S.O.S. Mathematics CyberBoard

The equation is 2y"+y'+2y=g(t).

I know that the Laplace transform of the equation is 2s^2Y(s)-2sy(0)-2y'(0)+sY(s)-y(0)+2Y(s) = (e^(-5s)-e^(-20s))/s;
however, I am having trouble following this: can you break it up and tell me what part of the transform goes with the 2y" part, which goes with y', and which part goes with the 2y?

How would the answer change if we looked at a slightly different equation? For example what would be different if we had y"+2y'+2y=g(t). The only difference would be the coefficients right? But since I am having a hard time "breaking" the transform apart, I am not sure which coefficients would be different.

I should mention that I know the transform of the first equation because it is an example in my book; however, the book did not break it up into components...
Thanks.

Use Formula 11 on http://www.sosmath.com/tables/laplace/laplace.html repeatedly.

_________________
The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low, and achieving our mark. - Michelangelo Buonarroti

Well I have been looking at that equation. Let's see if I understand it: I guess first I would integrate f'(t) and then take the Laplace transform of the result. Then I would take that and take the dot product with "s" and then subtract f(0) from that right? And I suppose I could either do that to the entirety of 2y"+y'+2y or to each individual part right? I'm not sure, but this seems like quite a painful process, or is it simple because the components are relatively "simple"? I suppose there is no trick you use to do this is there - or do I just have to grind through the process?
Thanks.

What? No. derivatives turn into powers of s multiplying your original transform, and the initial conditions determine the rest of the bits.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

huh? I think I better study this some. Does this Laplace transformation have a "name" so I can look it up and study it? Thanks.

Yes, it is called "the Laplace Transform"

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

huh, ummm, uh... Guess I better really study more then. I guess I have never really understood the Laplace Transform too well. So far, I've really just been relying on the tables, but I guess it is time I really learn and understand it.

Ok, well I studied some more. I will try to solve the problem I gave earlier.
y"+2y'+2y=g(t)

We need some conditions to solve this I guess so lets say y(0)=0, an y'(0)=1.

Then how does this look:

L{y"}+2L{y'}+2L{y}=0

sL{y'}-y'(0)+2L{y'}+2L{y}=0

s(sL{y}-y(0))-y'(0)+2(L{y}-y(0))+2L{y}=0

L{y}-s-1+2(sL{y})+2L{y}=0 (*1)

L{y}(+2s+2)-s-1=0

L{y}= (*2)

How does this look? Unfortunately I am stuck again however as I do not see anything in the table that looks like this...

Please note equation (*1) is supposed to be (s^2)L{y}-s-1+2(sL{y})+2L{y}=0
an equation (*2) is supposed to be L{y}= (s+1)/(s^2+2s+2)

Do you know what went wrong with my Latex?

You don't want \{}{}, you want { numerator \over denominator}

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

... or \frac{numerator}{denominator}.

_________________

Or more properly, the unilateral Laplace transform.

_________________