Laplace transform with heavyside step function

In summary, Fiona changed the first t in her homework equation to t-1 in order to be consistent with the process our lecturer used. She then calculated the transform of t and continued from there.
  • #1
fiona.young
4
0

Homework Statement



I need to solve the ODE y''-y = t - tH(t-1); y(0)=y'(0)=0

Homework Equations



-

The Attempt at a Solution



I'm fine with the process of solving the ODE, but I need a little help regarding the first t. From what I understand from lectures, all of the 't' terms need to be in the form of (t-a). So I've done:

y''-y = (t-1) - (t-1)H(t-1)+H(t-1)
L{y''-y} = L{t-1} - L{(t-1)H(t-1)} + L{H{t-1)}

I just wanted to check that it's correct to change the first 't' into (t-1) and then treat the RHS as:

L{y''-y} = L{t} - L{1} - L{(t-1)H(t-1)} + L{H{t-1)}

Thanks :)
Fiona
 
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  • #2
Why change the first t to t-1? And you need to check your arithmetic because if you expand (t-1) - (t-1)H(t-1)+H(t-1) back out I don't think you get the t - tH(t-1) you started with.
 
  • #3
Actually, that's what I want to know; whether or not I need to change the first t to t-1... Are you suggesting that I don't need to change this first t?

For the second half, my full working was:

tH(t-1) = [(t-1)+1]H(t-1) --> expand...
= (t-1)H(t-1) + H(t-1)

This is the same process used by our lecturer.
 
  • #4
LCKurtz said:
Why change the first t to t-1? And you need to check your arithmetic because if you expand (t-1) - (t-1)H(t-1)+H(t-1) back out I don't think you get the t - tH(t-1) you started with.

fiona.young said:
Actually, that's what I want to know; whether or not I need to change the first t to t-1... Are you suggesting that I don't need to change this first t?

Yes. You can calculate the transform of t can't you?

For the second half, my full working was:

tH(t-1) = [(t-1)+1]H(t-1) --> expand...
= (t-1)H(t-1) + H(t-1)

This is the same process used by our lecturer.

That step is OK. I was just saying t - tH(t-1) isn't equal to (t-1) - (t-1)H(t-1)+H(t-1), which it isn't.
 
  • #5
Oh, I understand! If I change the first t to t-1, it changes the equation completely. So I just leave it as is and calculate its Laplace transform as 1/s^2, and continue from there, correct?
 
  • #6
fiona.young said:
Oh, I understand! If I change the first t to t-1, it changes the equation completely. So I just leave it as is and calculate its Laplace transform as 1/s^2, and continue from there, correct?

Yes. Just be careful with your signs.
 
  • #7
Thanks for helping me see that silly mistake - very helpful :)
 

1. What is a Laplace transform with a Heaviside step function?

A Laplace transform with a Heaviside step function is a mathematical tool used to convert a function from the time domain to the frequency domain. The Heaviside step function, also known as the unit step function, is a function that is equal to 1 for positive values and 0 for negative values.

2. What is the purpose of using a Laplace transform with a Heaviside step function?

The purpose of using a Laplace transform with a Heaviside step function is to simplify the mathematical analysis of a system. It allows for the representation of complex functions as a simple algebraic expression, making it easier to solve differential equations and other mathematical problems.

3. How is a Laplace transform with a Heaviside step function calculated?

A Laplace transform with a Heaviside step function is calculated by first applying the Laplace transform to the original function, and then multiplying the result by the Heaviside step function. This essentially "turns on" the function for positive values and "turns off" the function for negative values.

4. What are the properties of a Laplace transform with a Heaviside step function?

The properties of a Laplace transform with a Heaviside step function include linearity, time shifting, and differentiation. These properties make it easier to manipulate and analyze the transformed function.

5. In what applications is a Laplace transform with a Heaviside step function commonly used?

A Laplace transform with a Heaviside step function is commonly used in fields such as engineering, physics, and mathematics. It is particularly useful in solving differential equations, analyzing control systems, and studying electronic circuits.

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