added lec 20 & 21

content/Cauchy-Euler equations (lec 10-11).md

@@ -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:
 $\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
 $\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
-plugging those derivatives in we get:
+plugging those derivatives in we get: #remember
 $$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$
 ^ this is a constant coefficient second order non-homogenous equation now! We can solve it now using prior tools.
 

content/Convolution (lec 19).md

@@ -1,15 +0,0 @@
-# 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

content/Convolution (lec 19-20).md

@@ -0,0 +1,84 @@
+# 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

content/Dirak δ-function (lec 21).md

@@ -0,0 +1,48 @@
+#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

content/Laplace transform (lec 14-16).md

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

content/Things to remember.md

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

content/_index.md

@@ -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.)
 [(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.)
-[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>
 [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>