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So I was working on problem 16 in Elementary Differential Equations 9th edition by DiPrima and I get to the point where I'm using partial fraction's to separate :

$\displaystyle {1 \over s(s^2+\frac1 4s+1)}$

I did some research and found that one way to set it up was

$\displaystyle{\frac1 s - {s+\frac14 \over s^2+\frac14s+1}}$ and then to solve for the second fraction as such $\displaystyle {s+\frac14 \over (s+\frac18)^2-\frac{1}{64}+1}$

However, I'm not sure why this is. Could anyone help me out? Thanks.

To clear up my question, what I'm asking for is how and why they chose to solve it this way. I'm also unsure as to how they got from 2 to 3.

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    $\begingroup$ Are you asking (1) how to obtain the second expression, (2) how to obtain the third expression, or (3) why get the third expression? $\endgroup$ Commented Apr 8, 2013 at 20:58
  • $\begingroup$ Sorry about that, added to the question $\endgroup$ Commented Apr 8, 2013 at 21:00
  • $\begingroup$ From the second to third result, you are just completing the square of the denominator: $$s^2+\frac{1}{4}s+1=\left(s+\frac{1}{8}\right)^2-\frac{1}{64}+1$$ $\endgroup$ Commented Apr 8, 2013 at 21:03
  • $\begingroup$ O I can't believe I didn't see that! Awesome $\endgroup$ Commented Apr 8, 2013 at 21:03

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You want to setup:

$\displaystyle {1 \over s(s^2+\frac1 4s+1)} = \frac{A}{s} + \frac{Bs + C}{s^2+\frac1 4s+1}$

Now you need to solve for the constants ($A, B, C$). You can cross multiply and then equate like powers to solve this, for example (there are other ways too).

Your final result should be (you can show this in various forms of course):

$\displaystyle {1 \over s(s^2+\frac1 4s+1)} = \frac{-4 s-1}{(4 s^2+s+4)} + \frac{1}{s}$

If we are looking to do the Inverse Laplace Transform, we would need to massage that result to get into one of the forms that simplifies finding the inverse.

For example, we could write the result as:

$\displaystyle -\frac{4 s}{4 s^2+s+4}-\frac{1}{4 s^2+s+4} + \frac{1}{s}$

Then, we would massage each term a bit to get into ILT format.

They chose their form in order to start setting you up for using forms in order to do the ILT.

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  • $\begingroup$ Awesome, thanks so much! I'll accept as soon as it lets me $\endgroup$ Commented Apr 8, 2013 at 21:02
  • $\begingroup$ @BrandonLile: Welcome to MSE and you are very welcome! Regards $\endgroup$
    – Amzoti
    Commented Apr 8, 2013 at 21:08
  • $\begingroup$ $\checkmark^{+1}$ $\endgroup$
    – amWhy
    Commented Apr 9, 2013 at 0:12

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