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Find inverse laplace transform of $f(s) = \frac{1}{(s-2)^2+9}$

Here is what I have gotten from partial fractions. I observed that $s^2-4s + 13$ is irreducible (doesn't have real roots). $$\frac{1}{(s-2)^2 + 9} = \frac{1}{s^2-4s + 13} = \frac{As + B}{s^2-4s + 13}$$A=0, B=1

The corresponding value in my table is $e^{at}\sin(bt)$ for the corresponding laplace transform $\frac{b}{(s-a)^2 + b^2}$.

so plugging in A and B, I get $e^{0t}\sin(1\cdot t) = \sin(t)$. However my textbook answer is $\frac{1}{3}e^{2t}\sin{3t}$.

I am not sure how to trouble shoot this problem. I am pretty sure I did the partial fractions correctly, but the textbook answer implies that my approach to finding the inverse laplace transform is very wrong because I have no idea where the $\frac{1}{3}$ comes from in the solution.

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Remember that

$\color{blue}{\mathcal{L}^{-1}\{F(s-a)\} = e^{at} \mathcal{L}^{-1} \{F(s)\}}$ and

$\color{blue}{\mathcal{L}^{-1}\left\{\frac{b}{s^2+b^2}\right\} = \sin(bt)}$

So,

$\mathcal{L}^{-1}\left\{\frac{1}{(s-2)^2+9}\right\} = \frac13\mathcal{L}^{-1}\left\{\frac{3}{(s-2)^2+3^2}\right\} = \frac13e^{2t}\sin(3t)$

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  • $\begingroup$ I don't recall what you're saying to remember. Or at least I don't recognize it in its entirety $\endgroup$
    – Evan Kim
    Apr 10 '20 at 17:37
  • $\begingroup$ Do you remember this formula $\mathcal{L}^{-1}\{F(s-a)\} = e^{at} \mathcal{L}^{-1} \{F(s)\}$? $\endgroup$
    – 19aksh
    Apr 10 '20 at 17:38
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$$F(s)=\frac{1}{(s-2)^2 + 3^2} \implies 3F(s)=\frac{3}{(s-2)^2 + 3^2}$$ Substitute $s-2=s'$: $$3F(s'+2)=\frac{3}{(s')^2 + 3^2}$$ $$3\mathcal{L^{-1}}F(s'+2)=\mathcal{L^{-1}} \left (\frac{3}{(s')^2 + 3^2} \right )$$ $$3\mathcal{L^{-1}}F(s'+2)=\sin (3t)$$ Finally $\mathcal {L^{-1}}(F(s+a))=e^{-at}f(t):$ $$3e^{-2t}f(t)=\sin (3t) \implies f(t)= \dfrac {e^{2t}}3\sin (3t)$$

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You've used the values of $A,\,B$ where you should use the values of $a,\,b$. Use $\mathcal{L}^{-1}\frac{1}{(s-a)^2+b^2}=\frac1be^{at}\sin bt$ with $a=2,\,b=3$.

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