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PostPosted: Thu, 3 Nov 2011 02:16:45 UTC 
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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.


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PostPosted: Thu, 3 Nov 2011 03:53:24 UTC 
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Use Formula 11 on http://www.sosmath.com/tables/laplace/laplace.html repeatedly.

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PostPosted: Thu, 3 Nov 2011 17:45:21 UTC 
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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.


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PostPosted: Thu, 3 Nov 2011 19:00:35 UTC 
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goteamusa wrote:
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.

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PostPosted: Thu, 3 Nov 2011 19:10:39 UTC 
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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.


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PostPosted: Thu, 3 Nov 2011 19:14:25 UTC 
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goteamusa wrote:
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"

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PostPosted: Thu, 3 Nov 2011 19:58:10 UTC 
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huh, ummm, uh... :oops: 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.


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PostPosted: Thu, 3 Nov 2011 20:31:18 UTC 
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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

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

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

L{y}=\{s+1}/{s^{2}+2s+2}} (*2)

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


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PostPosted: Thu, 3 Nov 2011 20:37:22 UTC 
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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?


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PostPosted: Fri, 4 Nov 2011 05:58:33 UTC 
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goteamusa wrote:
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}

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PostPosted: Fri, 4 Nov 2011 06:12:19 UTC 
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Shadow wrote:
goteamusa wrote:
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}


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

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 4 Nov 2011 06:13:12 UTC 
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Shadow wrote:
goteamusa wrote:
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"


Or more properly, the unilateral Laplace transform.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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