# Homework Help: Laplace transform with heavyside step function

1. Jun 7, 2012

### fiona.young

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

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

2. Relevant equations

-

3. 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

2. Jun 7, 2012

### LCKurtz

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. Jun 7, 2012

### fiona.young

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. Jun 7, 2012

### LCKurtz

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

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. Jun 7, 2012

### fiona.young

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. Jun 7, 2012

### LCKurtz

Yes. Just be careful with your signs.

7. Jun 7, 2012

### fiona.young

Thanks for helping me see that silly mistake - very helpful :)