added lec 20 & 21

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Sasserisop 2023-10-25 14:31:49 -06:00
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@ -17,7 +17,7 @@ $\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Right
compute 2nd derivative of y wrt to x: compute 2nd derivative of y wrt to x:
$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$ $\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
$\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$ $\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
plugging those derivatives in we get: plugging those derivatives in we get: #remember
$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$ $$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$
^ this is a constant coefficient second order non-homogenous equation now! We can solve it now using prior tools. ^ this is a constant coefficient second order non-homogenous equation now! We can solve it now using prior tools.

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# Convolution
A convolution is an operation of function, we take two functions, convolute them and get a new function.
Definition of convolution between f and g:
$$(f*g)(t):=\int _{0} ^t f(t-v)g(v)\, dv$$
property 1) $f*g=g*f$
proof:
$f*g=\int _{0} ^t f(t-v)g(v)\, \underset{ t-v=u }{ dv }=-\int _{t} ^0 f(u)g(t-u) \, du$
$=\int _{0} ^t g(t-u)f(u)\, du=g*f \quad \Box$
property 2) $(f+g)*h=f*h+g*h$
property 3) $(f*g)*h=f*(g*h)$
property 4) $f*0=0$
property 5) $\mathcal{L}\{f*g\}=F(s)G(s)$
he will see us tomorrow at 10oclock. ;)
#end of lec 19

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# Convolution
A convolution is an operation of function, we take two functions, convolute them and get a new function.
Definition of convolution between f and g:
$$(f*g)(t):=\int _{0} ^t f(t-v)g(v)\, dv$$
property 1) $f*g=g*f$
proof:
$f*g=\int _{0} ^t f(t-v)g(v)\, \underset{ t-v=u }{ dv }=-\int _{t} ^0 f(u)g(t-u) \, du$
$=\int _{0} ^t g(t-u)f(u)\, du=g*f \quad \Box$
property 2) $(f+g)*h=f*h+g*h$
property 3) $(f*g)*h=f*(g*h)$
property 4) $f*0=0$
property 5) $\mathcal{L}\{f*g\}=F(s)G(s)$
he will see us tomorrow at 10oclock. ;)
#end of lec 19
#start of lec 20
lets try proving property 5:
recall property 5: $\mathcal{L}\{f*g\}=F(s)G(s)$
$\mathcal{L}\{f*g\}=\int _{0} ^t \left( e^{-st}\int_{0} ^t f(t-v)g(v) \, dv \right)\, dt$
$\mathcal{L}\{f*g\}=\int _{0} ^t \left( e^{-st}\int_{0} ^\infty u(t-v)f(t-v)g(v) \, dv \right)\, dt$
two nested integrals!
using math 209, if both integrals exist, we can exchange the two integrals:
$=\int _{0}^\infty ( g(v)\underbrace{ \int _{0}^\infty e^{-st}f(t-v)u(t-v)\, dt }_{ \mathcal{L}\{f(t-v)u(t-v)\}=e^{-vs}F(s) } )\, dv$
$=F(s)\int _{0} ^\infty e^{-rs}g(v)\, dv=F(s)G(s) \quad \Box$
This is a very useful fact. We will see how it helps us solve differential equations.
ex:
$$\mathcal{L}^{-1}\left\{ \frac{1}{s^2+1}\frac{1}{s^2+1} \right\}$$
we know the inverse of 1/s^2+1 is sin(t):
then:
$=(\sin*\sin)(t)$
$=\int _{0}^t \sin(t-v)\sin(v)\, dv$
use identity: $\sin \alpha \sin \beta=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\beta-\alpha)$ DOUBLE CHECK!
$=\frac{1}{2}\int _{0} ^t (\cos(t-2v)-\cos(t))\, dv$
$=\frac{1}{2}\left( -\frac{1}{2}\sin(t-2v)|^t_{0}-t\cos t \right)=\frac{1}{2}\left( \frac{1}{2}\sin(t)+\frac{1}{2}\sin(t)-t\cos t \right)$
$$=\frac{1}{2}(\sin t-t\cos t)$$
#ex
solve the problem:
$$y'+y-\int _{0} ^t y(v)\sin(t-v) \, dv =-\sin t,\qquad y(0)=1$$
this is called an integral-differential equation.
we can convert it to a differential equation by taking the derivative of both sides (wrt to dt.):
$y''+y'-y\sin(t-v)=-\cos t$
ew thats a gross second order linear equation. lets do it another way
$sY-1+Y-\mathcal{L}\{(y*\sin)(t)\}=-\frac{1}{s^2+1}$
$\left( s+1-\frac{1}{s^2+1}\right)Y(s)=1-\frac{1}{s^2+1}=\frac{s^2}{s^2+1}$
$\frac{(s^2+1)(s+1)-1}{s^2+1}Y(s)=\frac{s^2}{s^2+1}$
$\frac{s^3+s^2+s+\cancel{ 1 }-\cancel{ 1 }}{s^2+1}Y(s)=\frac{s^2}{s^2+1}$
$Y(s)=\frac{s}{s^2+s+1}$
$y(t)=\mathcal{L}^{-1}\left\{ \frac{s}{s^2+s+1} \right\}=\mathcal{L}^{-1}\{\frac{s}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2}\}$
$=\mathcal{L}^{-1}\left\{ \frac{s+\frac{1}{2}-\frac{1}{2}}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2} \right\}$
$=e^{-t/2}\cos\left( \frac{\sqrt{ 2 }}{2}t \right)-\frac{1}{2} \frac{2}{\sqrt{ 3 }}\mathcal{L}^{-1}\left\{ \frac{\frac{\sqrt{ 3 }}{2}}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2} \right\}$
$$y(t)=e^{-t/2}\left( \cos \frac{\sqrt{ 3 }}{2}t-\frac{1}{\sqrt{ 3 }} \sin \frac{\sqrt{ 3 }}{2}t\right)$$
this is a good algorithmic method now for solving differential equations in software, for example solving circuits.
## Transfer function
imagine we have the equation:
$$ay''+by'+cy=g(t), \qquad y(0)=y_{0},\ y'(0)=y_{1}$$
1)
$ay''+by'+cy=g(t)$
$y(0)=y'(0)=0$
gives a solution $y_{*}$
2)
$ay''+by'+cy=0$
$y(0)=y_{0},\ y'(0)=y_{1}$
gives a soltuion $y_{**}=c_{1}y_{1}+c_{2}y_{2}$
then by principle of super position:
$y=y_{*}+y_{**}$
solving 1) gives us:
$as^2Y+bsY+cY=G(s)$
$Y(s)=\frac{1}{as^2+bs+c}G(s)$ the limit approaches 0 for large s so its a legitimate Laplace transform
let $Y(s)=H(s)G(s)$
where $H(s)=\frac{1}{as^2+bs+c}$ and called the transfer function
we put in $g(t)$ and we get out $Y(s)$. So it "transfers".
$H(s)=\frac{Y(s)}{G(s)}$
$\mathcal{L}^{-1}\{H\}=h(t)$ called the impulse response function. We will see why its called that later.
$y_{*}(t)=(h*g)(t)$
$y(t)=(h*g)(t)+c_{1}y_{1}+c_{2}y_{2}$
he's finished 8 minutes early, lets go!
#end of lec 20

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#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 the )
the definition of 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 this RHS came from the impulse delta and the effect it has 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]]
$0.2I_{1}+0.1I_{3}'+2I_{1}=g(t)$
$-I_{2}+0.1I_{3}'=0$
$I_{1}=I_{2}+I_{3}$
voila, a system of three equations.
#end of lec 21

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@ -17,7 +17,7 @@ compute the LT of this funny function:
$f(t)=\begin{cases}1 &\text{if } 0\leq t\leq 1 \\ 2 &\text{if } 1<t\leq 2 \\0 & \text{if } 2<t \end{cases}$ $f(t)=\begin{cases}1 &\text{if } 0\leq t\leq 1 \\ 2 &\text{if } 1<t\leq 2 \\0 & \text{if } 2<t \end{cases}$
$F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0$ $F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0$
$F(s)=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}$ $F(s)=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}$
$$F(s)=-\frac{1}{s}(e^{-s}-1)-\frac{2}{5}(e^{-2s}-e^{-s})$$ $$F(s)=-\frac{1}{s}(e^{-s}-1)-\frac{2}{s}(e^{-2s}-e^{-s})$$
We have shown how to compute the LT of a choppy function. We have shown how to compute the LT of a choppy function.
$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$ $\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$

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@ -6,11 +6,18 @@ Also remember the following:
## derivatives of trigs ## derivatives of trigs
$\frac{d}{dx}\tan(x)=sec^2(x)$ $\frac{d}{dx}\tan(x)=sec^2(x)$
$\frac{d}{dx}sec(x)=sec(x)\tan(x)$ $\frac{d}{dx}sec(x)=sec(x)\tan(x)$
... $\frac{d}{dx}\csc=-\csc x\cot x$
$\frac{d}{dx}\cot (x)=-\csc^2(x)$
## integrals of trigs ## integrals of trigs
$\int \tan(x) \, dx=\ln\mid \sec(x)\mid+C$ $\int \tan(x) \, dx=\ln\mid \sec(x)\mid+C$
$\int sec(x) \, dx=\ln\mid sec(x)+\tan(x)\mid+C$ $\int sec(x) \, dx=\ln\mid sec(x)+\tan(x)\mid+C$
... $\int \csc x \, dx=-\ln|\csc x+\cot x|+C$
$\int \cot x \, dx=-\ln|\csc x|+C$
## Integrals:
$\int \frac{1}{1+x^2} \, dx=\arctan(x)+C$
## integration by parts ## integration by parts
LIATE -> log, inv trig, algebraic, trig, exp LIATE -> log, inv trig, algebraic, trig, exp
set u to the first in the list above set u to the first in the list above

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@ -20,7 +20,8 @@ Good luck on midterms! <3 -Oct 18 2023
[Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html) (raw notes, not reviewed or revised yet.) [Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html) (raw notes, not reviewed or revised yet.)
[(Heaviside) Unit step function (lec 18)](heaviside-unit-step-function-lec-18.html) (raw notes, not reviewed or revised yet.) [(Heaviside) Unit step function (lec 18)](heaviside-unit-step-function-lec-18.html) (raw notes, not reviewed or revised yet.)
[Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.) [Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.)
[Convolution (lec 19)](convolution-lec-19.html) [Convolution (lec 19-20)](convolution-lec-19-20.html) (raw notes, not reviewed or revised yet.)
[Dirak δ-function (lec 21)](dirak-function-lec-21.html) (raw notes, not reviewed or revised yet.)
</br> </br>
[How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.) [How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
</br> </br>