add lec 15 LT stuff

content/Free vibrations (lec 11-12).md

@@ -15,8 +15,8 @@ then: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$
 then: $y_{h}(x)=x^2(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$
 now find one particular solution for a non homogenous soultion, using variation of parameters, combine the y_h and y_p to get y(x).
 
-not all equations can fall into cauchy euler type.
-$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procudure always
+# Reduction of order
+$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procedure always
 but, if $y_{1}(x)$ solves $y''+p(x)y'+q(x)y=0$
 then we can find the general solution to the non homogenous equation (1) by guessing it in the form $y(x)=v(x)y_{1}(x)$
 $y'=v'y_{1}+vy_{1}'$
@@ -29,10 +29,10 @@ $v'=u$
 $u'+\left( \frac{2y_{1}'}{y_{1}}+p \right)u=\frac{f}{y_{1}}$<- this is a linear first order equation
 how to solve linear first order equation? we  compute the integrating factor $\mu$
 $\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1})^2}e^{\int P(x) \, dx}=y_{1}^2e^{\int p(x) \, dx}$
-isnt this nice? some kind of magic. We made some guesses and we arrived somewhere.
+Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere.
 
-#ex find the general solution to the equation:
-$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$
+#ex #reduction_of_order find the general solution to the equation:
+$$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$$
 if $y_{1}(x)=e^{-x^2}$ is one solution.
 therefore were finding the solution of the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$
 $v'=u$

content/Laplace transform (lec 14).md

@@ -1,14 +0,0 @@
-# Laplace transform
-From now on, LT is short for Laplace Transform
-What is LT? It's denoted as $\mathcal{L}$ and it's an operator defined as the following:
-$$\underset{ =F(s) }{ \mathcal{L}\{f(t)\}(s) }:=\int _{0}^\infty e^{-st}f(t) dt=\lim_{ T \to \infty }\int _{0} ^T e^{-st}f(t)\, dt$$
-This doesn't look like anything useful, but later on we will learn how it is.
-
-$\mathcal{L}\{0\}=0$ Look at your bank account, integrate 0 you still get 0 :D
-$\mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s}$ if s>0
-$\mathcal{L}\{e^{at}\}$=$\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}$ if $s-a>0$
-$\mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2}$ by integration by parts
-similarly can be done for cos, but we have run out of time.
-#end of lec 14
-
-

content/Laplace transform (lec 14-15).md

@@ -0,0 +1,63 @@
+# Laplace transform
+From now on, LT is short for Laplace Transform
+What is LT? It's denoted as $\mathcal{L}$ and it's an operator defined as the following:
+$$\underset{ =F(s) }{ \mathcal{L}\{f(t)\}(s) }:=\int _{0}^\infty e^{-st}f(t) dt=\lim_{ T \to \infty }\int _{0} ^T e^{-st}f(t)\, dt$$
+This doesn't look like anything useful, but later on we will learn how it is.
+
+Here are some trivial example computations:
+$\mathcal{L}\{0\}=0$ Look at your bank account, integrate 0 you still get 0 :D
+$\mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s}$ if s>0
+$\mathcal{L}\{e^{at}\}$=$\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}$ if $s-a>0$
+$\mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2}$ by integration by parts
+similarly can be done for cos, but we have run out of time.
+#end of lec 14 #start of lec 15
+compute the LT of this funny function:
+$f(t)=\begin{pmatrix}1, & 0\leq t\leq 1 \\ 2, & 1<t\leq 2 \\0, & 2<t \end{pmatrix}$
+$F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0$
+$=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}=-\frac{1}{s}(e^{-s}-1)-\frac{2}{5}(e^{-2s}-e^{-s})$
+We have shown how to compute the LT of a choppy function.
+
+$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$
+the LT is a linear operator. as shown above. Proof:
+$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st}  \, dt+\beta \int_{0}^{\infty}g(t)e^{-st}  \, dt \quad \Box$
+
+Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a /finite/ number of points of /jump/ discontinuity
+what is continuouity? the limit exists and equals the value at that point.
+![[Drawing 2023-10-11 13.17.32.excalidraw]]
+Def: f(t) is of exponential order $\alpha$ if there $\exists\ T,\ M, T\geq0$ such that $\forall\ t>T : f(t)\leq Me^{\alpha t}$
+this is important so that f(t)e^{-st} doesn't go into infinity,
+this can be proven, but not necessary
+Theorem: If $f(t)$ is piecewise continuous and of an exponential order $\alpha$ on $[0, \infty)$, then $\mathcal{L}\{f\}$ exists $s>\alpha$
+and we are guarenteed that $s>\alpha$
+$e^{\alpha t}$ is this of exponential order? Of course, $M=1,\alpha=\alpha$
+what about $\sin(t)$? yes, sin is bounded b/w -1 and 1 
+What about $e^{t^2}$? No, t^2 always outgrows $\alpha t$ eventually. This function does not have a LT.
+
+![[Drawing 2023-10-11 13.21.18.excalidraw]]
+Theorem if $F(s)$ is a LT then $\lim_{ s \to \infty }F(s)=0$
+
+properties:
+assume LT of f(t) exists:
+first property:  $\mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)$
+
+$\mathcal{L}\{e^{\alpha t}\cdot 1\}=\frac{1}{s-\alpha}$
+$\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ these properties are essential for the midterm.
+
+using integration by parts:
+$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+s \underbrace{ \int e^{-st}f(t) \, dt }_{ F(s) }$$=sF(s)-f(0)$
+$\mathcal{L}{f''}=s^2F(s)-sf(0)-f'(0))$
+using proof by induction (try at home!):
+$\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$
+
+trig stuff:
+$\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$
+$\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}$
+$\cos(bt)=\frac{1}{b}(\sin(bt))'$
+
+$\frac{dF}{ds}(s)=\frac{d}{ds}\int _{0 } ^\infty e^{-st}f(t)\, dt=\int _{0} ^\infty f(t) \frac{d}{ds}(e^{-st}) \, dt$
+$=-\int _{0} ^\infty e^{-st}tf(t)\, dt$
+$-\frac{dF}{ds}=\mathcal{L}\{tf(t)\}$; again we can use induction to prove:
+$\mathcal{L}\{t''f(t)\}=(-1)^{n} \frac{d^nF}{ds^n}$
+$\mathcal{L}\{t 1\}=-(s^-1)^1=s^-2$; $\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$ (can be proven by induction)
+today covers all midterm material. Yay!
+#end of lec 15

content/Method of undetermined coefficients (lec 8-9).md

@@ -105,6 +105,6 @@ s=2 if r is a double root.
 case ii) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+Q_{n}(t)e^{\alpha t}\sin(\beta t)$
 Then we guess the particular solution is of the form: $y_{p}(t)=t^s[(A_{k}t^k+A_{k-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
 where:
-k=max(m,n)
-s=0 if $\alpha+i\beta$ is not a root,
-s=1 if $\alpha+i\beta$ is a root.  
+$k=\max(m,n)$
+$s=0$ if $\alpha+i\beta$ is not a root,
+$s=1$ if $\alpha+i\beta$ is a root.  

content/_index.md

@@ -13,7 +13,7 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.html)
 [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
 [Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
-[Laplace transform (lec 14)](laplace-transform-lec-14.html)
+[Laplace transform (lec 14-15)](laplace-transform-lec-14-15.html) (raw notes, not reviewed or revised yet.)
 
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png)