Find the inverse Laplace transforms of the following functions:

Note that as soon as we step inside the exam hall we write down the following rule that is on the tables but not nearly in as nice a form:

$\mathcal{L}^{-1}\left\{\frac{1}{s^n}\right\}=\frac{t^{n-1}}{(n-1)!}$.

(i)

$\frac{(s+1)^3}{s^4}$.

Solution

This looks quite tricky but all it needs is a little trick. Multiply out $(s+1)^3$ by hand (or by memory or by the Binomial Theorem (in tables)) to get:

$\frac{s^3+3s^2+3s+1}{s^4}=\frac{s^3}{s^4}+3\frac{s^2}{s^4}+3\frac{s}{s^4}+\frac{1}{s^4}=\frac{1}{s}+3\frac{1}{s^2}+3\frac{1}{s^3}+\frac{1}{s^4}$.

Now taking advantage of the linearity of the inverse Laplace transform (I’m just going to write $K$ instead of $\mathcal{L}^{-1}$. To get a nice inverse Laplace transform symbol I have to type \mathcal{L}^{-1} — along with some dollar signs and I’m sick of it! — no keyboard shortcuts on this page — the notes have shortcuts.)

$K\left\{\frac{1}{s}\right\}+3 K\left\{\frac{1}{s^2}\right\}+3K\left\{\frac{1}{s^3}\right\}+K\left\{\frac{1}{s^4}\right\}$.

Now using the tables;

$=1+3t+3\frac{t^2}{2}+\frac{t^3}{6}$.

(ii)

$\frac{1}{4s+1}$

Solution

Now this looks like

$\frac{1}{s-a}$,

which we know how to deal with. It is our job to get it looking a bit more like this. We can do this by taking out a $4$ out of the bottom line:

$\frac{1}{4(s+1/4)}=\frac{1}{4}\cdot\frac{1}{s-(-1/4)}$.

Now using linearity, and the tables, this has inverse transform:

$\frac{1}{4}e^{-t/4}$.