# Find inverse Laplace Transform of Reciprocal Quadratic Function I(s)=(6)/(Ls^2+Rs+1/C). Find what i(t) is by doing the inverse Laplace transform.

Find inverse Laplace Transform of Reciprocal Quadratic Function
$I\left(s\right)=\frac{6}{L{s}^{2}+Rs+\frac{1}{C}}$
Find what i(t) is by doing the inverse Laplace transform.
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Jimena Torres
$I\left(s\right)=\frac{6}{L{s}^{2}+Rs+\frac{1}{C}}$
Having fixed R, C and L (for example $R=6$, $C=1/5$, $L=1$), we have:
$I\left(s\right)=\frac{6}{{s}^{2}+6s+5}$
First of all, find the zeros of denominator:
${s}^{2}+4s+1=0⇒s=-5\vee s=-1$
Then ${s}^{2}+4s+1=\left(s+5\right)\left(s+1\right)$
We can write
$I\left(s\right)=\frac{a}{s+5}+\frac{b}{s+1}=\frac{a\left(s+1\right)+b\left(s+5\right)}{\left(s+1\right)\left(s+5\right)}=\frac{s\left(a+b\right)+\left(a+5b\right)}{{s}^{2}+6s+5}=\frac{6}{{s}^{2}+6s+5}$
Then:
$\left\{\begin{array}{lcl}a+b& =& 0\\ a+5b& =& 6\end{array}⇒\left\{\begin{array}{lcl}a& =& -b\\ -b+5b& =& 6\end{array}⇒\left\{\begin{array}{lcl}a& =& -\frac{3}{2}\\ b& =& \frac{3}{2}\end{array}$
Then:
$I\left(s\right)=-\frac{\frac{3}{2}}{s+5}+\frac{\frac{3}{2}}{s+1}$
The antitransformation yield to:
$i\left(t\right)=-\frac{3}{2}{e}^{-5t}+\frac{3}{2}{e}^{-t}$