the Laplace transform of x' - kx . I'm betting that the problem is with the step function. Do you know how to transform a step function when it is written in the standard form of H(t - a) ? I think it's usually covered in an introductory chapter to Laplace transforms, along with the Dirac delta function.

If hH(44 - t) is giving you the trouble, try rewriting it like this: hH(44 - t) = h - hH(t - 44). You should see that they both have the same shape: h when t < 44 ; 0 when t > 44 . With this modification, you have the Heaviside function in the right form to perform a Laplace transform on it.

If you haven't got any idea what to do with any step function at all, let me know and I will try to help a little more. Also, you should include the initial condition x(0) = ? for a Laplace transform problem.

7. May 19, 2004

### angel

ok, i know how to find the laplace of h(t-a) the answer is:

e^-as/s

ok, i understand how u got the values but what im really confused on is to apply the laplace on everything, even the first bit.

i know what the answer should be, its:

x(t) = (x0 - h/k)*e^kt + h/k * H(44-t) + h/k * (e^k(t-44))H(t-44)

but i havent got a clue how to get to the answer.

I can either prove it going forward to the laplace or going from the laplace to the actual x'(t).

Last edited: May 19, 2004
8. May 19, 2004

### angel

initila condition is not provided, but you can assume it to be 16.

9. May 19, 2004

### TALewis

May I ask if this question is from a certain course, or are you doing this on your own?

The differential equation again:

$$x' - kx + h - hH(t-44) = 0$$

You can perform a Laplace transform on each of the terms of the equation separately and then add them together:

$$\mathcal{L}\{x' - kx + h - hH(t-44)\} = \mathcal{L}\{x'\} - \mathcal{L}\{kx\} + \mathcal{L}\{h\} - \mathcal{L}\{hH(t-44)\}$$
$$\mathcal{L}\{x'\} = sX - x(0)$$
$$\mathcal{L}\{kx\} = kX$$
$$\mathcal{L}\{h\} = h/s$$
$$\mathcal{L}\{hH(t-44)\} = he^{-44s}/s$$

Therefore:

$$\mathcal{L}\{\mathrm{d.e.}\} = sX - x_0 - kX + h/s -he^{-44s}/s = 0$$

Does that help?

Edited, because LaTeX is hard and whatnot.

Last edited: May 19, 2004
10. May 19, 2004

### angel

thanks alot

Last edited: May 19, 2004
11. May 19, 2004