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revise cauchy euler page

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content/Cauchy-Euler equations (lec 10).md
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we know how to solve second order equations where a,b,c are constants. Even if they're not constant some can be expressed as a linear equation. But not always will they be solvable. However, there is one class of second order equations with non constant coefficients that are always solvable. we know how to solve second order equations where a,b,c are constants. Even if they're not constant some can be expressed as a linear equation. But not always will they be solvable. However, there is one class of second order equations with non constant coefficients that are always solvable.
# Cauchy-Euler equations # Cauchy-Euler equations
*if it has a name in it, its very important, if it has 2 names its very important!* *If it has a name in it, its very important, if it has 2 names, its very very important!*
$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$ #cauchy-euler equations are equations in the form:
$$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$$
where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$ where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
note if x=0 is not interesting as the derivative terms disappear. Note: x=0 is not interesting as the derivative terms disappear.
how to solve? two approaches: How to solve? There are two approaches:
textbook only use 2nd method. prof doesn't like this. Textbook only use 2nd method, prof doesn't like this. You can find both methods in the profs notes. Btw, do you know Stewart? Multimillionaire, he's living in a mansion in Ontario.
you can find both methods in the profs notes.
you know Stewart? multimillionaire, he's living in a mansion in Ontario.
introduce change of variables: introduce change of variables:
$x=e^t\Rightarrow t=\ln x$ (x is always +) $x=e^t\Rightarrow t=\ln x$ (x is always +)
(do $x=-e^t$ if you need it to be negative.) (do $x=-e^t$ if you need it to be negative.)
find derivatives with respect to t now. y is a function of t which is a function of x. find derivatives with respect to t now. y is a function of t which is a function of x.
$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ Impor\tan t }{ x\frac{dy}{dx}=\frac{dy}{dt} }$ $\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ \text{Important} }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
compute 2nd derivative of y wrt to x: compute 2nd derivative of y wrt to x:
$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\left( \frac{dt}{dx} \right)^2+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$ $\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
$\underset{ \mathrm{Im}por\tan t }{ x^22{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$ $\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
plugging those derivatives in we get:
$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$ $$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$
^ this is a constant coefficient equation now! We can solve it now using prior tools. ^ this is a constant coefficient equation now! We can solve it now using prior tools.
#ex ## Example:
solve: #ex #second_order #second_order_nonhomogenous #cauchy-euler
Find the general solution for:
$$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$ $$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$
$x=e^t$ substitute: $x=e^t$
transform using the technique we showed just earlier: transform using the technique we showed just earlier:
$\frac{d^2y}{dt^2}+2{\frac{dy}{dt}}+y=e^{-t}$ $\frac{d^2y}{dt^2}+2{\frac{dy}{dt}}+y=e^{-t}$
1) $r^2+2r+1=0$ 1) $r^2-2r+1=0$
$r_{1,2}=-1$ $r_{1,2}=-1$
$y_{h}(t)=c_{1}e^{-t}+c_{2}te^{-t}$ $y_{h}(t)=c_{1}e^{-t}+c_{2}te^{-t}$
2) $y_{p}(t)=At^2e^{-t}$ <- using method of undetermined coefficients 2) $y_{p}(t)=At^2e^{-t}$ <- using method of undetermined coefficients
$A=\frac{1}{2}$ $A=\frac{1}{2}$
general solution in terms of t: general solution in terms of t:
$y_{1}(t)=c_{1}e^t+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$ $y(t)=c_{1}e^{-t}+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
bottom line: solution in terms of t, but we want solution wrt to x: but we want solution in terms of x:
$y_{1}(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$ $y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$
$=c_{1}x^-1+c_{2}\ln(x)x^-1+\frac{1}{2}{\ln(x)^2}x^-1$ $$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$
We are done.
#end of lecture 10 #end of lecture 10

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content/Free vibrations (lec 11-12).md
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@ -99,7 +99,7 @@ $A=\frac{5}{4}$
$\tan \phi=\frac{4}{3}$ $\tan \phi=\frac{4}{3}$
$\phi \approx 0.9273\dots$ $\phi \approx 0.9273\dots$
$$y(t)=\frac{5}{4}e^{-3t}\sin(4t+\phi)$$ $$y(t)=\frac{5}{4}e^{-3t}\sin(4t+\phi)$$
"I know engineers loves calculators, I know mathematicians hates calculators, and that's probably the only difference between mathematicians and engineers." -Peter (referring to calculating arctan(4/3) on an exam) "I know engineers love calculators, I know mathematicians hate calculators, and that's probably the only difference between mathematicians and engineers." -Peter (referring to calculating arctan(4/3) on an exam)
3) b=10 3) b=10
$r_{1,2}=-5$ $r_{1,2}=-5$
$y(t)=(c_{1}+c_{2}t)e^{-5t}$ $y(t)=(c_{1}+c_{2}t)e^{-5t}$

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content/Resonance & AM (lec 13-14).md
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#start of lec 13 #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: He has good news. he's excited to tell us about electric currents! In particular, how a radios use resonance to selectively listen to a particular frequency:
# Resonance # 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): 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):
$my''+by'+ky=F_{o}\cos(\gamma t)$ $my''+by'+ky=F_{o}\cos(\gamma t)$
@ -35,10 +35,11 @@ take the derivative of $\mu$ wrt to $\gamma$:
$\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$ $\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
case one: $\gamma=0$ not interesting, because then the force applied would just a constant force. case one: $\gamma=0$ not interesting, because then the force applied would just a constant force.
case two: case two:
$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$ where the r means resonance $$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$$
By plugging in $\gamma_{r}$ into $\mu()$ we can know the maximum amplitude at resonance: $\mu_{max}(\gamma_{r})=\mu(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$ 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 $b^2\geq 4mk>2mk$ (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: $\gamma_{r}=\sqrt{\frac{ 2mk-b^2}{2m^2}}$ if $b^2<2mk<4mk$ (assumed from the very beginning above) then we get a resonant frequency $\gamma_{r}$ which can be calculated as shown above.
what if b=0? (no resistance): what if b=0? (no resistance):
$my''+ky=F_{o}\cos(\gamma t)$ $my''+ky=F_{o}\cos(\gamma t)$
@ -48,8 +49,7 @@ $=A\sin(\omega t+\phi)$
$y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$ $y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
assume $\gamma=\omega$ with zero resistance we get: 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.) $y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
#end of lec 13 #end of lec 13 #start of lecture 14
#start of lecture 14
# Amplitude modulation # Amplitude modulation
Last lecture we showed we can selectively listen to a specific signal by using resonance. 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: He has something else he's excited to show us; amplitude modulation! Lets recall from last lecture:
@ -74,7 +74,7 @@ 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! 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. >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 # Shortcut for solving DE of a mass-spring system
![[Drawing 2023-10-06 13.24.11.excalidraw]] ![[Drawing 2023-10-06 13.24.11.excalidraw]]
$my''+by'+ky=mg+F$ $my''+by'+ky=mg+F$
move $mg$ to LHS and replace $y$ with $y_{new}$ (remember, the $\frac{mg}{k}$ is a constant, its derivative is 0): move $mg$ to LHS and replace $y$ with $y_{new}$ (remember, the $\frac{mg}{k}$ is a constant, its derivative is 0):

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content/Separable equations (lec 1).md
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@ -44,4 +44,4 @@ $C=-1$
So, the answer is: So, the answer is:
$$y=\arctan(x^2+1)$$ $$y=\arctan(x^2+1)$$
#end of Lecture 1 #end of Lecture 1

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content/Variation of parameters (lec 9-10).md
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# Variation of parameters # Variation of parameters
#voparam
>[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: >[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:
## Variation of parameters is a method to solve: ## Variation of parameters is a method to solve:
## $$ay''+by'+cy=f(t)$$ ## $$ay''+by'+cy=f(t)$$
#voparam
Variation of parameters is generally known to be a more powerful replacement for method of undetermined coefficients.
First, find the homogenous solution: First, find the homogenous solution:
$y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
Now we need the particular solution, let $y_{p}$ be in the following form: Now we need the particular solution, let $y_{p}$ be in the following form:
@ -34,7 +35,8 @@ Alternatively, you could memorize the system of equations and solve for $v_{1}'$
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. 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 #ex #second_order #IVP #voparam #mouc
$y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$ Solve the IVP:
$$y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$$
Can we use undetermined coefficients? Yes and no. 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. 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.
First, find general solution to homogenous counterpart: First, find general solution to homogenous counterpart:
$y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$ $y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
@ -77,6 +79,7 @@ $$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin
#end of lecture 9 #start of lecture 10 #end of lecture 9 #start of lecture 10
#ex #second_order #voparam #mouc #ex #second_order #voparam #mouc
Find the general solution for:
$$y''-2y'+y=e^t\ln(t)+2\cos(t)$$ $$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
Find homogenous solution first: Find homogenous solution first:
$r^2-2r+1=0$ $r^2-2r+1=0$

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content/_index.md
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@ -10,11 +10,11 @@ I have written these notes for myself, I thought it would be cool to share them.
[Second order homogenous linear equations (lec 5-7)](second-order-homogenous-linear-equations-lec-5-7.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) [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) [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.) [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.) [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) [Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
[Laplace transform (lec 14)](laplace-transform-lec-14.html) [Laplace transform (lec 14)](laplace-transform-lec-14.html)
</br> </br>
[How to solve any DE, a flow chart](Solve-any-DE.png) [How to solve any DE, a flow chart](Solve-any-DE.png)
</br> </br>