added lec 14 and added cauchy euler page and revised voparam page

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

@@ -46,6 +46,7 @@ general solution:
 $$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$
 
 ## Free vibrations
+Free vibrations are when there are no externally applied forces acting upon an oscillatory system. RHS=0.
 $mr^2+br+k=0$ characteristic polynomial
 (i) $r_{1}\ne r_{2}$ $b^2-4mk>0$
 $y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$

content/Laplace transform (lec 14).md

@@ -0,0 +1,14 @@
+# 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/Method of undetermined coefficients (lec 8-9).md

@@ -67,9 +67,9 @@ We wanna solve this IVP! We know from earlier that it must have a unique solutio
  $y_{p}(t)=y_{p_{1}}(t)+y_{p_{2}}(t)$
  where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
  $y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
- lets try $y_{p_{1}}=Ae^{-t}$ Does this work? look at it, A must be zero but if A is zero you still get problems.
+ lets try $y_{p_{1}}=Ae^{-t}$ Does this work?
  $y_{p_{1}}'=-Ae^{-t}$
- $y_{p_{1}}''=Ae^{-t}$ plug in these three and we find that A=2
+ $y_{p_{1}}''=Ae^{-t}$ plug in these three and we find that A=2. Yes it works!
  
 second equation, not so easy:
 solution of $\cos(t)$ doesn't quite work because the LHS will obtain a $\sin$ term. Lets try this instead:

content/Prof variation of parameters.md

@@ -0,0 +1,22 @@
+
+# Variation of parameters
+$ay''+by'+cy=f(t)$
+1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ <- h is homogenous, ie: $f(t)=0$
+Lagrange proposed this method to find the particular solution $y_{p}$:
+$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
+we put $y_p$ into the equation and make it equal to the RHS. To do so, find the derivatives first:
+$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
+to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
+1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
+$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
+so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
+now we plug into the second order equation:
+$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
+$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
+Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
+2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
+we now have a system of two equations:
+$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$ 
+by definition $y_1$ and $y_2$ are linearly independent solutions so the above can never be 0!
+$v_{1}'=-\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
+Finally, your solution is: $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ where you get $y_{1},y_{2}$ from your homogenous solution.

content/Resonance & AM (lec 13-14).md

@@ -0,0 +1,83 @@
+#start of lec 13
+He has good news. he's excited to tell us about electric currents! In particular, how a radio uses resonance to selectively listen to a particular frequency:
+# Resonance
+Let's imagine a mass spring system which has an applied force with a forcing frequency of $\gamma$ and an amplitude of $F_{o}$ (a constant):
+$my''+by'+ky=F_{o}\cos(\gamma t)$
+(Think of the driving force being the radio transmitter, and the mass-spring system is an LC tank circuit in a old-style radio)
+In order to study the phenomenon of resonance, we need an underdamped system.
+so we let: $b^2-4mk<0$ (ie: complex roots)
+then the homogenous solution becomes:
+$y_{h}(t)=e^{-bt/2m}\left( c_{1}\cos\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right)+c_{2}\sin\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right) \right)$
+$y_{h}(t)=Ae^{-bt/2m}\sin(\omega t+\phi)$ 
+where $\omega$ is called the angular frequency and equals $\frac{\sqrt{ 4mk-b^2 }}{2m}$
+and where $Ae^{-bt/2m}$ is called the transient part of the equation (goes to 0 as t->$\infty$).
+For particular solution, we use method of undetermined coefficients #mouc
+We guess: $y_p=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$ where $\gamma\ne \omega$ because if $\gamma=\omega$ and b=0 then we would have to multiply our guess by t.
+$A_{1}=\frac{F_{o}(k-m\gamma)}{(k-m\gamma^2)^2+b^2\gamma^2}$
+$A_{2}=\frac{F_{o}b\gamma}{(k-m\gamma^2)^2+b^2\gamma^2}$
+$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)$
+$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
+$k-m\gamma^2=A\sin \theta$
+$br=A\cos \theta$
+$A=\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }$
+$\tan \theta=\frac{k-m\gamma^2}{b\gamma}$
+$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)=\frac{F_{o}\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}{(k-m\gamma^2)^2+b^2\gamma^2}\sin(\gamma t+\theta)$
+$=\frac{F_{o}}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}\sin(\gamma t+\theta)$
+where we define $\mu(\gamma)=\frac{1}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}$ called the gain factor
+and the general solution is $y(t)=y_{h}(t)+y_{p}(t)$
+
+See how $y_h$ goes to zero as time progresses but $y_p$ stays? $y_p$ is the steady state part of the solution. If you were to graph $y(t)$ you would see a "beating" effect due to the sum of the two sins that eventually decays off.
+
+If we make the value in the denominator of the gain factor small, the amplitude goes to a very high value, higher than $F_{o}$! This is the equivalent of tuning a radio receiving circuit to resonate to a certain transmitted frequency, the amplitude grows to a great value in the receiver when excited with the right frequency. This is resonance! as b (friction) is decreased to zero we get stronger and stronger resonance.
+
+Lets find the maximum of the amplitude (resonance point)
+take the derivative of $\mu$ wrt to $\gamma$:
+$\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
+case one: $\gamma=0$ not interesting, because then the force applied would just a constant force.
+case two:
+$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$ where the r means resonance
+By plugging in $\gamma_{r}$ into $\mu()$ we can know the maximum amplitude at resonance: $\mu_{max}(\gamma_{r})=\mu(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$
+if $b^2\geq 4mk>2mk$ (overly damped/critically damped case) (no resonance, imaginary numbers) $\gamma=0$
+if $b^2<2mk<4mk$ (assumed from the very beginning above) then we get a resonant frequency: $\gamma_{r}=\sqrt{\frac{  2mk-b^2}{2m^2}}$
+
+what if b=0? (no resistance):
+$my''+ky=F_{o}\cos(\gamma t)$
+$y_{h}(t)=c_{1}\cos \omega t+c_{2}\sin \omega t$ , $\omega=\sqrt{ \frac{k}{m} }$
+$=A\sin(\omega t+\phi)$
+
+$y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
+assume $\gamma=\omega$ with zero resistance we get:
+$y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
+#end of lec 13
+#start of lecture 14
+# Amplitude modulation
+Last lecture we showed we can selectively listen to a specific signal by using resonance.
+He has something else he's excited to show us; amplitude modulation! Lets recall from last lecture:
+$my''+ky=F_{o}\cos(\gamma t)$ (undamped driven mass-spring system)
+solving characteristic equation:
+$\omega=\sqrt{ \frac{k}{m} }$
+$y_{h}(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t)$ ; $\gamma\ne\omega$
+$y_{p}=A\cos(\gamma t)$ we can guess the particular solution is a constant times $\cos$. There will be no $\sin$ term on the LHS as there's no first derivative  (aka no friction)
+$\Rightarrow A=\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)$
+$y(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t)+\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)$
+assume $y(0)=0=y'(0)$
+$c_{2}=0$ $c_{1}=-\frac{F_{o}}{m(\omega^2-\gamma)}$
+$y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}(\cos(\gamma t)-\cos(\omega t))$
+Hmm, It's not very easy to visualize a cos minus cos term.
+Use trig identity to make equation easier to visualize: $2\sin(\alpha)\sin(\beta)=\cos\left( \frac{{\alpha-\beta}}{2} \right)-\cos\left( \frac{{\alpha+\beta}}{2} \right)$
+$2\gamma t=\alpha-\beta$
+$2\omega t=\alpha+\beta$
+$\Rightarrow \alpha=\frac{(\omega+\gamma)t}{2}$ $\beta=\frac{(\omega-\gamma)t}{2}$
+$$y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}\sin\frac{(\omega-\gamma)t}{2}\sin\frac{(\omega+\gamma)t}{2}$$
+This is an amplitude modulated signal! also can be seen as a "beating frequency".
+
+To recap, taking a frictionless mass spring system which is initially at rest, and applying a driving frequency that is strictly different from the system's natural frequency, results in a displacement that follows an AM modulated signal. How cool is that! 
+>Sometimes I feel this effect when I'm sitting in the back of the bus, where the bus is stopped at a red light. I wonder if this modulated vibration pattern is attributed to this exact phenomenon.
+
+# Shortcut for solving DE of a mass spring system
+![[Drawing 2023-10-06 13.24.11.excalidraw]]
+$my''+by'+ky=mg+F$
+move $mg$ to LHS and replace $y$ with $y_{new}$ (remember, the $\frac{mg}{k}$ is a constant, its derivative is 0):
+$m\left( y-\frac{mg}{k} \right)''+b\left( y-\frac{mg}{k} \right)'+k\left( y-\frac{mg}{k} \right)=F$
+$my_{new}''+by_{new}'+ky_{new}=F$
+This simplifies our approach to solving a mass spring system. We could do it without this rearrangement, but it's more complex as the RHS has a sum of two terms. Either way works though, pick what you like.

content/Resonance in free vibrations (lec 13).md

@@ -1,41 +0,0 @@
-#start of lec 13
-He has good news. he's excited to tell us about electric currents!
-$my''+by'+ky=F_{o}\cos(\gamma t)$ the system has an applied force with a forcing frequency of $\gamma$ and an amplitude of $F_{o}$ (a constant)
-$b^2-4mk<0$ (complex roots)
-$y_{h}(t)=e^{-bt/2m}\left( c_{1}\cos\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right)+c_{2}\sin\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right) \right)$
-$y_{h}(t)=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\omega$ is angular frequency and equals $\frac{\sqrt{ 4mk-b^2 }}{2m}$
-where $Ae^{-bt/2m}$ is called the transient part of the equation (makes equation go to 0 as t->$\infty$)
-We guess: $y_p=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$ where $\gamma\ne \omega$ because if $\gamma=\omega$ and b=0 then we would have to multiply our guess by t.
-$A_{1}=\frac{F_{o}(k-m\gamma)}{(k-m\gamma^2)^2+b^2\gamma^2}$
-$A_{2}=\frac{F_{o}b\gamma}{(k-m\gamma^2)^2+b^2\gamma^2}$
-$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)$
-$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
-$k-m\gamma^2=A\sin \theta$
-$br=A\cos \theta$
-$A=\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }$
-$\tan \theta=\frac{k-m\gamma^2}{b\gamma}$
-$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)=\frac{F_{o}\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}{(k-m\gamma^2)^2+\beta^2\gamma^2}\sin(\gamma t+\theta)$
-$=\frac{F_{o}}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}\sin(\gamma t+\theta)$
-where we define $\mu(\gamma)=\frac{1}{(k-m\gamma^2)^2+b^2\gamma^2}$ called the gain factor
-and the general solution is $y(t)=y_{h}(t)+y_{p}(t)$
-
-see how y_h goes to zero but y_p stays? y_p is the steady state part of the solution. if you were to graph it you would see a wah wah effect that decays to zero.
-
-if we make the value in the denominator of the gain factor small the amplitude goes to a very high value, higher than $F_{o}$ this is equivalent to tuning a radio circuit to resonate to a certain transmitted frequency, the amplitude grows to a great value in the receiver when excited with the right frequency. This is resonance! as b (friction) decreases to zero we get stronger and stronger resonance.
-lets find the maximum of the amplitude (resonance point)
-$\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
-cases: $\gamma=0$ not interesting, beacuse then the force applied is a constant force.
-$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$ where the r means resonance
-by taking second derivative$\mu_{max}(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$
-if $b^2\geq 4mk>2mk$ (overly damped/critically damped case) (no resonance, imaginary numbers) $\gamma=0$
-if $b^2<2mk<4mk$ (assumed from the beginning above) then we get a resonant frequency $\gamma_{r}=\sqrt{\frac{  2mk-b^2}{2m^2}}$
-
-what if b=0? (no resistance):
-$my''+ky=F_{o}\cos(\gamma t)$
-$y_{h}(t)=c_{1}\cos \omega t+c_{2}\sin \omega t$ , $\omega=\sqrt{ \frac{k}{m} }$
-$=A\sin(\omega t+\phi)$
-
-$y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
-assume $\gamma=\omega$ with zero resistance we get:
-$y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
-#end of lec 13

content/Variation of parameters (lec 9-10).md

@@ -1,90 +1,105 @@
 # Variation of parameters 
-$ay''+by'+cy=f(t)$
+#voparam
-1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}t$ <- h is homogenous, ie: $f(t)=0$
+>[Professors definition/derivations during lecture](prof-variation-of-parameters.html) <- I found this to be too big brain for me. Here is a simplified definition:
-Lagrange proposed: find a particular solution of $y_{p}$
-$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
-we put $y_p$ into the equation and make it equal to the RHS. To do so, find the derivatives first:
-$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
-to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
-1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
-$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
-so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
-now we plug into the second order equation:
-$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
-$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
-Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
-2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
-$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$ 
-by definition $y_1$ and $y_2$ are linearly independent solutions so the above can never be 0!
-$v_{1}'=\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=-\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
 
-#ex #second_order #IVP 
+## Variation of parameters is a method to solve:
+## $$ay''+by'+cy=f(t)$$
+First, find the homogenous solution:
+$y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
+Now we need the particular solution, let $y_{p}$ be in the following form:
+$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair. They are obtained from your homogenous solution.
+Impose the following: 
+1) $v_{1}'y_{1}+v_{2}'y_{2}=0$
+Compute the derivatives and simplify:
+$y'_{p}=v_{1}y_{1}'+v_{2}y_{2}'$
+$y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
+Now we plug those into the second order equation and simplify:
+2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
+We now have a system of two equations (1 and 2). Now we can solve for $v_{1}$ and $v_{2}$:
+Using Cramer's rule, we can solve for the system of equations and obtain the solutions:
+$v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}$; $v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
+also, $W[y_1,y_{2}]$ is the Wrońskian, and it equals to: $\det \begin{pmatrix}y_{1}&y_{2}\\ y_{1}' &y_{2}'\end{pmatrix}=y_{1}y_{2}'-y_{2}y_{1}'$
+Finally, the general solution is: 
+$$y(t)=y_{h}+y_{p}\qquad \text{where}\qquad y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$$
+## What you need to remember
+So, what do you need to commit to memory? I believe memorizing these three is a good tradeoff between memory allocated and speed for when you're solving a #voparam problem:
+1) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
+2) $v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}$
+3) $v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$
+Alternatively, you could memorize the system of equations and solve for $v_{1}'$ and $v_{2}'$. Ie:
+1) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
+2) $v_{1}'y_{1}+v_{2}'y_{2}=0$
+3) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
+This is what the prof likes. I love you Dr. Minev, but this I personally disagree with. I think I'll stick with the formulas.
+---
+#ex #second_order #IVP #voparam #mouc
 $y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
-can we use undetermined coefficients? yes and no
+Can we use undetermined coefficients? Yes and no. We can use it on the $e^t$ term. However, guessing and checking $y_{p}$ for the $2\tan(2t)$ term might take a really, really long time.
-find general solution to homogenous counterpart
+First, find general solution to homogenous counterpart:
-1) $y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
+$y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
-$y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin(2t)$
+$y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin(2t)$ Done. Easy peasy.
-2) $y''+4y=-e^t$ <- use method of undetermined coefficients
+For $-e^t$ lets use method of undetermined coefficients:
-$y_{p}^1(t)=Ae^{t}$
+$y''+4y=-e^t$ 
+$y_{p_{1}}(t)=Ae^{t}$
 $5Ae^t=-e^t$
 $A=-\frac{1}{5}$
-$y_{p}^1(t)=-\frac{1}{5}e^t$
+$y_{p_{1}}(t)=-\frac{1}{5}e^t$
-(ii) $y''+4y=2\tan(2t)$ <- cant use method of undetermined coefficients
-$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
-plugging in:
-we get a system of eq:
-	$\cos(2t)v_{1}'+\sin(2t)v_{2}'=0$
-	$-2\sin(2t)v_{1}'+2\cos(2t)v_{2}'=2\tan(2t)$
-> we know these two will give a unique solution.
->to solve system of eq multiply each by:
->$2\cos(2t)$
->$\sin(2t)$
 
-$2(\sin^2(2t)+\cos^2(2t))v_{2}'=2\tan(2t)\cos(2t)$
+Now for $2\tan(2t)$, we cannot realistically use method of undetermined coefficients.
-$v_{2}'=\sin(2t)$
+Let's use variation of parameters instead:
-$v_{2}(t)=-\frac{1}{2}\cos(2t)$ no constant of integration, we want one solution only
+$y''+4y=2\tan(2t)$
-$v_{1}'=-{\frac{\sin^2(2t)}{\cos(2t)}}$
+$y_{p_{2}}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
+where $y_{1}=\cos(2t)$, $y_{2}=\sin(2t)$
+recall:
+$v_{1}'=-\frac{f(t)y_{2}}{aW[y_{1},y_{2}]}$
+$v_{2}'=\frac{f(t)y_{1}}{aW[y_{1},y_{2}]}$
+plugging in:
+$v_{1}'=-\frac{2\tan(2t)\sin(2t)}{(1)(\cos2t(2)\cos2t-\sin2t(-2)\sin2t}=-\frac{2\tan(2t)\sin(2t)}{2\cos^22t+2\sin^22t}=-\tan(2t)\sin(2t)$
+$v_{2}'=\frac{2\tan(2t)\cos(2t)}{2}=\sin(2t)$ <- isn't it nice how we can reuse our computation for the denominator? :D
+Now we integrate.
+$v_{2}=-\frac{1}{2}\cos(2t)$ <- Don't add a constant of integration, we want one solution only.
 $v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dt$
 $v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt$
 $v_1=-\int sec(2t) \, dt+\int \cos(2t) \, dt$
 $v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t)$
-$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
+$y_{p_{2}}(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
-$y(t)=y_{h}(t)+y_{p}^1(t)+y_{p}^2(t)$
+> ^Now you can start to see how guessing $y_{p_{2}}$ would take a really, really long time.
+
+$y(t)=y_{h}(t)+y_{p_{1}}(t)+y_{p_{2}}(t)$
 =$c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t$
 is the general answer.
 
 IVP solution:
-$y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5}\Rightarrow c_{1}=\frac{1}{5}$
+$y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5} \implies c_{1}=\frac{1}{5}$
-skipping some differentiation: $y'(0)=2c_{2}+y_{p}'(0)=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5}=\frac{4}{5}\Rightarrow c_{2}=1$
+$y'(0)=\frac{4}{5}=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5} \implies c_{2}=1$
 $$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$$
 
 #end of lecture 9
 #start of lecture 10
-# Variation of parameters
+
-last lec we did some variation of parameters
+#ex #second_order #voparam #mouc
-$ay''+by'+cy=f(t)$
-1) $y_{h}(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
-2) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
-	$y_{1}v_{1}'+y_{2}v_{2}'=0$
-	$y_{1}v_{1}'+y_{2}'v_{2}'=\frac{b}{a}$ or f/a?
-is the system of equations we will need to solve. You can also memorize a formula but peter likes remembering this system of equations and moving on from there.
-#ex #variation_of_parameters
 $$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
-i) $y_{h}(t)=?$
+Find homogenous solution first:
 $r^2-2r+1=0$
-$r_{1,2}=1$
+$r_{1,2}=1$ (repeated root)
 $y_{h}(t)=c_{1}e^t+c_{2}te^t$
 2) $y_{p}(t)=?$
 $y''-2y'+y=2\cos (t)$
-$y_{p}''=A\cos(t)+B\sin(t)$ is our first guess. but it does not solve the homogenous eq.
+let's use method of undetermined coefficients:
-$y_{p}'=-\sin(t)$ (obtained by using method of undetermined coefficients, computation not shown.)
+$y_{p_{1}}=A\cos(t)+B\sin(t)$ is our guess
+$y_{p_{1}}'=-A\sin t+B\cos t$
+$y_{p_{1}}''=-A\cos t-B\sin t$
+$-A\cos t-B\sin t+2A\sin t-2B\cos t+A\cos t+B\sin t=2\cos t$
+$-2B\cos t+2A\sin(t)=2\cos t(t)$
+$\implies A=0,\ B=-1$
+$y_{p_{1}}=-\sin(t)$ 
 $y''-2y'+y=e^t\ln(t)$ cant use undetermined coefficients, use variation of parameters
 $y''_{p}(t)=v_{1}y_{1}+v_{2}y_{2}$
 $=v_{1}e^t+v_{2}te^t$
-compute v1 and v2, using the linear system:
+Compute v1 and v2. This time lets do it using the linear system for funzies:
-eq1) $e^t+v_{1}'+te^tv_{2}'=0$
+eq1) $e^tv_{1}'+te^tv_{2}'=0$
 eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$
-subtract eq1 from eq2 $v_{2}'=\ln(t)$
+subtract eq1 from eq2: $v_{2}'=\ln(t)$
 $v_{2}(t)=\int \ln(t) \, dt$
 integrate by parts
 $=t\ln(t)-\int t\frac{1}{t} \, dt$

content/_index.md

@@ -9,9 +9,12 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Exact equations (lec 4-5)](exact-equations-lec-4-5.html)
 [Second order homogenous linear equations (lec 5-7)](second-order-homogenous-linear-equations-lec-5-7.html) 
 [Method of undetermined coefficients (lec 8-9)](method-of-undetermined-coefficients-lec-8-9.html)
-[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) (raw notes, not reviewed or revised yet.)
+[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) 
+[Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.html) (raw notes, not reviewed or revised yet.)
 [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
-[Resonance in free vibrations (lec 13)](resonance-in-free-vibrations-lec-13.html) (raw notes, not reviewed or revised yet.)
+[Resonance in free vibrations (lec 13-14)](resonance-in-free-vibrations-lec-13.html)
+[Laplace transform (lec 14)](laplace-transform-lec-14.html)
+
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png)
 </br>