# Laplace Transforms to solve Initial-Value problem

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• Dec 9th 2010, 10:29 AM
jegues
Quote:

Originally Posted by TheEmptySet
I think you need to have

$\displaystyle \frac{A}{(s+2)}+\frac{B}{(s+2)^2}$

So the correct PF expansion would be,

$Y=\dfrac{A}{s+2}+\dfrac{B}{(s+2)^{2}} + \dfrac{C(s+1)+D}{(s+1)^{2}+1}$

?
• Dec 9th 2010, 04:58 PM
TheEmptySet
Quote:

Originally Posted by jegues
So the correct PF expansion would be,

$Y=\dfrac{A}{s+2}+\dfrac{B}{(s+2)^{2}} + \dfrac{C(s+1)+D}{(s+1)^{2}+1}$

?

It should be

$\displaystyle \dfrac{A}{s+2}+\dfrac{B}{(s+2)^{2}} + \dfrac{Cs+D}{(s+1)^{2}+1}$
• Dec 9th 2010, 08:14 PM
Ackbeet
I, for one, am not seeing why there has to be a

$\dfrac{B}{(s+2)^{2}}$

term when there isn't one in the LT. Is there one in the LT? If so, why didn't it cancel? Here's what I have from the original DE:

$y'' + 4y' + 4y = e^{-t}(\sin(t) + \cos(t)), y(0) = 0, y'(0) = 0.$

LT gives

$s^{2}Y+4sY+4Y=\dfrac{1}{(s+1)^{2}+1}+\dfrac{s+1}{( s+1)^{2}+1}=\dfrac{s+2}{(s+1)^{2}+1}.$ This implies

$Y(s+2)^{2}=\dfrac{s+2}{(s+1)^{2}+1},$ or

$Y(s+2)=\dfrac{1}{(s+1)^{2}+1},$ and hence

$Y=\dfrac{1}{(s+2)((s+1)^{2}+1)}.$

There's no $1/(s+2)^{2}$ in there.

Did I do something wrong?
• Dec 9th 2010, 09:02 PM
TheEmptySet
Quote:

Originally Posted by Ackbeet
I, for one, am not seeing why there has to be a

$\dfrac{B}{(s+2)^{2}}$

term when there isn't one in the LT. Is there one in the LT? If so, why didn't it cancel? Here's what I have from the original DE:

$y'' + 4y' + 4y = e^{-t}(\sin(t) + \cos(t)), y(0) = 0, y'(0) = 0.$

LT gives

$s^{2}Y+4sY+4Y=\dfrac{1}{(s+1)^{2}+1}+\dfrac{s+1}{( s+1)^{2}+1}=\dfrac{s+2}{(s+1)^{2}+1}.$ This implies

$Y(s+2)^{2}=\dfrac{s+2}{(s+1)^{2}+1},$ or

$Y(s+2)=\dfrac{1}{(s+1)^{2}+1},$ and hence

$Y=\dfrac{1}{(s+2)((s+1)^{2}+1)}.$

There's no $1/(s+2)^{2}$ in there.

Did I do something wrong?

No, I did something wrong, I have apparently forgotten how to add :/
• Dec 10th 2010, 02:40 AM
Ackbeet
Well, you know you're a mathematician if you can do vector calculus but not long division. (Wink)
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