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Damnit! I am terrible at Partial Fractions!

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve y"+4y'=sin 3t subject to y(0)=y'(0)=0 using Laplace Transform




    3. The attempt at a solution

    So I got:

    [tex]s^2Y(s)-sy(0)-y'(0)+4[sY(s)-y(0)]=\frac{3}{s^2+9}[/tex]

    [tex]\Rightarrow Y(s)=\frac{3}{(s^2+9)(s^2+4)}[/tex]

    Now it looks like two irreducible quadratics, which I know should not be too bad, but I have never dealt with more than one.

    Now am I correct to say that

    [tex]\frac{3}{(s^2+9)(s^2+4)}=\frac{Ax+B}{s^2+9}+\frac{Cx+D}{s^2+4}[/tex]

    This is where I think I have the problem... the notation.

    Thanks!
     
  2. jcsd
  3. Mar 31, 2008 #2

    dynamicsolo

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    Homework Helper

    Given this transform:

    [tex]s^2Y(s)-sy(0)-y'(0)+4[sY(s)-y(0)]=\frac{3}{s^2+9}[/tex] ,

    shouldn't this be

    [tex]\Rightarrow Y(s)=\frac{3}{(s^2+9)(s^2+4*s*)}[/tex]


    Also, when you go to solve the partial fractions, you want to have 's' in the numerators:

    [tex]\frac{3}{(s^2+9)(s^2+4s)}=\frac{As+B}{s^2+9}+...[/tex][/QUOTE]
     
  4. Mar 31, 2008 #3
    Oh crap...

    Yes, so I get:

    [tex]\frac{As+B}{s^2+9}+\frac{C}{s}+\frac{D}{s+4}[/tex]

    Thanks!!!! !
     
  5. Mar 31, 2008 #4

    Anyone know of a quick way to do this? I let s=0 and -4 to solve for C and D..... but what about A and B? Do I have to distribute this whole mess out? Or is there a more expedient way?
     
  6. Mar 31, 2008 #5
    Guess not.

    So now I have

    [tex]Y(s)=\frac{-4/75s-3/25}{s^2+9}+\frac{1}{12s}+\frac{3}{100(s+4)}[/tex]

    How do I simplify the 1st term?

    I can see that it looks like cosine. But...How do I get rid of all the crap?

    Hmm I guess I could....Oh!!! Break it up! I think that will work!
     
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