#start of lec 21
From $ma=F$
$m\frac{dv}{dt}=f(t)$
integrate both sides:
$m\int_{t_{0}} ^{t_{1}} \frac{dv}{dt}dt =\int _{{t_{0}}} ^{t_{1}}f(t) \, dt$
$mv(t_{1})-mv(t_{0})=\int _{t_{0}}^{t_{1}}f(t) \, dt$
that is, change in momentum on the LHS equates to an impulse on the RHS.
(picture shown, you can have the same impulse, the same area under the graph if you squish down f(t) to be narrower, as long as you make it taller. If we take it to the extreme we get the Dirak delta function.)

The definition of the Dirak delta function:
$\delta(t-a)=\begin{cases}0,  & t\ne a \\''\infty'',  & t=a\end{cases}$
however, a more useful definition is:
$\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$

properties:
$\int_{{-\infty}}^{\infty} \delta(t-a)\, dt=1$
![[Drawing 2023-10-25 13.16.20.excalidraw]]
$\int _{\infty} ^t \delta(x-a)\, dx=\begin{cases}0, & t<a \\ 1,  & a\leq t\end{cases}=u(t-a)$
$u'(t-a)=\delta(t-a)$

What is $\mathcal{L}\{\delta(t-a)\}$?
$\mathcal{L}\{\delta(t-a)\}=\int _{0} ^\infty \delta(t-a)e^{-st} \, dt$ for $a>0$
$=\int _{-\infty} ^{\infty} e^{-st} \delta(t-a) \, dt=e^{-as}$ using the definition earlier

#ex 
$$w''+6w'+5w=e^t\delta(t-1) \qquad w(0)=0 \qquad w'(0)=4$$
solving this using something like #voparam would be very difficult, using LT should be very easy!
$s^2W-sw(0)-w'(0)+6sW+5W=\int _{0} ^\infty e^{-st}e^t\delta(t-1)\, dt$
$=s^2-4+6sW+5W=\int _{-\infty} ^\infty e^{-st}e^t\delta(t-1)\, dt$
(we can extend the range of the integral as the integrand is 0 for t<1)
this allows us to use the earlier definition of the delta function ie:
$=e^{-(-s-1)}$ using definition of delta function earlier.
$W(s^2+6s+5)=4+ee^{-s}$
$W(s)=\frac{4}{(s+1)(s+5)}+\frac{ee^{-s}}{(s+1)(s+5)}$
$w(t)=\frac{1}{4}\mathcal{L}^{-1}\left\{ \frac{1}{s+1} - \frac{1}{s+5} \right\}+e\mathcal{L}^{-1}\{e^{-s}\left( \frac{1}{s+1} - \frac{1}{s+5} \right)\}$
$=\underbrace{ \frac{1}{4}(e^{-t}-e^{ -5t }) }_{ \text{this came from initial conditions} }+\dots$
$\mathcal{L}^{-1}\{e^{as}F(s)\}=f(t-a)u(t-a)$
$$y(t)= \frac{1}{4}(e^{-t}-e^{ -5t }) +\frac{e}{4}u(t-1)(e^{ -(t-1) }-e^{ -5(t-1) })$$
notice that the right most term came from the impulse and the effect it had on the system.
side note: delta functions are useful for quantum physics.

Lets start modelling some electric circuits again:
![[Drawing 2023-10-25 13.43.26.excalidraw]]
the circuit is starts switched on  and is then switched off at $t=1$
Applying KVL:
$0.2I_{1}+0.1I_{3}'+2I_{1}=g(t) \qquad \text{where } g(t)=\begin{cases}6, & 0\leq t\leq 1 \\0, & 1<t\end{cases}$
$-I_{2}+0.1I_{3}'=0$
$I_{1}=I_{2}+I_{3}$
voila, a system of three equations.
#end of lec 21
#start of lec 22
The prof shared some really cool story of George Green and how he lived on his father's grain mill and derived Green's Theorem and how this theorem eventually came to Stokes desk and Stoke decided to put this new, never before seen theorem on his exam and how it's really not like nowadays where everything is Mcdonalds style, Mcdonalds style education.
Another cool story of Ramanujan, he was very poor, but he managed to enter a school, an accomplishment in and of it self. His teacher saw his magical ability and told him to write to Hardy, a British mathematician. Hardy was impressed and he flew Ramanujan there and worked with him and he saw they things Ramanujan showed and it was inhuman, but he had few mathematical skills so he taught him how to make proofs and now Hardy said that if we were to rate all the mathematicians in the world he would give:
myself 20/100
Hilbert 40/100
Ramanujan 80/100

OKAY get back to the lecture, that's 6 minutes worth of stories. (I really live for Petar's stories. Idk why but they just feel like jewels and I love saving them. I hope he's okay with me doing that.)
We continue on the last lecture problem:
$0.2I_{1}+0.1I_{3}'+2I_{1}=g(t) \qquad \text{where } g(t)=\begin{cases}6, & 0\leq t\leq 1 \\0, & 1<t\end{cases}$
$-I_{2}+0.1I_{3}'=0$
$I_{1}=I_{2}+I_{3}$
two differential equations, one algebraic equation.
express g in terms of unit step funtion:
$g(t)=6-6u(t-1)$
applying this on the second equation:
$-I_{1}+I_{3}+0.1I_{3}'=0$

now we have two equations:
$0.2I_{1}+0.1I_{3}'+2I_{1}=6-6u(t-1)$
$-I_{1}+I_{3}+0.1I_{3}'=0$
where $I_1(0)=I_{2}(0)=I_{3}=0$
multiply eqations by 10:
$2I_{1}+1I_{3}'+20I_{1}=60(1-u(t-1))$
$-10I_{1}+10I_{3}+I_{3}'=0$
hit it with the LT!
$J_{1}=\mathcal{L}\{I_{1}(t)\}(s)$
$J_3=\mathcal{L}\{I_{3}\}$
$2sJ_{1}+sJ_{3}+20J_{1}=\mathcal{L}\{g(t)\}$
$2(s+10)J_{1}+sJ_{3}=\dots$
$-10J_{1}+(s+10)J_{3}=0$
where $\dots$ is:
using:
$\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$
$\mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$

so:
$2(s+10)J_{1}+sJ_{3}=60 \frac{1-e^{-s}}{s}$
multiply 2 by s+10
multiply 1 by s:

$J_{1}(s)=30 \frac{s+10}{s(s+5)(s+20)}(1-e^{-s})$ 
use partial fraction:
$30 \frac{s+10}{s(s+5)(s+20)}=\frac{A}{s}+\frac{B}{s+5}+\frac{C}{s+20}$
$=\frac{{A(s+5)(s+20)+B(s^2+20s)+C(s^2+5s)}}{s(s+5)(s+20)}$
then: $A+B+C=0$
$25A+20B+5C=30$
$100A=300 \implies A=3$
$\implies B=-2 \qquad C=-1$
$J_{1}(s)=\left( \frac{3}{s}-\frac{2}{s+5}- \frac{1}{s+20} \right)(1-e^{-s})$ 
invert the LT: (use $\mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$)
$I_{1}=3-2e^{-5t}-e^{-20t}-u(t-1)(3-2e^{-5(t-1)}-e^{-20(t-1)})$
whats $I_{3}$? use: $-10I_{1}+10I_{3}+I_{3}'=0$ (i think)
$J_{3}=\frac{10}{s+10}J_{1}$
$J_{3}=300 \frac{1}{s(s+5)(s+20)}(1-e^{ -s })$
partial fraction it so we can eventuall take the inverse LT:
skip some steps:
$J_{3}=\left( \frac{3}{s}-\frac{4}{s+5}+\frac{1}{s+20} \right)(1-e^{ -s })$
$I_{3}=3-4e^{ -5t }+e^{ -20t }-(3-4e^{ -5(t-1) }+e^{ -20(t-1) })u(t-1)$
$I_{2}=I_{1}-I_{3}$

$ay''+by'+cy=f$
$y'=z$
$\begin{cases}y'-z=0 \\az'+bz+cy=f\end{cases}$

last example of the chapter:
#ex
$x'+y=0, \qquad x(0)=0$
$x+y'=1-u(t-z) \qquad y(0)=0$
this is review problem in chapter 7 of the textbook.
hit equation 1 and 2 with the LT:
$sX+Y=0$
$X+sY=\frac{{1-e^{ -2s }}}{s}$
multiply equation 1 by s:
$-(s^2-1)X=\frac{{1-e^{ -2s }}}{2}$
$X(s)=e^{=2s} \frac{1}{s(s-1)(s+1)}-\frac{1}{s(s-1)(s+1)}$
use partial fractions:
$\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s+1}=\frac{1}{s(s-1)(s+1)}$
$A(s^2-1)+B(s^2+s)+C(s^2-s)=1$
$A+B+C=0$
$B-C=0$
$-A=1\implies A=-1$
$\implies B=\frac{1}{2} \qquad C=\frac{1}{2}$
$X(s)=\left( -\frac{1}{s}+\frac{1}{2} \frac{1}{s-1}+\frac{1}{2} \frac{1}{s+1} \right)(e^{ -2s }-1)$
inverse laplace:
$x(t)=1-\frac{1}{2}e^t-\frac{1}{2}e^{-t}-u(t-2)\left( 1-\frac{1}{2}e^{t-2} -\frac{1}{2}e^{-(t-2)}\right)$
$Y=-sX=-s (\frac{-1}{s(s-1)(s+1)}+ \frac{e^{ -2s }}{s(s-1)(s+1)})$
$=\frac{1}{(s-1)(s+1)}- \frac{e^{-2s}}{(s-1)(s+1)}$
$=\frac{1}{2}(1-e^{-2s})\left( \frac{1}{s-1}-\frac{1}{s+1} \right)$
$\implies y(t)=\frac{1}{2}(e^t-e^{-t})-\frac{1}{2}u(t-2)(e^{t-2}-e^{-(t-2)})$
we are done
#end of lec 22