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I had a test yesterday and couldn't figure out the answer to this question. Was wondering if someone could walk me through the solution. $$\dfrac{(13s+3)}{(s^2+2s+5)}$$

I figured I had to complete the square in the denominator then split the fraction up into $$\dfrac{13s}{(s+1)^2+4} + \dfrac3{(s+1)^2+4}$$

That would allow me to solve the second fraction as $3/2 \exp(-t) \sin(2t)$.

I couldn't figure how to solve for the first fraction unless I shouldn't have split them up at all.

So what is the proper way to solve it?

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The first fraction will transform into a $\cos ()$ term. –  Tpofofn Jun 9 '12 at 18:58
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hint: $13s+3=13(s+1)-10$ –  Robert Mastragostino Jun 9 '12 at 19:33
    
you can use the model solution and just plug in the constants –  Valentin Jun 9 '12 at 19:51
    
Hmmm...13e^(-t) x cos(2t) - 5e^(-t) x sin(2t) ???? –  Chef Flambe Jun 9 '12 at 22:53
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Since you noticed that you needed to complete the square in the denominator, let that dictate the form of the new, preferred numerator as follows: $$ {13s+3\over s^2+2s+5}={13s+3\over (s+1)^2+2^2}={A\color{blue}{(s+1)}+B\cdot \color{red}{2}\over \color{blue}{(s+1)}^2+\color{red}{2}^2}, $$ where equating coefficients between the first and last versions of the numerator yields $A=13$, $B=-5$.

This last form is quite helpful since a standard table of Laplace transforms shows \begin{align} \mathscr{L}\left\{e^{-at}\cos(bt)\right\}={s+a\over (s+a)^2+b^2},\\ \mathscr{L}\left\{e^{-at}\sin(bt)\right\}={b\over (s+a)^2+b^2}, \end{align} so \begin{align} \mathscr{L}^{-1}\left\{{13s+3\over s^2+2s+5}\right\}&=\mathscr{L}^{-1}\left\{{A(s+1)+B\cdot 2\over (s+1)^2+2^2}\right\}\\ &=A\,\mathscr{L}^{-1}\left\{{s+1\over (s+1)^2+2^2}\right\}+B\,\mathscr{L}^{-1}\left\{{2\over (s+1)^2+2^2}\right\}\\ &=Ae^{-t}\cos(2t)+Be^{-t}\sin(2t),\\ &=13e^{-t}\cos(2t)-5e^{-t}\sin(2t). \end{align}

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