We got a joke! Heaviside (British electrician) was told during a restaurant that mathematicians are finally using his formula, he replied: I don't need the chef to tell me the food was good. (a lil bit cynical!)
# (Heaviside) Unit step function
#heaviside
The Heaviside step function is defined as:

$u(t-a)=\begin{cases}0,\quad t<a \\1,\quad a\leq t\end{cases}$
graph it, it just "switches on" when $t=a$
If we use laplace transforms we will see that solving equations with this Heaviside function is very natural. Using #mouc wont work, #voparam might work but it's messy.
Let's derive some fundamental LT properties involving the unit step function:

$$\mathcal{L}\{u(t-a)f(t-a)\}$$
$\int _{0} ^\infty e^{-st}u(t-a)f(t-a)\, dt$
$=\int _{0}^a 0 \, dt+\int _{a}^\infty e^{-st}\cdot 1f(t-a)\, dt$
let $x=t-a$
$=\int _{0} ^\infty e^{-s(x+a)}f(x)\, dx$
$=e^{-as}\int _{0} ^\infty e^{-sx}f(x)\, dx$
the right side is nothing but the LT of f. How nice!
$$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)\quad\leftarrow\text{Very useful formula!}$$
Same process can be done to show the following:
$$\mathcal{L}\{u(t-a)f(t)\}=e^{-as}\mathcal{L}\{f(t+a)\} \leftarrow\text{Also useful}$$
We can express a case-function in terms of unit step functions.
Consider this function where multiple functions are being switched:
$f(t)=\begin{cases} 1, \quad t<1 \\ t, \quad 1\leq t\leq \pi \\ \sin(t), \quad \pi\leq t\end{cases}$
We express it using heaviside functions:
$f(t)=1-u(t-1)+tu(t-1)-tu(t-\pi)+\sin(t)u(t-\pi)$
think about it switching on and off certain parts of the equation at certain times t.

#ex #heaviside 
current in electric circuit $I(t)$ is defined by:
$$I''+I=g(t),\quad I(0)=I'(0)=0$$
where:
$$g(t)=\begin{cases}1,\quad 0\leq t<\pi \\ -1,\quad \pi\leq t\end{cases}$$
find $I(t)$
The capacitance here is huge, 1F! (I believe it comes from the coefficient of $y$ term)
and we are switching it instantaneously, lets imagine we don't blow anything up.
We can express $g$ using just one heaviside unit step function:
$I''+I=1-2u(t-\pi)$
$\mathcal{L}\{I(t)\}=J(s)$
$s\cdot I(0)=0 \qquad I'(0)=0$
Hit it with the LT! 
$s^2J(s)+J=\frac{1}{s}-\mathcal{L}\{2u(t-\pi)\}$
using formula we derived earlier, ie: $\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$ 
$s^2J(s)+J=\frac{1}{s}-e^{-\pi s}\frac{2}{s}$
$J=\underbrace{ \frac{1}{s(s^2+1)} }_{ =F(s) }- \frac{2}{s(s^2+1)}e^{-\pi s}$
$\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2+1)} \right\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^2+1}\}$ <- using partial fractions
$\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2+1)} \right\}=1-\cos t$
$J(s)=F(s)-2e^{-\pi s}F(s)$
Take the inverse LT of both sides:
$I(s)=1-\cos t-2\mathcal{L}^{-1}\{e^{-\pi s}F(s)\}$
How do we take the inverse of $e^{-\pi s}F(s)$?
Recall that we derived: $\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$
So: $u(t-a)f(t-a)=\mathcal{L}^{-1}\{e^{-as}F(s)\}$
Applying this formula gives:
$I(t)=1-\cos(t)-2u(t-\pi)(1-\cos(t-\pi))$
$$I(t)=1-\cos(t)-2u(t-\pi)(1+\cos(t))$$
#end of lec 18