# I'm having difficulties calculating a simple Laplace inverse

I'm having difficulties calculating a simple Laplace inverse :
$\frac{S-4}{{S}^{2}-2S-11}$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

enchantsyseq
When Laplace transforms are rational functions, the typical line of attack is to put it into the form of partial fractions. Note that ${S}^{2}-2S-11={\left(S-1\right)}^{2}-12$ has roots $1±\sqrt{12}=1±2\sqrt{3}$, so:
$\frac{S-4}{{S}^{2}-2S-11}=\frac{S-4}{\left(S-1-2\sqrt{3}\right)\left(S-1+2\sqrt{3}\right)}$
So, we try to write it in the form $\frac{A}{S-1-2\sqrt{3}}+\frac{B}{S-1+2\sqrt{3}}$ , with A,B constants. One way of solving for these constants is the cover-up method. We apply this here:
$A=\frac{S-4}{S-1+2\sqrt{3}}{\mid }_{S=1+2\sqrt{3}}=\frac{-3+2\sqrt{3}}{4\sqrt{3}}=\frac{2-\sqrt{3}}{4}$
$B=\frac{S-4}{S-1-2\sqrt{3}}{\mid }_{S=1-2\sqrt{3}}=\frac{-3-2\sqrt{3}}{-4\sqrt{3}}=\frac{2+\sqrt{3}}{4}$
Thus, our function's Laplace transform can be written as $\frac{1}{4}\left(\frac{2-\sqrt{3}}{S-1+2\sqrt{3}}+\frac{2+\sqrt{3}}{S-1-2\sqrt{3}}\right)$
We now recall that ${e}^{at}$ has Laplace transform $\frac{1}{s-a}$ , and this allows us to recover our function as:
$f\left(t\right)=\frac{1}{4}\left(\left(2-\sqrt{3}\right){e}^{\left(1-2\sqrt{3}\right)t}+\left(2+\sqrt{3}\right){e}^{\left(1+2\sqrt{3}\right)t}\right)$