594 lines
14 KiB
HTML
594 lines
14 KiB
HTML
<!DOCTYPE html>
|
|
<html lang="en">
|
|
|
|
<head>
|
|
<meta charset="utf-8" />
|
|
<meta http-equiv="X-UA-Compatible" content="IE=edge">
|
|
<meta name="viewport" content="width=device-width, initial-scale=1">
|
|
<meta name="description" content="#start of lecture 2
|
|
2 new tricks: homogenous and linear DE
|
|
Homogenous equations:
|
|
$\frac{dy}{dt}=f\left( \frac{y}{t} \right)$ (I’m calling this #de_h_type1)
|
|
let $u=\frac{y}{t}$ $y=tu \quad \frac{dy}{dt}=u+t\frac{du}{dt}$
|
|
so $\frac{dy}{dt}=f(u)=u+t{\frac{du}{dt}}$
|
|
The homogenous equation has …">
|
|
<meta name="apple-mobile-web-app-capable" content="yes">
|
|
<meta name="mobile-web-app-capable" content="yes">
|
|
<meta name="apple-mobile-web-app-status-bar-style" content="default">
|
|
|
|
|
|
<link rel="manifest" href="./manifest.json"><meta property="og:title" content="" />
|
|
<meta property="og:description" content="#start of lecture 2
|
|
2 new tricks: homogenous and linear DE Homogenous equations: $\frac{dy}{dt}=f\left( \frac{y}{t} \right)$ (I’m calling this #de_h_type1) let $u=\frac{y}{t}$ $y=tu \quad \frac{dy}{dt}=u+t\frac{du}{dt}$ so $\frac{dy}{dt}=f(u)=u+t{\frac{du}{dt}}$ The homogenous equation has been converted into a separable DE! $\frac{du}{dt}=\frac{f(u)-u}{t}$ $\frac{du}{f(u)-u}=\frac{dt}{t}$
|
|
Another way you can write a homogenous equation: $\frac{dy}{dx}=G(ax+by)\quad \text{where a, b }\in \mathbb{R}$ (I’m calling this #de_h_type2) Then, let $u=ax+by$ $\frac{du}{dx}=a+b{\frac{dy}{dx}}$ $\frac{dy}{dx}=\frac{1}{b}{\frac{du}{dx}}-\frac{a}{b}=G(u)$ Again, the homogenous equation has been converted to a separable DE!" />
|
|
<meta property="og:type" content="article" />
|
|
<meta property="og:url" content="/homogenous-de-lec-2.html" />
|
|
|
|
<meta name="twitter:card" content="summary"/>
|
|
<meta name="twitter:title" content=""/>
|
|
<meta name="twitter:description" content="#start of lecture 2
|
|
2 new tricks: homogenous and linear DE Homogenous equations: $\frac{dy}{dt}=f\left( \frac{y}{t} \right)$ (I’m calling this #de_h_type1) let $u=\frac{y}{t}$ $y=tu \quad \frac{dy}{dt}=u+t\frac{du}{dt}$ so $\frac{dy}{dt}=f(u)=u+t{\frac{du}{dt}}$ The homogenous equation has been converted into a separable DE! $\frac{du}{dt}=\frac{f(u)-u}{t}$ $\frac{du}{f(u)-u}=\frac{dt}{t}$
|
|
Another way you can write a homogenous equation: $\frac{dy}{dx}=G(ax+by)\quad \text{where a, b }\in \mathbb{R}$ (I’m calling this #de_h_type2) Then, let $u=ax+by$ $\frac{du}{dx}=a+b{\frac{dy}{dx}}$ $\frac{dy}{dx}=\frac{1}{b}{\frac{du}{dx}}-\frac{a}{b}=G(u)$ Again, the homogenous equation has been converted to a separable DE!"/>
|
|
|
|
|
|
|
|
|
|
|
|
<title>Homogenous DE (lec 2) - My New Hugo Site</title>
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
<link rel="stylesheet" href="./css/main.min.203106d73d4370d04c60441691746dd8e021e38bbbc83f65f636dc8ae886a9f3.css" />
|
|
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script>
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
<script src="./js/main.min.2dd2f7073384163751d1886bcb921097bc2af8ec60cb37deebf49f61a0eca5c3.js" integrity="sha256-LdL3BzOEFjdR0Yhry5IQl7wq+Oxgyzfe6/SfYaDspcM="></script>
|
|
|
|
|
|
|
|
</head>
|
|
|
|
<body>
|
|
|
|
|
|
<style>
|
|
search-menu {
|
|
display: block;
|
|
}
|
|
|
|
#search {
|
|
height: 100%;
|
|
width: 0;
|
|
position: fixed;
|
|
background: var(--background-search);
|
|
z-index: 1;
|
|
top: 0;
|
|
left: 0;
|
|
border-right: 1px solid var(--separator-color);
|
|
overflow-x: hidden;
|
|
overflow-y: auto;
|
|
opacity: 0;
|
|
|
|
-ms-overflow-style: none;
|
|
scrollbar-width: none;
|
|
}
|
|
|
|
#search::-webkit-scrollbar { display: none; }
|
|
|
|
#search-header {
|
|
padding: 12px;
|
|
position: fixed;
|
|
padding-left: 12px;
|
|
padding-right: 12px;
|
|
background: var(--background-search);
|
|
width: 250px;
|
|
opacity: 1;
|
|
height: 50px;
|
|
z-index: 2;
|
|
border-bottom: 1px solid var(--separator-color);
|
|
}
|
|
|
|
#search .input-container {
|
|
position: relative
|
|
}
|
|
|
|
#search-input {
|
|
width: 100%;
|
|
height: 24px;
|
|
border: 1px solid var(--separator-color);
|
|
border-radius: 4px;
|
|
padding-left: 16px;
|
|
background-color: white;
|
|
display: inline-block;
|
|
}
|
|
|
|
#search-input:focus {
|
|
border: 1px solid var(--search-field-focused-color);
|
|
}
|
|
|
|
#search-header .input-container .search-icon {
|
|
position: absolute;
|
|
top: 6px;
|
|
left: 8px;
|
|
fill: darkGray;
|
|
}
|
|
|
|
#search-results img {
|
|
width: 122px;
|
|
height: 76px;
|
|
border: 1px solid var(--separator-color);
|
|
object-fit: cover;
|
|
}
|
|
|
|
#search-results {
|
|
margin-top: 50px;
|
|
overflow: auto;
|
|
height: 100%;
|
|
}
|
|
|
|
#search-results a {
|
|
width: 100%;
|
|
padding-left: 25px;
|
|
padding-right: 25px;
|
|
padding-top: 12px;
|
|
padding-bottom: 12px;
|
|
display: inline-block;
|
|
|
|
color: var(--text-base-color);
|
|
border-bottom: 1px solid var(--separator-color);
|
|
border-left: 6px solid var(--background-search);
|
|
|
|
}
|
|
|
|
#search-results a:first-child:hover, a:first-child:focus, .selected {
|
|
outline: 0;
|
|
background-color: var(--note-table-cell-selected-color);
|
|
border-left: 6px solid var(--note-table-cell-ribbon-color) !important;
|
|
}
|
|
|
|
|
|
#search-results li { text-indent: 0; }
|
|
#search-results li:before,
|
|
#search-results h1:before,
|
|
#search-results h2:before,
|
|
#search-results h3:before,
|
|
#search-results h4:before,
|
|
#search-results h5:before,
|
|
#search-results h6:before {
|
|
content: "";
|
|
visibility: hidden;
|
|
display: none;
|
|
}
|
|
</style>
|
|
|
|
|
|
<search-menu id="search" data-turbolinks-permanent>
|
|
<header id="search-header">
|
|
<div class="input-container">
|
|
<svg aria-hidden="true" style="" class="search-icon" width="12" height="12" viewBox="0 0 18 18">
|
|
<path d="M18 16.5l-5.14-5.18h-.35a7 7 0 10-1.19 1.19v.35L16.5 18l1.5-1.5zM12 7A5 5 0 112 7a5 5 0 0110 0z">
|
|
</path>
|
|
</svg>
|
|
|
|
<input type="search" autocomplete="off" id="search-input" onkeyup="performSearch()" tabindex="0" placeholder="Search note">
|
|
|
|
|
|
|
|
</div>
|
|
</header>
|
|
|
|
<ul id="search-results"></ul>
|
|
</search-menu>
|
|
<script>
|
|
</script>
|
|
|
|
<style>
|
|
#toolbar {
|
|
position: fixed;
|
|
top: 0;
|
|
right: 0;
|
|
|
|
width: 60px;
|
|
height: 100%;
|
|
|
|
display: flex;
|
|
flex-direction: column;
|
|
justify-content: flex-start;
|
|
align-items: center;
|
|
|
|
transition: 1s;
|
|
opacity: 0.5;
|
|
|
|
padding: 18px 0px 18px 0px;
|
|
}
|
|
|
|
#toolbar:hover {
|
|
opacity: 1;
|
|
}
|
|
|
|
#close-nav-icon {
|
|
display: none;
|
|
}
|
|
|
|
</style>
|
|
|
|
<aside id="toolbar">
|
|
<span style="cursor:pointer" id="open-nav-icon" onclick="handleNavVisibility()">
|
|
<svg width="18" height="18" viewBox="0 0 20 20" xmlns="http://www.w3.org/2000/svg"><circle fill="none" stroke="var(--text-base-color)" stroke-width="1.1" cx="9" cy="9" r="7"></circle><path fill="none" stroke="var(--text-base-color)" stroke-width="1.1" d="M14,14 L18,18 L14,14 Z"></path></svg>
|
|
</span>
|
|
|
|
<span onclick="imageMode()" style="cursor:pointer;margin-top:16px;">
|
|
<svg width="20" height="20" viewBox="0 0 20 20" xmlns="http://www.w3.org/2000/svg"><circle cx="16.1" cy="6.1" r="1.1"></circle><rect fill="none" stroke="var(--text-base-color" x=".5" y="2.5" width="19" height="15"></rect><polyline fill="none" stroke="var(--text-base-color" stroke-width="1.01" points="4,13 8,9 13,14"></polyline><polyline fill="none" stroke="var(--text-base-color)" stroke-width="1.01" points="11,12 12.5,10.5 16,14"></polyline></svg>
|
|
</span>
|
|
</aside>
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
<main id="main">
|
|
|
|
<div id="note-wrapper" class="note-wrapper">
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
<p>#start of lecture 2</p>
|
|
<h1 id="2-new-tricks-homogenous-and-linear-de">2 new tricks: homogenous and linear DE</h1>
|
|
<h2 id="homogenous-equations">Homogenous equations:</h2>
|
|
<p>$\frac{dy}{dt}=f\left( \frac{y}{t} \right)$ (I’m calling this <a class="hashtag" onclick="focusTag(this)">de_h_type1)</a>
|
|
let $u=\frac{y}{t}$ $y=tu \quad \frac{dy}{dt}=u+t\frac{du}{dt}$
|
|
so $\frac{dy}{dt}=f(u)=u+t{\frac{du}{dt}}$
|
|
The homogenous equation has been converted into a separable DE!
|
|
$\frac{du}{dt}=\frac{f(u)-u}{t}$
|
|
$\frac{du}{f(u)-u}=\frac{dt}{t}$</p>
|
|
<h2 id="another-way-you-can-write-a-homogenous-equation">Another way you can write a homogenous equation:</h2>
|
|
<p>$\frac{dy}{dx}=G(ax+by)\quad \text{where a, b }\in \mathbb{R}$ (I’m calling this <a class="hashtag" onclick="focusTag(this)">de_h_type2)</a>
|
|
Then, let $u=ax+by$
|
|
$\frac{du}{dx}=a+b{\frac{dy}{dx}}$
|
|
$\frac{dy}{dx}=\frac{1}{b}{\frac{du}{dx}}-\frac{a}{b}=G(u)$
|
|
Again, the homogenous equation has been converted to a separable DE!
|
|
$dx=\frac{du}{b{G(u)+\frac{a}{b}}}$
|
|
Just integrate both sides as usual and you’re chilling.</p>
|
|
<h2 id="examples-of-homogenous-equations">Examples of homogenous equations:</h2>
|
|
<h3 id="heading"></h3>
|
|
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
|
|
<a class="hashtag" onclick="focusTag(this)">de_h_type1</a>:$\frac{dy}{dx}=\frac{{x+y}}{x-y} \quad x>y\quad\text{This condition is added so the denominator}\ne 0$</p>
|
|
<p>but $\frac{{x+y}}{x-y}\ne f(\frac{y}{x})$… Or is it? How can this be written as a homogenous equation?
|
|
divide the top and bottom by x:
|
|
$\frac{dy}{dx}=\frac{{1+\frac{y}{x}}}{1-\frac{y}{x}}$
|
|
Yay! now it’s a function of $\frac{y}{x}$
|
|
let $u=\frac{y}{x} \quad \frac{dy}{dx}=u+x{\frac{du}{dx}}$
|
|
$\frac{dy}{dx}=\frac{1+u}{1-u}=u+x{\frac{du}{dx}}$
|
|
$\frac{dx({f(u)-u})}{x}=du$
|
|
$\frac{dx}{x}=\frac{du}{{f(u)-u}}$</p>
|
|
<blockquote>
|
|
<p>That’s odd, why is it not $\frac{du}{f(u)-u}=\frac{x}{dx}$? I got this by moving the top over.
|
|
(it’s because you must move all multiplicative factors when using this technique of moving the top. Be careful!)</p>
|
|
</blockquote>
|
|
<p>$\int\frac{dx}{x}=\int\frac{du}{{f(u)-u}}$
|
|
$\ln\mid x\mid=\int \frac{du}{\frac{{1+u}}{1-u}-u}$
|
|
$\ln\mid x\mid=\int \frac{du}{\frac{{1+u-u+u^2}}{1-u}}$
|
|
$\ln\mid x\mid=\int \frac{1-u}{{1+u^2}}du$</p>
|
|
<blockquote>
|
|
<p>let $1+u^2=v \quad dv=2udu$
|
|
$=\int \frac{{1-u}}{v} , du$ Gah, doesn’t work. I didn’t notice I could split the integral up first.</p>
|
|
</blockquote>
|
|
<p>$\ln\mid x\mid=\int \frac{1}{{1+u^2}},du-\int \frac{u}{1+u^2} , du=\arctan\left( \frac{y}{x} \right)+C-I_{0}$
|
|
for $I_{0}$ let $v=1+u^2 \quad dv=2udu$
|
|
$I_{0}=\int \frac{u}{v} , \frac{dv}{2u}=\frac{1}{2}\int \frac{dv}{v}=\frac{1}{2}\ln(1+u^2)$</p>
|
|
<blockquote>
|
|
<p>^Note no abs value needed in the $\ln()$ as $1+u^2$ is always +</p>
|
|
</blockquote>
|
|
<p>$\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln(1+u^2)$
|
|
$\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln\left( 1+\frac{y^2}{x^2} \right)$
|
|
$\mid x\mid=e^{\arctan(\frac{y}{x})+C-\ln(\sqrt{ 1+y^2/x^2 })}$
|
|
$x=\frac{e^{\arctan(y/x)}A}{\sqrt{ 1+\frac{y^2}{x^2} }}$
|
|
$x\sqrt{ 1+\frac{y^2}{x^2}} ={e^{\arctan(y/x)}A}$
|
|
So the final general solution to the problem is:</p>
|
|
<h4 id="sqrt-x2y2-earctanleft-fracyx-righta">$$\sqrt{ x^2+y^2 }=e^{\arctan\left( \frac{y}{x} \right)}A$$</h4>
|
|
<hr>
|
|
<h3 id="heading-1"></h3>
|
|
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
|
|
<a class="hashtag" onclick="focusTag(this)">de_h_type2</a>: $$(2x-2y-1)dx+(x-y+1)dy=0$$
|
|
Can we write it in the form $\frac{dy}{dx}=G(ax+by)$?
|
|
$(x-y+1)dy=-(2x-2y-1)dx$
|
|
$\frac{dy}{dx}=\frac{{2y+1-2x}}{x-y+1}$
|
|
factor out a -2?
|
|
$\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$
|
|
Yep! looks like a
|
|
<a class="hashtag" onclick="focusTag(this)">de_h_type2</a>
|
|
let $u=x-y$
|
|
$\frac{du}{dx}=1-\frac{dy}{dx}$
|
|
$1-\frac{du}{dx}=\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$</p>
|
|
<blockquote>
|
|
<p>Obviously we don’t work with x and y as I was entailing above, substitute $u=x-y$ in you silly goose.</p>
|
|
</blockquote>
|
|
<p>$1-\frac{du}{dx}=-2\frac{{u-\frac{1}{2}}}{u+1}$
|
|
$\frac{du}{dx}=2\frac{{u-\frac{1}{2}}}{u+1}+1$
|
|
$\frac{du}{dx}=\frac{2u-1}{u+1}+1$
|
|
$\frac{du}{dx}=\frac{{2u-1+u+1}}{u+1}$
|
|
$\frac{du}{dx}=\frac{3u}{u+1}$
|
|
$\frac{(u+1)du}{3u}=dx$
|
|
$\int \frac{(u+1)du}{3u}=\int dx$</p>
|
|
<blockquote>
|
|
<p>$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=\ln\mid x\mid+C$
|
|
Ah, I made a mistake. $\int dx \ne \ln\mid x\mid+C$</p>
|
|
</blockquote>
|
|
<p>$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=x+C$</p>
|
|
<blockquote>
|
|
<p>Okay, now that we have integrated, we can start talking in terms of x and y again</p>
|
|
</blockquote>
|
|
<p>$\frac{x-y}{3}+\frac{1}{3}\ln\mid x-y\mid = x+C$
|
|
$x-y+\ln\mid x-y\mid=3x+C$
|
|
$\ln\mid x-y\mid=C+y+2x$ < this is where he moved the C to the left
|
|
$\mid x-y\mid=e^Ce^ye^{2x}$
|
|
$x-y=Ae^ye^{2x}$
|
|
$A(x-y)=e^{y+2x}$</p>
|
|
<blockquote>
|
|
<p>I know that above step looks illegal, but the prof did this (indirectly, he moved C to the LHS in a prior step without regarding it’s sign). I wonder what happens if A was 0 though? Do we get divide by zero errors? Thinking about it more, we are changing $x-y=0$ to $e^{y+2x}=0$ when $A=0$ The first one has a solution (y=x) the second loses that solution because of ln(0) issues (gives a function that’s undefined for all x). when checking y(x)=x in the DE, it is a valid solution. So it is an illegal step! Because we lost a valid solution. I’ll have to check with the prof.
|
|
Interestingly, if we act like $e^{y+2x}=0$ is defined, we get $\frac{dy}{dx}=-2$</p>
|
|
</blockquote>
|
|
<p>Proof:
|
|
$\lim_{ n \to 0 }e^{y+2x}=n$
|
|
$\lim_{ n \to 0 }\ln(n)=y+2x$
|
|
$\lim_{ n \to 0 }\frac{d}{dx}\ln(n)=0=\frac{dy}{dx}+2$
|
|
$\frac{dy}{dx}=-2$</p>
|
|
<blockquote>
|
|
</blockquote>
|
|
<p>so from $\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$ we get:
|
|
$-2=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$
|
|
$x-y+1=x-y-\frac{1}{2}$
|
|
$1=-\frac{1}{2}$
|
|
So what does this all mean? I honestly have no idea. I think it means we assumed that $e^{y+2x}=0$ is defined and because we arrived at a contradiction, our assumption was wrong. That didn’t really get us to show if it was a valid solution or not like I imagined.</p>
|
|
<p>We can rearrange to our liking, but we have found the general solution to the DE:</p>
|
|
<h4 id="x-yae2xy">$$x-y=Ae^{2x+y}$$</h4>
|
|
<hr>
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
<h3>Referenced in</h3>
|
|
|
|
<ul>
|
|
<li>No backlinks found</li>
|
|
|
|
</ul>
|
|
|
|
|
|
|
|
</div>
|
|
</main>
|
|
|
|
<script type="text/javascript">
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
</script>
|
|
|
|
</body>
|
|
</html>
|