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Math Help - Where did my 2 go doing inverse Laplace?

  1. #1
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    Where did my 2 go doing inverse Laplace?

    Find inverse Laplace of \frac{3s+2}{s^2+2s+10}. Book's answer (using partial fractions) 3^{-t}\cos 3t - \frac 13 e^{-t} \sin 3t. My answer 3^{-t}\cos 3t - \frac 23 e^{-t} \sin 3t (I have an extra 2 in the top of the second term). Where did my 2 go?

    My work. Since my first term agrees with the book's answer, I'll just do inverse Laplace of
    \frac{2}{s^2+2s+10} = \frac{2}{(s+1)^2+9}.

    Now, to understand this I do it a little differently than is explained in the book (and various web pages).
    If F(s+1) = \frac{2}{(s+1)^2+9}, then F(s) = \frac{2}{(s)^2+9}.

    I want the inverse Laplace of F(s+1), but first I find the inverse L of F(s).

    The Laplace of \sin 3t = \frac 3{s^2+9}. Multiply both sides by \frac 23 and I get that the inverse Laplace of \frac 23 \sin 3t = \frac 2{s^2+9} = F(s). Thus the inverseL of \frac 2{s^2+9} is \frac 23 \sin 3t.

    But I want to find the inverse Laplace of F(s+1) which gives me e^{-t}\frac 23 \sin 3t, and I still have an extra 2 in the numerator. What'd I do wrong?
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  2. #2
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    Re: Where did my 2 go doing inverse Laplace?

    $3^{-t}$ is that a mistake? Should it instead be $3e^{-t}$?

    Here's what I get;

    $\mathcal{L}^{-1} \{ \frac{3s+2}{(s+1)^2 +3^2} \}$

    add $+1$, $-1$ to the numerator

    $\mathcal{L}^{-1} \{ \frac{3s+2+1-1}{(s+1)^2 +3^2} \}$

    Group and carry out the division

    $\mathcal{L}^{-1} \{ \frac{3s+3}{(s+1)^2 +3^2} \} - \mathcal{L}^{-1} \{ \frac{1}{(s+1)^2 +3^2} \}$

    $ 3 \mathcal{L}^{-1} \{ \frac{s+1}{(s+1)^2 +3^2} \} - \frac{1}{3} \mathcal{L}^{-1} \{ \frac{3}{(s+1)^2 +3^2} \}$

    $ 3 e^{-t}\mathcal{L}^{-1} \{ \frac{s}{(s)^2 +3^2} \} - \frac{1}{3}e^{-t} \mathcal{L}^{-1} \{ \frac{3}{(s)^2 +3^2} \}$

    Since $\mathcal{L} \{ cos(\omega t) \}= \frac{s}{(s)^ +\omega^2}$ and $\mathcal{L} \{ sin(\omega t) \}= \frac{\omega}{(s)^2 +\omega^2} $ we have,

    $3e^{-t} cos(3t) -\frac{1}{3}e^{-t}sin(3t)$
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    Re: Where did my 2 go doing inverse Laplace?

    That is a typo (but it won't let me edit the post to fix). You did it the same way as the book. So I still don't see why I have an extra 2 in my answer.
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    Re: Where did my 2 go doing inverse Laplace?

    Quote Originally Posted by mathDad View Post
    That is a typo (but it won't let me edit the post to fix). You did it the same way as the book. So I still don't see why I have an extra 2 in my answer.
    I think the ability to edit times out after a couple of hours.

    It looks like the problem is with the term you said matched exactly with the book. So the problem could be with how you carried out the division or in inversion.

    Perhaps something false like $\mathcal{L}^{-1} \{ \frac{s}{(s+1)^2 + 3^2} \}= e^{-t} \mathcal{L}^{-1} \{ \frac{s}{(s)^2 + 3^2} \} $ occurred?
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    Re: Where did my 2 go doing inverse Laplace?

    OK. I found it. All the work in my OP is correct (except the missing e).

    But I split the original question like this \frac{3s+2}{s^2+2s+10} = \frac{3s}{s^2+2s+10} + \frac{2}{s^2+2s+10}, which is allowed, but not useful. Then, somehow, I got the inverseL of first term to come out to the first term of the correct answer. That is, I got \mathcal{L}^{-1} \{\frac{3s}{s^2+2s+10}\} = 3e^{-t}\cos 3t, which is not quite right. I don't have that scratch work anymore (my classmate took it to study), so I don't recall what I did. So, if I can't get the second term to resolve to the second term of the correct answer, I am not off by one term, I am way off base. Anyway, I understand how to do it now.

    BTW, I also liked the style you notated the shift. It's easier to see what's happening there.
    Last edited by mathDad; May 12th 2014 at 07:20 PM.
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    Re: Where did my 2 go doing inverse Laplace?

    Not useful? It's exactly what you should be doing.

    Now complete the square on the denominator. The first term needs s on the top. The second term needs 3. What can you do to ensure this?
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    Re: Where did my 2 go doing inverse Laplace?

    Completing the square of the first term (of the right hand side) leaves me with (s+1)^2+9 on the bottom. Since that term has only an s on top, you can't do the inverse laplace. The way bowser did it is right.
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    Re: Where did my 2 go doing inverse Laplace?

    You can also look up the answers on WolframAlpha. For inverse Laplace Transforms, it doesn't show you how to get the answer, but you can verify if you are correct (or if there is a typo in the book) rather quickly.

    inverse laplace transform - Wolfram|Alpha
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    Re: Where did my 2 go doing inverse Laplace?

    Quote Originally Posted by mathDad View Post
    Completing the square of the first term (of the right hand side) leaves me with (s+1)^2+9 on the bottom. Since that term has only an s on top, you can't do the inverse laplace. The way bowser did it is right.
    Yes, and this is why you're getting the 2 when you shouldn't.

    You will need to write $\displaystyle \begin{align*} \frac{s}{(s+1)^2 + 9} = \frac{s + 1 - 1}{(s + 1)^2 + 9} = \frac{s+1}{(s+1)^2 + 9} - \frac{1}{(s+1)^2 + 9} \end{align*}$

    So this extra 1 needs to be subtracted from the other term.
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