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

I've tried to decompose the fraction using

$$\frac{As+B}{s^2+9}+\frac{Cs+D}{(s^2+9)^2}$$

$$1=(As+B)(s^2+9)+Cs+D$$

yet D=1, still giving me the same exact equation

$$\frac{1}{(s^2+9)^2}$$

any help?

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Rewrite it as follows

$$\frac{1}{(s^2+9)^2}=-\frac{1}{2s}\frac{-2s}{(s^2+9)^2}=-\frac{1}{2s}\frac{\mathrm d}{\mathrm ds}\left( \frac{1}{s^2+9}\right)$$ Now use the following properties:

$$tf(t)\stackrel{\mathcal{L}}\longleftrightarrow-\frac{\mathrm d}{\mathrm ds}F(s)$$ $$\int_{0}^{t}g(\tau)d\tau\stackrel{\mathcal{L}}\longleftrightarrow\frac{1}{s}G(s)$$

as well as

$$\frac{1}{3}\sin(3t)\stackrel{\mathcal{L}}\longleftrightarrow\frac{1}{s^2+9}$$

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You can pull this straight from the table

$$\mathcal{L}^{-1} \left(\frac{1}{(s^2 + k^2)^2}\right) = \frac{1}{2k^3}(\sin(kt) - kt\cos(kt)).$$

The answer is

$$\frac{1}{54}(\sin{3t} - 3t\cos{3t}).$$

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  • 1
    $\begingroup$ Is there any way you can make the following fraction to look like one of the forms in Laplace Transform Table? $$\frac{1}{(s^2+16)^2} $$ $\endgroup$ – socrates May 17 '17 at 0:17
  • $\begingroup$ @socrates Why not use $k=\pm 4$? $\endgroup$ – KYHSGeekCode Sep 21 '18 at 11:23

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