Inverse laplace transform of a function

[Drain Brain]

find the inverse Laplace of the ff:

1. $\frac{n\pi L}{L^2s^2+n^2 \pi^{2}}$

2. $\frac{18s-12}{9s^2-1}$
for the 2nd prob

I did partial fractions

$\frac{18s-12}{9s^2-1}=\frac{9}{3s+1}-\frac{3}{3s-1}$

$\mathscr{L}^{-1}\{\frac{18s-12}{9s^2-1}\} = \frac{9}{3}\left(\frac{1}{s+\frac{1}{3}}\right)-\frac{3}{3}\left(\frac{1}{s-\frac{1}{3}}\right)$

=$3e^{-\frac{1}{3}t}-e^{\frac{1}{3}t}$ please check

any hint for prob 1?

 

[Prove It]

find the inverse Laplace of the ff:

1. $\frac{n\pi L}{L^2s^2+n^2 \pi^{2}}$

2. $\frac{18s-12}{9s^2-1}$
for the 2nd prob

I did partial fractions

$\frac{18s-12}{9s^2-1}=\frac{9}{3s+1}-\frac{3}{3s-1}$

$\mathscr{L}^{-1}\{\frac{18s-12}{9s^2-1}\} = \frac{9}{3}\left(\frac{1}{s+\frac{1}{3}}\right)-\frac{3}{3}\left(\frac{1}{s-\frac{1}{3}}\right)$

=$3e^{-\frac{1}{3}t}-e^{\frac{1}{3}t}$ please check

any hint for prob 1?

Your second looks correct. As for the first, take out $\displaystyle \begin{align*} \frac{1}{L^2} \end{align*}$ as a factor and then the denominator will be a $\displaystyle \begin{align*} s^2 + a^2 \end{align*}$ form...

 

[Drain Brain]

please check my work

pulling out a factor of$\frac{1}{L^2}$

$\frac{1}{L^2}\left[\frac{n\pi L}{s^2+\left(\frac{n\pi L}{L}\right)^2} \right]$
$\mathscr{L}^{-1}\{\frac{n\pi L}{L^2s^2+n^2 \pi^{2}}\} =\frac{n\pi L}{L^2}\cdot \frac{L}{n\pi}\sin\left(\frac{n\pi}{L}\right)= \sin\left(\frac{n\pi}{L}\right)t$

did I get the correct answer?