fixed a bug that interprets *'s into bold/italic inside equations
This commit is contained in:
parent
3663c2779a
commit
5042277a35
|
@ -1,7 +1,7 @@
|
||||||
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 very important!*
|
<i>If it has a name in it, its very important, if it has 2 names, its very very important!</i>
|
||||||
#cauchy-euler equations are equations in the form:
|
#cauchy-euler equations are equations in the form:
|
||||||
$$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$$
|
$$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$$
|
||||||
where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
|
where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
|
||||||
|
@ -50,13 +50,13 @@ substitute: $y=x^r$
|
||||||
after calculating derivatives, plugging in, and simplifying we obtain the polynomial equation:
|
after calculating derivatives, plugging in, and simplifying we obtain the polynomial equation:
|
||||||
$ar^2+(b-a)r+C=0$
|
$ar^2+(b-a)r+C=0$
|
||||||
Three cases:
|
Three cases:
|
||||||
**(i)** $r_1\ne r_{2}$ then:
|
<b>(i)</b> $r_1\ne r_{2}$ then:
|
||||||
$y_{h}(t)=c_{1}e^{rt}+c_{2}e^{rt}$
|
$y_{h}(t)=c_{1}e^{rt}+c_{2}e^{rt}$
|
||||||
$y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$ (lousy notation, because the two $y_{h}$ do not equal each other)
|
$y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$ (lousy notation, because the two $y_{h}$ do not equal each other)
|
||||||
**(ii)** $r_{1}=r_{2}=r$ then:
|
<b>(ii)</b> $r_{1}=r_{2}=r$ then:
|
||||||
$y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$
|
$y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$
|
||||||
$y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$ (derived by reduction of order.)
|
$y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$ (derived by reduction of order.)
|
||||||
**(iii)** $r_{1,2}=\alpha\pm i\beta$ then:
|
<b>(iii)</b> $r_{1,2}=\alpha\pm i\beta$ then:
|
||||||
$y_{h}=e^\alpha(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
|
$y_{h}=e^\alpha(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
|
||||||
$y_{h}(x)=x^\alpha(c_{1}\cos(\beta\ln x)+c_{2}\sin(\beta \ln x))$
|
$y_{h}(x)=x^\alpha(c_{1}\cos(\beta\ln x)+c_{2}\sin(\beta \ln x))$
|
||||||
Now compute your particular solution, $y_{p}$, and combine with $y_{h}$ to obtain your general solution.
|
Now compute your particular solution, $y_{p}$, and combine with $y_{h}$ to obtain your general solution.
|
|
@ -104,7 +104,7 @@ $I_{1}=3-2e^{-5t}-e^{-20t}-u(t-1)(3-2e^{-5(t-1)}-e^{-20(t-1)})$
|
||||||
whats $I_{3}$? use: $-10I_{1}+10I_{3}+I_{3}'=0$ (i think)
|
whats $I_{3}$? use: $-10I_{1}+10I_{3}+I_{3}'=0$ (i think)
|
||||||
$J_{3}=\frac{10}{s+10}J_{1}$
|
$J_{3}=\frac{10}{s+10}J_{1}$
|
||||||
$J_{3}=300 \frac{1}{s(s+5)(s+20)}(1-e^{ -s })$
|
$J_{3}=300 \frac{1}{s(s+5)(s+20)}(1-e^{ -s })$
|
||||||
partial fraction it so we can eventuall take the inverse LT:
|
partial fraction it so we can eventually take the inverse LT:
|
||||||
skip some steps:
|
skip some steps:
|
||||||
$J_{3}=\left( \frac{3}{s}-\frac{4}{s+5}+\frac{1}{s+20} \right)(1-e^{ -s })$
|
$J_{3}=\left( \frac{3}{s}-\frac{4}{s+5}+\frac{1}{s+20} \right)(1-e^{ -s })$
|
||||||
$I_{3}=3-4e^{ -5t }+e^{ -20t }-(3-4e^{ -5(t-1) }+e^{ -20(t-1) })u(t-1)$
|
$I_{3}=3-4e^{ -5t }+e^{ -20t }-(3-4e^{ -5(t-1) }+e^{ -20(t-1) })u(t-1)$
|
||||||
|
|
|
@ -48,7 +48,7 @@ $\phi$ is the phase shift in radians.
|
||||||
$\frac{\omega}{2\pi}$ is the natural frequency (oscillations/second).
|
$\frac{\omega}{2\pi}$ is the natural frequency (oscillations/second).
|
||||||
Taking the inverse, $\frac{2\pi}{\omega}$ is the period (seconds/oscillation).
|
Taking the inverse, $\frac{2\pi}{\omega}$ is the period (seconds/oscillation).
|
||||||
This behavior is all a result of classical mechanics, but rather beautifully, electronic circuits composed of resistors, inductors, and capacitors can also be described by this equation. You'll see it in biology too.
|
This behavior is all a result of classical mechanics, but rather beautifully, electronic circuits composed of resistors, inductors, and capacitors can also be described by this equation. You'll see it in biology too.
|
||||||
*"... Nature is beautiful and harmonic."* -Prof (not an exact quotation, I didn't catch what he precisely said.)
|
<i>"... Nature is beautiful and harmonic."</i> -Prof (not an exact quotation, I didn't catch what he precisely said.)
|
||||||
btw we know $A=\sqrt{ c_{1}^2+c_{2}^2 }$
|
btw we know $A=\sqrt{ c_{1}^2+c_{2}^2 }$
|
||||||
and $\tan \phi=\frac{c_{1}}{c_{2}}$
|
and $\tan \phi=\frac{c_{1}}{c_{2}}$
|
||||||
so we can compute $A$ and $\phi$ from $c_1$ and $c_2$.
|
so we can compute $A$ and $\phi$ from $c_1$ and $c_2$.
|
||||||
|
|
|
@ -1,5 +1,5 @@
|
||||||
|
>
|
||||||
*The world is non-linear, many solutions, many paths to the solution. It's why linear equations play so nice. We just look down it's path and we will know that it's a straight line for eternity.*
|
<i>The world is non-linear, many solutions, many paths to the solution. It's why linear equations play so nice. We just look down it's path and we will know that it's a straight line for eternity.</i>
|
||||||
|
|
||||||
# Linear equation:
|
# Linear equation:
|
||||||
$$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$
|
$$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$
|
||||||
|
|
|
@ -6,9 +6,9 @@ in the case when $f(t)=0$ :
|
||||||
1) $ay''+b'y+cy=0$
|
1) $ay''+b'y+cy=0$
|
||||||
then $ar^2+br+c=0$ and solve with quadratic formula
|
then $ar^2+br+c=0$ and solve with quadratic formula
|
||||||
general solutions are:
|
general solutions are:
|
||||||
if $r_{1}\ne r_{2}\Rightarrow y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ <- **overdamped**
|
if $r_{1}\ne r_{2}\Rightarrow y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ <- <u>overdamped</u>
|
||||||
if $r_{1}=r_{2}\Rightarrow y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ <- **critically damped**
|
if $r_{1}=r_{2}\Rightarrow y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ <- <u>critically damped</u>
|
||||||
if $r_{1,2}\in \mathbb{C}\Rightarrow y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$ <- **underdamped**
|
if $r_{1,2}\in \mathbb{C}\Rightarrow y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$ <- <u>underdamped</u>
|
||||||
where h means homogenous, (when $f(t)=0$ the equation is homogenous.)
|
where h means homogenous, (when $f(t)=0$ the equation is homogenous.)
|
||||||
|
|
||||||
But what about the case when $f(t)\ne 0$ ?
|
But what about the case when $f(t)\ne 0$ ?
|
||||||
|
@ -92,10 +92,10 @@ final solution to IVP:
|
||||||
$$y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$$
|
$$y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$$
|
||||||
|
|
||||||
#end of lec 8 #start of lec 9
|
#end of lec 8 #start of lec 9
|
||||||
*remember in a previous example when we had to guess that $y_{p}=At^2e^{2t}$? Here is a generalized algorithm that can find $y_{p}$ when the RHS falls under the following form. Reducing the guess work to zero:*
|
<i>remember in a previous example when we had to guess that $y_{p}=At^2e^{2t}$? Here is a generalized algorithm that can find $y_{p}$ when the RHS falls under the following form. Reducing the guess work to zero:</i>
|
||||||
# Generalized guesses for undetermined coefficients:
|
# Generalized guesses for undetermined coefficients:
|
||||||
case i) $ay''+by'+cy=P_m(t)e^{rt}$
|
case i) $ay''+by'+cy=P_m(t)e^{rt}$
|
||||||
where $P_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$ *i.e. P is a polynomial degree m.*
|
where $P_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$ <i>i.e. P is a polynomial degree m.</i>
|
||||||
Then we guess the particular solution is of the form: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
|
Then we guess the particular solution is of the form: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
|
||||||
where:
|
where:
|
||||||
s=0, if r is not a root,
|
s=0, if r is not a root,
|
||||||
|
|
|
@ -6,6 +6,7 @@ Heat equation not only describes thermodynamics but it can also model the diffus
|
||||||
Strikingly, it can also model option prices in the stock market. However, using it as a strategy to make money is not so simple, because if it worked then everyone would try to use it to make money, which would cause the overall strategy to be less effective as the option prices start to get priced to accommodate for the prediction (🤯).
|
Strikingly, it can also model option prices in the stock market. However, using it as a strategy to make money is not so simple, because if it worked then everyone would try to use it to make money, which would cause the overall strategy to be less effective as the option prices start to get priced to accommodate for the prediction (🤯).
|
||||||
![[Drawing 2023-11-08 13.07.19.excalidraw.png]]
|
![[Drawing 2023-11-08 13.07.19.excalidraw.png]]
|
||||||
>I'm sorry the image doesn't display properly :( I'm trying to get images to work on my notes. For now you can see the relevant .png files in the github repo under content/drawings/
|
>I'm sorry the image doesn't display properly :( I'm trying to get images to work on my notes. For now you can see the relevant .png files in the github repo under content/drawings/
|
||||||
|
|
||||||
We assume that the tube is perfectly insulating along its surface, this helps reduce the problem into a one dimensional problem. Heat can only travel inside and along the x axis.
|
We assume that the tube is perfectly insulating along its surface, this helps reduce the problem into a one dimensional problem. Heat can only travel inside and along the x axis.
|
||||||
Fourier figured out that:
|
Fourier figured out that:
|
||||||
$\text{Heat flux} = -k(x)a\frac{\partial u}{\partial x}(t,x) \Delta t$
|
$\text{Heat flux} = -k(x)a\frac{\partial u}{\partial x}(t,x) \Delta t$
|
||||||
|
|
|
@ -1,3 +1,4 @@
|
||||||
|
|
||||||
#start of lec 19
|
#start of lec 19
|
||||||
This lecture we will learn about periodic functions, specifically, non-sinusoidal periodic functions.
|
This lecture we will learn about periodic functions, specifically, non-sinusoidal periodic functions.
|
||||||
# Periodic function
|
# Periodic function
|
||||||
|
|
|
@ -149,18 +149,18 @@ $a_{4}=\frac{2^2}{4!}a_{1}$
|
||||||
the pattern leads us to:
|
the pattern leads us to:
|
||||||
$a_{3k+1}=\frac{(2\cdot 5 \dots(3k-1))^2}{(3k+1)!}$ where k=1,2,3, ...
|
$a_{3k+1}=\frac{(2\cdot 5 \dots(3k-1))^2}{(3k+1)!}$ where k=1,2,3, ...
|
||||||
$a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ k=1,2,...
|
$a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ k=1,2,...
|
||||||
$z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1*4*\dots(3k-2))^2}{(3k)!}x^{3k} \right)$
|
$z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1\cdot4\cdot\dots(3k-2))^2}{(3k)!}x^{3k} \right)$
|
||||||
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$
|
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$
|
||||||
there we go, $z$ is a linear combination of those two expressions
|
there we go, $z$ is a linear combination of those two expressions
|
||||||
class done at 1:56 (a lil late but the journey is worth it)
|
class done at 1:56 (a lil late but the journey is worth it)
|
||||||
#end of lec 23 #start of lec 24
|
#end of lec 23 #start of lec 24
|
||||||
*midterms have been marked and returned today.*
|
<i>midterms have been marked and returned today.</i>
|
||||||
|
|
||||||
we consider:
|
we consider:
|
||||||
$$y''+p(x)y'+q(x)y=0$$
|
$$y''+p(x)y'+q(x)y=0$$
|
||||||
this is in standard form, it's a second order linear equation
|
this is in standard form, it's a second order linear equation
|
||||||
Definition:
|
Definition:
|
||||||
if $p(x)$ and $q(x)$ are **analytic** functions in a vicinity of $x_{0}$ then $x_0$ is **ordinary**. Otherwise, $x_{0}$ is **singular**.
|
if $p(x)$ and $q(x)$ are <u>analytic</u> functions in a vicinity of $x_{0}$ then $x_0$ is <u>ordinary</u>. Otherwise, $x_{0}$ is <u>singular</u>.
|
||||||
we expect that the solution y can be represented by a power series. This is true according to the following theorem:
|
we expect that the solution y can be represented by a power series. This is true according to the following theorem:
|
||||||
Theorem: If $x_{0}$ is ordinary point then the differential equation above has two linearly independent solution of the form $\sum_{n=0} ^\infty a_{n}(x-x_{0})^n, \qquad\sum_{n=0}^\infty b_{n}(x-x_{0})^n$.
|
Theorem: If $x_{0}$ is ordinary point then the differential equation above has two linearly independent solution of the form $\sum_{n=0} ^\infty a_{n}(x-x_{0})^n, \qquad\sum_{n=0}^\infty b_{n}(x-x_{0})^n$.
|
||||||
The radius of convergence for them is at least as large as the distance between $x_{0}$ and the closest singular point (which can be real or complex).
|
The radius of convergence for them is at least as large as the distance between $x_{0}$ and the closest singular point (which can be real or complex).
|
||||||
|
@ -173,7 +173,7 @@ put it in standard form:
|
||||||
$y''-\frac{3xy'}{x+1}+\frac{2y}{x+1}=0$
|
$y''-\frac{3xy'}{x+1}+\frac{2y}{x+1}=0$
|
||||||
the only singular point for this equation is $x=-1$
|
the only singular point for this equation is $x=-1$
|
||||||
so the minimum value of radius convergence is $\rho=2$ (distance between -1 and x_0)
|
so the minimum value of radius convergence is $\rho=2$ (distance between -1 and x_0)
|
||||||
we are guaranteed that the power series will converge *at least* in $(-1,3)$, possibly more. You can try solving for y as a power series.
|
we are guaranteed that the power series will converge <i>at least</i> in $(-1,3)$, possibly more. You can try solving for y as a power series.
|
||||||
|
|
||||||
#ex
|
#ex
|
||||||
$$y''-\tan xy'+y=0 \quad x_{0}=0$$
|
$$y''-\tan xy'+y=0 \quad x_{0}=0$$
|
||||||
|
|
|
@ -1,5 +1,5 @@
|
||||||
most of these "models" in EE are based on these DE. You'll see how important DE are in chemical, electrical, mechanical, engphys, civil (very important for civil!), (mining? idk what's in mining :D -prof)
|
most of these "models" in EE are based on these DE. You'll see how important DE are in chemical, electrical, mechanical, engphys, civil (very important for civil!), (mining? idk what's in mining :D -prof)
|
||||||
*DE are important* -prof
|
<i>DE's are important</i> -prof
|
||||||
## Second order linear equations
|
## Second order linear equations
|
||||||
Second order equations arise from very simple problems many engineers face, for instance a pendulum can be described by a second order equation.
|
Second order equations arise from very simple problems many engineers face, for instance a pendulum can be described by a second order equation.
|
||||||
#second_order
|
#second_order
|
||||||
|
|
|
@ -1,6 +1,6 @@
|
||||||
|
|
||||||
#start of lec 17
|
#start of lec 17
|
||||||
*Lecture starts with a 25 minute long midterm review.*
|
<i>Lecture starts with a 25 minute long midterm review.</i>
|
||||||
He spends most of the time giving us a summary of all the equations (separable, linear, homogenous, exact, ...) The review is available on the eclass page.
|
He spends most of the time giving us a summary of all the equations (separable, linear, homogenous, exact, ...) The review is available on the eclass page.
|
||||||
|
|
||||||
So, why did we learn all this stuff about Laplace transforms? We will now see how its useful:
|
So, why did we learn all this stuff about Laplace transforms? We will now see how its useful:
|
||||||
|
|
|
@ -30,6 +30,10 @@
|
||||||
{{ end }}
|
{{ end }}
|
||||||
|
|
||||||
{{ $content := .Scratch.Get "content" }}
|
{{ $content := .Scratch.Get "content" }}
|
||||||
|
|
||||||
|
{{/* $content := $content | replaceRE "(\\$.*)\\*(.*\\$)" "$1$2" */}}
|
||||||
|
{{ $content := $content | replaceRE "\\*" "\\*"}}
|
||||||
|
<!-- The above command replaces * with \* so that *'s inside equations dont get interpreted as emphasis or bold tags. Needed for the page on convolutions.>
|
||||||
{{ $content := $content | replaceRE "\n" "\n\n" }}
|
{{ $content := $content | replaceRE "\n" "\n\n" }}
|
||||||
{{ $content := $content | replaceRE $matrixBreak "\\\\\\\\\\\\\\\\" }}
|
{{ $content := $content | replaceRE $matrixBreak "\\\\\\\\\\\\\\\\" }}
|
||||||
<!--The above command replaces a \\ in content/*.md files with a \\\\ yes it looks silly-->
|
<!--The above command replaces a \\ in content/*.md files with a \\\\ yes it looks silly-->
|
||||||
|
|
Loading…
Reference in New Issue