 # Find ccL^(-1)(F(s)), if given F(s)=(2)/(s(s^2+4)). Kaleigh Ayers 2022-09-13 Answered
Find ${\mathcal{L}}^{-1}\left(F\left(s\right)\right)$, if given $F\left(s\right)=\frac{2}{s\left({s}^{2}+4\right)}$
I have tried as below.
To find inverse of Laplace transform, I want to make partial fraction as below.
$\begin{array}{r}\frac{2}{s\left({s}^{2}+4\right)}=\frac{A}{s}+\frac{Bs+C}{{s}^{2}+4}=\frac{\left(A+B\right){s}^{2}+Cs+4A}{s\left({s}^{2}+4\right)}.\end{array}$
After that, we have system of linear equation
$\begin{array}{rl}A+B& =0\\ C& =0\\ 4A& =2.\end{array}$
Thus we have A=2, B=−2, and C=0. Now, substituting A,B,C and we have
$\begin{array}{r}\frac{2}{s\left({s}^{2}+4\right)}=\frac{2}{s}+\frac{-2s}{{s}^{2}+4}.\end{array}$
But the fact is
$\begin{array}{r}\frac{2}{s\left({s}^{2}+4\right)}\ne \frac{2}{s}+\frac{-2s}{{s}^{2}+4}=\frac{8}{{s}^{2}+4}.\end{array}$
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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You made a mistake because $A\ne 2$
$4A=2$ then $A=\frac{1}{2}$

We have step-by-step solutions for your answer! Damon Cowan
Note that
$\frac{2}{s\left({s}^{2}+4\right)}=\frac{1}{2s}-\frac{s}{2\left({s}^{2}+4\right)}$
So, you have
${\mathcal{L}}^{-1}\left\{\frac{2}{s\left({s}^{2}+2\right)}\right\}=\frac{1}{2}{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}{\mathcal{L}}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}$
Finally, you can use
${\mathcal{L}}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}=\mathrm{cos}\left(2t\right)$
and
${\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=1$

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