# Thread: Solving an I.V.P. with Laplace transforms.

1. ## Re: Solving an I.V.P. with Laplace transforms.

Originally Posted by romsek
I'm just going to show you how it's done.

$\frac{1}{(s+3)^2}\Longleftrightarrow \left(t e^{-3 t} u(t)\right)$

$s F(s)\Longleftrightarrow \left(\frac{d}{dt}f(t)\right)$

$\frac{d}{dt}t e^{-3 t} u(t)=\delta (t) t e^{-3 t}+\left(e^{-3 t}-3 t e^{-3 t}\right) u(t) = \left(e^{-3 t}-3 t e^{-3 t}\right) u(t)$

I'll let you mess with getting the constants right.
I'm still not really following you. I don't have a rule lying around that says;
$s F(s)\Longleftrightarrow \left(\frac{d}{dt}f(t)\right)$

The closest I have is;

${\Lapl}{\mathcal{L}}{(f^\prime)}(s) = s{\Lapl}{\mathcal{L}}(f)(s) - f(0$)

And I know that $f(0)=-1$

Do you know a general name for that property so i can dig up a proof? Or is it on one of the lists and I'm just not seeing it?

2. ## Re: Solving an I.V.P. with Laplace transforms.

I apologize I should have spelled out the full form of the property. You just get used to saying f'(t) <--> s F(s)

It's #4 on your 2nd page. The one you identified. You have f(0) so use it.

3. ## Re: Solving an I.V.P. with Laplace transforms.

Originally Posted by romsek
I apologize I should have spelled out the full form of the property. You just get used to saying f'(t) <--> s F(s)

It's #4 on your 2nd page. The one you identified. You have f(0) so use it.
Oh OK good.

Don't I need to dig up a -1 someplace? Do I need to completely rearrange the problem?

4. ## Re: Solving an I.V.P. with Laplace transforms.

Originally Posted by bkbowser
Oh OK good.

Don't I need to dig up a -1 someplace? Do I need to completely rearrange the problem?
no. Just compute f(t) the way I outlined below and then at the end subtract off f(0) = -1

Page 2 of 2 First 12