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Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.

I have tried as below.

To find inverse of Laplace transform, I want to make partial fraction as below. \begin{align*} \dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}. \end{align*} After that, we have system of linear equation \begin{align*} A+B&=0\\ C&=0\\ 4A&=2. \end{align*} Thus we have $A=2$, $B=-2$, and $C=0$. Now, substituting $A, B, C$ and we have \begin{align*} \dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}. \end{align*} But the fact is \begin{align*} \dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}. \end{align*}

I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.

Anyone can give me hint to give me hint for this problem?

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Note that $$\frac{2}{s(s^{2}+4)}=\frac{1}{2s}-\frac{s}{2(s^{2}+4)}$$ So, you have $$\mathcal{L}^{-1}\left\{\frac{2}{s(s^{2}+2)} \right\}=\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s} \right\}-\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+4} \right\}$$ Finally, you can use $$\mathcal{L}^{-1}\left\{ \frac{s}{s^{2}+4}\right\}=\cos(2t)$$ and $$\mathcal{L}^{-1}\left\{ \frac{1}{s}\right\}=1$$

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You made a mistake because $A \neq 2$
$4A=2$ then $A=\frac{1}{2}$

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$$F(s)=\dfrac{2}{s(s^2+4)}$$ $$F(s)=\dfrac{1}s\dfrac 2{(s^2+4)}= \mathcal {L}(1) \mathcal{L} ( \sin (2t)$$ You can also use the theorem of convolution : $$f(t)=1* \sin (2t)=\int_0^t 1 \times \sin(2\tau) d\tau$$ $$f(t)= \dfrac {-\cos (2 \tau)}{2} \bigg |_0^t$$ $$f(t)= \dfrac 12 -\dfrac {\cos (2 t)}{2} $$

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