General

Okay, I think I've got it.

Thanks a lot for the help, I didn't know there were so many smart people on this site.

[quote=SunsetDews]Okay, I tried out you guy's suggestions (sorry yeah in the bold part before the actual question it said x(0) = x'(0) = 0) and I got this far:

-4sF(s) - sF'(s) - s^2F'(s) = 0

Am I supposed to take the inverse laplace now or something?[/quote]

F'(s) (-s^2 - s) = 4sF(s)

It's a separable equation, just solve like you normally would. When you integrate instead of adding an arbitrary constant +C add +lnC and it'll make your life easier. Or at least that's what I always liked to do.

Okay, I tried out you guy's suggestions (sorry yeah in the bold part before the actual question it said x(0) = x'(0) = 0) and I got this far:

-4sF(s) - sF'(s) - s^2F'(s) = 0

Am I supposed to take the inverse laplace now or something?

I remember doing that, but I can't remember how I did that
I hated math lol

I just realized I made a pretty bad typo in my original post, when I said:

"In other words take the derivative of the Laplace transform of f(t), in your case f(t) is as simple as t itself.
So tx'' would become -d/ds[s^2 x(s) - sx(0) - x'(0)] and you'd do the same for tx'."

I meant to say to take the derivative of f(t) where f(t) is x'' in this case. Sorry about that, that was really misleading. The second line I said is correct though. I guess I'm used to writing out the second line but not used to really explaining it so I mixed it up or something

Lol I can't wait for Junior and Senior year. I'll love Calculus stuff.

tx'' + (t - 2)x' + x = 0

L{tx'' + (t - 2)x' + x} = (-d/ds)*L{x"} + (-d/ds)*L{x'} - 2L{x'} + L{x}.. Using a table you can transform this to:
= -d/ds [s^2 F(s) - s x(0) - x'(0) ] - d/ds [s F(s) - x(0) ] - 2s F(s) + F(s) = 0

This is a differential equation, even though it wasn't stated in your question, I believe they DID give you an initial value to work with, such as x(0) = 0.
If that's the case, then simply use that value and substitute in:
-d/ds [s^2 F(s) - x'(0) ] - d/ds [s F(s)] - 2s F(s) + F(s) = 0
-[2s F(s) + s^2 F'(s)] - [F(s) + s F'(s)] - 2s F(s) + F(s) = 0
- 2s F(s) - s^2 F'(s) - F(s) - s F'(s) - 2s F(s) + F(s) = 0 combining liked terms we would get:
-(s^2+s)F'(s) - 2sF(s) = 0

(s+1)F'(s) + 4 F(s) = 0..then solve it like an ordinary DE.
F'(s) + [4/(s+1)] F(s) = 0
F(s) = e^Integral{-4/(s+1) ds} = e^Ln[s+1]^-4 = (s+1)^-4 = 1/(s+1)^4

[b]x(t) = L{F(s)} = (1/6)*(e^-t)* t^3[/b]

Laplace transform table: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

[quote=SunsetDews]I saw what you wrote in my notes (the {-tf(t)} = F'(s) thing) but I'm not really sure if this applies? Is the negative in front of d/ds a typo?[/quote]

Assuming the derivatives are also a function of time, why wouldn't it apply? I think you're getting confused with notation.

Rewrite your problem as this:

tx''(t) + (t - 2)x'(t) + x = 0

is t equal to t? Check.
is x''(t) a function of t, in other words, can it be written as f(t)? Yes.

Consider x(t) = t^2
x'(t) = 2t is the derivative but it's still a function of t and can just be renamed as g(t) or f(t) or whatever.

The negative isn't a typo, it would be a typo if there was a -t in front of your x'', but there's a positive t. The theorem is L{-tf(t)} = F'(s) so you make up for the lack of a negative by just factoring out a -1.

If the derivatives in your question are functions other than time, then I have no clue and your question's way over my head

[quote=SunsetDews]I saw what you wrote in my notes (the {-tf(t)} = F'(s) thing) but I'm not really sure if this applies? Is the negative in front of d/ds a typo?[/quote]
No I believe tf(t) transformed = -F'(s)

[quote=ulti25]You need to take the Laplace transform of a product of a function t and a derivative

Use the fact that L{-tf(t)} = F'(s)

In other words take the derivative of the Laplace transform of f(t), in your case f(t) is as simple as t itself.
So tx'' would become -d/ds[s^2 x(s) - sx(0) - x'(0)] and you'd do the same for tx'.

I'm assuming you know what to do for the constant coefficients (just apply Laplace transform to a derivative)

iirc you should get a separable differential equation involving X(s) and X'(s) which you need to solve by integrating.[/quote]

I saw what you wrote in my notes (the {-tf(t)} = F'(s) thing) but I'm not really sure if this applies? Is the negative in front of d/ds a typo?

I took DE last year, but I forgot a ton of it sadly.

My friend was TA in the course this semester, I'm sure he could answer the question well...

Glad I could help?

[quote=SunsetDews]I tried that but I'm not sure exactly what I'd do after...?

I'd have to take the inverse laplace of t wouldn't I?[/quote]

If the question only wants a Laplace transform of it, just treat t as a constant for this question, I think.

[quote=HolyDragon]tx'' + (t - 2)x' + x

Laplace transform individually
First term
t(s^2X - sx(0) -x'(0))

That's my guess[/quote]

I tried that but I'm not sure exactly what I'd do after...?

I'd have to take the inverse laplace of t wouldn't I?

You need to take the Laplace transform of a product of a function t and a derivative

Use the fact that L{-tf(t)} = F'(s)

In other words take the derivative of the Laplace transform of f(t), in your case f(t) is as simple as t itself.
So tx'' would become -d/ds[s^2 x(s) - sx(0) - x'(0)] and you'd do the same for tx'.

I'm assuming you know what to do for the constant coefficients (just apply Laplace transform to a derivative)

iirc you should get a separable differential equation involving X(s) and X'(s) which you need to solve by integrating.

tx'' + (t - 2)x' + x

Laplace transform individually
First term
t(s^2X - sx(0) -x'(0))

That's my guess

lol that's my usual plan but I'm not going to be able to study this weekend...
Need to try to finish this by today since the class is on monday

[quote=QuantumFlux]Dude its Friday night. Do this 1 hour before you go to bed on Sunday.[/quote]
How about trying to be helpful?
[url=http://tutorial.math.lamar.edu/Classes/DE/LaplaceIntro.aspx]This might help[/url]

looks like a second order differential equation