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@ -13,5 +13,3 @@ I have written these notes for myself, I thought it would be cool to share them.
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</br>
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[How to solve any DE, a flow chart](Solve-any-DE.png)
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</br>
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I'd like to add a search by tag feature. I'm also thinking of hosting the source code for all this on a git server. That way, people can contribute and fix my notes for me :P
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It would also allow people to contribute or host their own notes, which would be pretty cool. (Side note: I am against sharing instructor materials without their approved consent.)
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public/404.html
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@ -1,396 +0,0 @@
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<meta charset="utf-8" />
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<meta http-equiv="X-UA-Compatible" content="IE=edge">
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<meta name="viewport" content="width=device-width, initial-scale=1">
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<meta name="description" content="">
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<meta name="apple-mobile-web-app-capable" content="yes">
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<meta name="mobile-web-app-capable" content="yes">
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<meta name="apple-mobile-web-app-status-bar-style" content="default">
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<link rel="manifest" href="./manifest.json"><meta property="og:title" content="404 Page not found" />
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<meta property="og:description" content="" />
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<meta property="og:type" content="website" />
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<meta property="og:url" content="/404.html" />
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<meta name="twitter:card" content="summary"/>
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<meta name="twitter:title" content="404 Page not found"/>
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<meta name="twitter:description" content=""/>
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<title> - My New Hugo Site</title>
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<link rel="stylesheet" href="./css/main.min.203106d73d4370d04c60441691746dd8e021e38bbbc83f65f636dc8ae886a9f3.css" />
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<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script>
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<script src="./js/main.min.2dd2f7073384163751d1886bcb921097bc2af8ec60cb37deebf49f61a0eca5c3.js" integrity="sha256-LdL3BzOEFjdR0Yhry5IQl7wq+Oxgyzfe6/SfYaDspcM="></script>
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</head>
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<body>
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<style>
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}
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#search-input:focus {
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border: 1px solid var(--search-field-focused-color);
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}
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#search-header .input-container .search-icon {
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position: absolute;
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top: 6px;
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border: 1px solid var(--separator-color);
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}
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#search-results {
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margin-top: 50px;
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#search-results li { text-indent: 0; }
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#search-results li:before,
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#search-results h1:before,
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#search-results h2:before,
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#search-results h3:before,
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#search-results h4:before,
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#search-results h5:before,
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#search-results h6:before {
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display: none;
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</style>
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<search-menu id="search" data-turbolinks-permanent>
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<header id="search-header">
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<div class="input-container">
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<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">
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</path>
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</svg>
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<input type="search" autocomplete="off" id="search-input" onkeyup="performSearch()" tabindex="0" placeholder="Search note">
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</div>
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</header>
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<ul id="search-results"></ul>
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</search-menu>
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<script>
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</script>
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<style>
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#toolbar {
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position: fixed;
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height: 100%;
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display: flex;
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flex-direction: column;
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justify-content: flex-start;
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transition: 1s;
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padding: 18px 0px 18px 0px;
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</style>
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<aside id="toolbar">
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<span style="cursor:pointer" id="open-nav-icon" onclick="handleNavVisibility()">
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<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>
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<span onclick="imageMode()" style="cursor:pointer;margin-top:16px;">
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</aside>
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<main id="main">
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<div id="note-wrapper" class="note-wrapper">
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<h1>404</h1>
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<p>Note not found</p>
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</div>
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</main>
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<script type="text/javascript">
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</script>
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</body>
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</html>
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@ -1,538 +0,0 @@
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<meta charset="utf-8" />
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<meta http-equiv="X-UA-Compatible" content="IE=edge">
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<meta name="viewport" content="width=device-width, initial-scale=1">
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<meta name="description" content="Bernoulli’s equation:
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$$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$
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(I’m calling this #de_b_type1) It looks almost like a linear equation! In fact if n=0 it is by definition. We will see further that if n=1 you also still get a linear equation.
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Bernoulli’s …">
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<meta name="apple-mobile-web-app-capable" content="yes">
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<meta name="mobile-web-app-capable" content="yes">
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<meta name="apple-mobile-web-app-status-bar-style" content="default">
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<link rel="manifest" href="./manifest.json"><meta property="og:title" content="" />
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<meta property="og:description" content="Bernoulli’s equation: $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$ (I’m calling this #de_b_type1) It looks almost like a linear equation! In fact if n=0 it is by definition. We will see further that if n=1 you also still get a linear equation. Bernoulli’s equations are important as you will see it in biology and in engineering.
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If y is + then y(x)=0 is a solution to the equation: $\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$ Let’s move the y to the LHS: $y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$ notice that y(x)=0 is no longer a solution!" />
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<meta property="og:type" content="article" />
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<meta property="og:url" content="/bernoulli-de-lec-3.html" />
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<meta name="twitter:card" content="summary"/>
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<meta name="twitter:title" content=""/>
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<meta name="twitter:description" content="Bernoulli’s equation: $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$ (I’m calling this #de_b_type1) It looks almost like a linear equation! In fact if n=0 it is by definition. We will see further that if n=1 you also still get a linear equation. Bernoulli’s equations are important as you will see it in biology and in engineering.
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If y is + then y(x)=0 is a solution to the equation: $\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$ Let’s move the y to the LHS: $y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$ notice that y(x)=0 is no longer a solution!"/>
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<title>Bernoulli DE (lec 3) - My New Hugo Site</title>
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|
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||||
|
||||
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<link rel="stylesheet" href="./css/main.min.203106d73d4370d04c60441691746dd8e021e38bbbc83f65f636dc8ae886a9f3.css" />
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<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script>
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<script src="./js/main.min.2dd2f7073384163751d1886bcb921097bc2af8ec60cb37deebf49f61a0eca5c3.js" integrity="sha256-LdL3BzOEFjdR0Yhry5IQl7wq+Oxgyzfe6/SfYaDspcM="></script>
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</head>
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<body>
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||||
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<style>
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||||
search-menu {
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display: block;
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||||
}
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||||
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||||
#search {
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||||
height: 100%;
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||||
width: 0;
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||||
position: fixed;
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||||
background: var(--background-search);
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||||
z-index: 1;
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top: 0;
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left: 0;
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border-right: 1px solid var(--separator-color);
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overflow-x: hidden;
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overflow-y: auto;
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opacity: 0;
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||||
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-ms-overflow-style: none;
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||||
scrollbar-width: none;
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}
|
||||
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||||
#search::-webkit-scrollbar { display: none; }
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||||
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||||
#search-header {
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||||
padding: 12px;
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||||
position: fixed;
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||||
padding-left: 12px;
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||||
padding-right: 12px;
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background: var(--background-search);
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width: 250px;
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||||
opacity: 1;
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||||
height: 50px;
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||||
z-index: 2;
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||||
border-bottom: 1px solid var(--separator-color);
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||||
}
|
||||
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||||
#search .input-container {
|
||||
position: relative
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||||
}
|
||||
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||||
#search-input {
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||||
width: 100%;
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||||
height: 24px;
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||||
border: 1px solid var(--separator-color);
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||||
border-radius: 4px;
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||||
padding-left: 16px;
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||||
background-color: white;
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||||
display: inline-block;
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||||
}
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||||
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||||
#search-input:focus {
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||||
border: 1px solid var(--search-field-focused-color);
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||||
}
|
||||
|
||||
#search-header .input-container .search-icon {
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||||
position: absolute;
|
||||
top: 6px;
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||||
left: 8px;
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||||
fill: darkGray;
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||||
}
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||||
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||||
#search-results img {
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||||
width: 122px;
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||||
height: 76px;
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||||
border: 1px solid var(--separator-color);
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||||
object-fit: cover;
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||||
}
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||||
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||||
#search-results {
|
||||
margin-top: 50px;
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||||
overflow: auto;
|
||||
height: 100%;
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||||
}
|
||||
|
||||
#search-results a {
|
||||
width: 100%;
|
||||
padding-left: 25px;
|
||||
padding-right: 25px;
|
||||
padding-top: 12px;
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||||
padding-bottom: 12px;
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||||
display: inline-block;
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||||
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||||
color: var(--text-base-color);
|
||||
border-bottom: 1px solid var(--separator-color);
|
||||
border-left: 6px solid var(--background-search);
|
||||
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||||
}
|
||||
|
||||
#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;
|
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display: none;
|
||||
}
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<h1 id="bernoullis-equation">Bernoulli’s equation:</h1>
|
||||
<h3 id="frac-dy--dx--pxyqxyn-quadquad-ninmathbbr">$$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$</h3>
|
||||
<p>(I’m calling this
|
||||
<a class="hashtag" onclick="focusTag(this)">de_b_type1)</a> It looks almost like a linear equation! In fact if n=0 it is by definition. We will see further that if n=1 you also still get a linear equation.
|
||||
Bernoulli’s equations are important as you will see it in biology and in engineering.</p>
|
||||
<p>If y is + then y(x)=0 is a solution to the equation:
|
||||
$\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$
|
||||
Let’s move the y to the LHS:
|
||||
$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$
|
||||
notice that y(x)=0 is no longer a solution! It was lost due to dividing by zero. So from here on out we will have to remember to add it back in our final answers.
|
||||
let $y^{1-n}=u$
|
||||
Differentiating this with respect to x gives us:
|
||||
$(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$</p>
|
||||
<blockquote>
|
||||
<p>notice that when n=1 the above turns into a linear equation:
|
||||
$0=\frac{ du }{ dx }$
|
||||
$y^{1-n}=u=0+C$
|
||||
1=C
|
||||
Hold on I dont think I did the above correctly. Anyways.
|
||||
So we consider that $n\ne 0,1$ for Bernoulli’s equations as we can solve those cases with earlier tools.</p>
|
||||
</blockquote>
|
||||
<p>$y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$
|
||||
substituting in we get:
|
||||
$y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$</p>
|
||||
<p>and we get a linear equation again: (Handy formula if you wanna solve specific Bernoulli equations quick.)
|
||||
$$\frac{1}{1-n}\frac{ du }{ dx }+P(x)=Q(x)\quad \Box$$</p>
|
||||
<hr>
|
||||
<h1 id="examples-of-bernoullis-equation">Examples of Bernoulli’s equation:</h1>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
|
||||
<a class="hashtag" onclick="focusTag(this)">de_b_type1</a> Find the general solution to:
|
||||
$y'+y=(xy)^2$
|
||||
Looks like a Bernoulli equation because when we distribute the $^2$ we get $x^2y^2$ on the RHS. This also tells us that n=2
|
||||
$y'+y=x^2y^2$
|
||||
$y’y^{-2}+y^{-1}=x^2$</p>
|
||||
<blockquote>
|
||||
<p>Note that we lost the y(x)=0 solution here, we will have to add it back in the end.</p>
|
||||
</blockquote>
|
||||
<p>let $u=y^{1-n}=y^{-1}$
|
||||
Differentiating wrt. x we get: $\frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}$
|
||||
$y^{-2}{\frac{dy}{dx}=-\frac{ du }{ dx }}$
|
||||
$y^{-2}{\frac{dy}{dx}+y^{-1}=-\frac{ du }{ dx }}+y^{-1}$
|
||||
${x^2=-\frac{ du }{ dx }}+y^{-1}$
|
||||
$x^2=-\frac{du}{dx}+u$
|
||||
$\frac{du}{dx}-u=-x^2$
|
||||
Yay we have a linear equation now! We can solve it using the techniques & formulas we learned for them.
|
||||
let $P(x)=-1 \quad Q(x)=-x^2 \qquad I(x)=e^{\int -1 , dx}=e^{-x}$
|
||||
$u=-e^{x}\int e^{-x}x^2 , dx$
|
||||
LIATE log, inv, alg, trig, exp
|
||||
$\int fg' , dx=fg-\int f’g , dx$
|
||||
let $f=x^2 \qquad f'=2x \qquad g'=e^{-x} \qquad g=-e^{-x}$
|
||||
$u=-e^{x}\left( x^2(-e^{-x})-\int 2x(-e^{-x}) , dx \right)$
|
||||
$u=-e^{x}\left( -x^2e^{-x}+2\int xe^{-x} , dx \right)$
|
||||
let $f=x \qquad f'=1 \qquad g'=e^{-x} \qquad g=-e^{-x}$
|
||||
$u=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-\int -e^{-x} , dx \right) \right)$
|
||||
$\frac{1}{y}=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-e^{-x} +C\right) \right)$
|
||||
$\frac{1}{y}=x^2+2(x+1+Ce^x)$
|
||||
$\frac{1}{y}=x^2+2x+2+Ce^x$
|
||||
The general solution to the DE is:
|
||||
$y(x)=\frac{1}{x^2+2x+2+Ce^x}$ as well as $y(x)=0$</p>
|
||||
<hr>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">end</a> of lecture 3</p>
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<h3>Referenced in</h3>
|
||||
|
||||
<ul>
|
||||
<li>No backlinks found</li>
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</ul>
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<meta name="description" content="#start of lecture 2
|
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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 $u=\frac{y}{t}$ $y=tu \quad \frac{dy}{dt}=u+t\frac{du}{dt}$
|
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so $\frac{dy}{dt}=f(u)=u+t{\frac{du}{dt}}$
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The homogenous equation has …">
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<meta property="og:description" content="#start of lecture 2
|
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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!" />
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<meta name="twitter:title" content=""/>
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<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!"/>
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<title>Homogenous DE (lec 2) - My New Hugo Site</title>
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<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>
|
||||
|
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<ul>
|
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<li>No backlinks found</li>
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I have written these notes for myself, I thought it would be cool to share them. These notes may be inaccurate, incomplete, or incoherent. No warranty is expressed or implied. Reader assumes all risk and liabilities.
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Seperable DE (lec 1)
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Homogenous DE (lec 2)
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<main id="main">
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<h1 id="this-is-the-main-index">This is the main index</h1>
|
||||
<p>I have written these notes for myself, I thought it would be cool to share them. These notes may be inaccurate, incomplete, or incoherent. No warranty is expressed or implied. Reader assumes all risk and liabilities.</p>
|
||||
<p><a href="seperable-de-lec-1.html">Seperable DE (lec 1)</a></p>
|
||||
<p><a href="homogenous-de-lec-2.html">Homogenous DE (lec 2)</a></p>
|
||||
<p><a href="linear-de-lec-23.html">Linear DE (lec 2&3)</a></p>
|
||||
<p><a href="bernoulli-de-lec-3.html">Bernoulli DE (lec 3)</a></p>
|
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|
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</div>
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</main>
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</script>
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</html>
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@ -1,178 +0,0 @@
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<meta name="description" content="two new equations
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Linear coefficients equations
|
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$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$
|
||||
imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!
|
||||
so can we make them 0?
|
||||
let $x=u+k$
|
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$y=v+l$
|
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where $k,l$ are constants …">
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<meta name="apple-mobile-web-app-capable" content="yes">
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<link rel="manifest" href="./manifest.json"><meta property="og:title" content="" />
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<meta property="og:description" content="two new equations Linear coefficients equations $$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$ imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!
|
||||
so can we make them 0? let $x=u+k$ $y=v+l$ where $k,l$ are constants $(a_{1}u+b_{1}vK+\underbrace{\cancel{ c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l } }_{ 0 })dv=0$ $a_{1}k+b_{1}l=-c_1$ $a_{2}k+b_{2}l=-c_{2}$ if $\det(a_{1},b_{1},a_{2},b_{2})\ne 0$ turn into homogenous if $\det(\dots)=0 \Rightarrow$ equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)
|
||||
Example #ex #de_LC_type1 $$(-3x+y+6)dx+(x+y+2)dy=0$$ let $x=u+k$ $y=v+l$ $(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$ we want $6-3k+l$ and $2+k+l$ to equal 0 so: $-3k+l=-6$ $k+l=-2$ $det(-3,1,1,1)=-4$ //he can call it a dinosaur if he wanted to :D solving gives us: $k=1,l=-3$ so $x=u+1 \quad y=v-3$ $(-3u+v)du+(u+v)dv=0$ //Beutiful1!" />
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<meta name="twitter:card" content="summary"/>
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<meta name="twitter:title" content=""/>
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<meta name="twitter:description" content="two new equations Linear coefficients equations $$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$ imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!
|
||||
so can we make them 0? let $x=u+k$ $y=v+l$ where $k,l$ are constants $(a_{1}u+b_{1}vK+\underbrace{\cancel{ c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l } }_{ 0 })dv=0$ $a_{1}k+b_{1}l=-c_1$ $a_{2}k+b_{2}l=-c_{2}$ if $\det(a_{1},b_{1},a_{2},b_{2})\ne 0$ turn into homogenous if $\det(\dots)=0 \Rightarrow$ equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)
|
||||
Example #ex #de_LC_type1 $$(-3x+y+6)dx+(x+y+2)dy=0$$ let $x=u+k$ $y=v+l$ $(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$ we want $6-3k+l$ and $2+k+l$ to equal 0 so: $-3k+l=-6$ $k+l=-2$ $det(-3,1,1,1)=-4$ //he can call it a dinosaur if he wanted to :D solving gives us: $k=1,l=-3$ so $x=u+1 \quad y=v-3$ $(-3u+v)du+(u+v)dv=0$ //Beutiful1!"/>
|
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|
||||
|
||||
|
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|
||||
|
||||
<title>(lec 4) - My New Hugo Site</title>
|
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<body>
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<style>
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display: block;
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#search {
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height: 100%;
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width: 0;
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position: fixed;
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background: var(--background-search);
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||||
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||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<h1 id="two-new-equations">two new equations</h1>
|
||||
<h2 id="linear-coefficients-equations">Linear coefficients equations</h2>
|
||||
<p>$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$
|
||||
imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!</p>
|
||||
<p>so can we make them 0?
|
||||
let $x=u+k$
|
||||
$y=v+l$
|
||||
where $k,l$ are constants
|
||||
$(a_{1}u+b_{1}vK+\underbrace{\cancel{ c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l } }_{ 0 })dv=0$
|
||||
$a_{1}k+b_{1}l=-c_1$
|
||||
$a_{2}k+b_{2}l=-c_{2}$
|
||||
if $\det(a_{1},b_{1},a_{2},b_{2})\ne 0$ turn into homogenous
|
||||
if $\det(\dots)=0 \Rightarrow$ equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)</p>
|
||||
<h3 id="example">Example</h3>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">ex</a> <a class="hashtag" onclick="focusTag(this)">de_LC_type1</a>
|
||||
$$(-3x+y+6)dx+(x+y+2)dy=0$$
|
||||
let $x=u+k$
|
||||
$y=v+l$
|
||||
$(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$
|
||||
we want $6-3k+l$ and $2+k+l$ to equal 0
|
||||
so:
|
||||
$-3k+l=-6$
|
||||
$k+l=-2$
|
||||
$det(-3,1,1,1)=-4$ //he can call it a dinosaur if he wanted to :D
|
||||
solving gives us:
|
||||
$k=1,l=-3$
|
||||
so $x=u+1 \quad y=v-3$
|
||||
$(-3u+v)du+(u+v)dv=0$ //Beutiful1! it’s homogenous now
|
||||
$\frac{ dv }{ du }=\frac{{3u-v}}{u+v}=\frac{{3-\frac{v}{u}}}{1+\frac{v}{u}}$
|
||||
$\frac{v}{u}=w \quad v=uw \quad \frac{ dv }{ du }=w+u\frac{ dw }{ du }$
|
||||
$w+u\frac{ dw }{ du }=\frac{{3-w}}{1+w}$ This is the equation we have to solve
|
||||
$u\frac{ dw }{ du }=\frac{{3-2w-w^2}}{1+w}$
|
||||
$-\frac{{w+1}}{w^2+2w-3}dw=\frac{du}{u}$
|
||||
$\int-\frac{{w+1}}{w^2+2w-3}dw=\int\frac{du}{u}$
|
||||
let $z=w^2+2w-3$
|
||||
$dz=2(w+1)dw$
|
||||
$\frac{1}{2}\int \frac{dz}{z}=\ln\mid u\mid^{-1}$
|
||||
$\ln\mid z\mid^{1/2}-\ln\mid u\mid^{-1}=C$
|
||||
$\ln(\mid z\mid^{1/2}\mid u\mid)=C$
|
||||
$\mid z\mid^{1/2}u=e^C$
|
||||
$\mid z\mid u^2=e^{2C}$
|
||||
$zu^2=A$
|
||||
$\left( \left( \frac{v}{u} \right)^2+\frac{2v}{u}-3 \right)u^2=A$
|
||||
remember $u=x-1 \quad v=y+3$
|
||||
$$\left( \left( \frac{{y+3}}{x-1} \right)^2+\frac{2(y+3)}{x-1}-3 \right)(x-1)^2=A$$
|
||||
you can “simplify” it to: $(y+3)^2+2(y+3)(x-1)-3(x-1)^2=A$</p>
|
||||
<hr>
|
||||
<h2 id="exact-equations">Exact equations</h2>
|
||||
<p>two variable equations
|
||||
$dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ suppose it equals to zero (as shown in the equation) you get a horizontal plane (a constant)
|
||||
so $F(x,y)=C$
|
||||
the solution to these exact equations is given by $F()$ but how do we get F from the derivatives?
|
||||
Equation of the form: $$M(x,y)dx=N(x,y)dy=0$$
|
||||
is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$
|
||||
then differentiating we get:
|
||||
$\frac{ \partial M }{ \partial y }=\frac{ \partial^{2} F }{ \partial y\partial x }$
|
||||
$\frac{ \partial N }{ \partial x }=\frac{ \partial^{2} F }{ \partial x\partial y }$ Order of going in x then y vs y then x doesn’t matter as it lands you on the same point (idk how this is related yet)
|
||||
Exact equation$\Rightarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ if it’s continuous (?)
|
||||
also: Exact equation$\Leftarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$
|
||||
Test for exactness:
|
||||
exact $\iff \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ (this can be proved, but it wasnt proved in class)
|
||||
<a class="hashtag" onclick="focusTag(this)">end</a> of lecture 4</p>
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<h3>Referenced in</h3>
|
||||
|
||||
<ul>
|
||||
<li>No backlinks found</li>
|
||||
|
||||
</ul>
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</main>
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</script>
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</body>
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</html>
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@ -1,553 +0,0 @@
|
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<!DOCTYPE html>
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Linear equation:
|
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$$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$ (I’m calling this …">
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<meta property="og:description" content="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.
|
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Linear equation: $$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$ (I’m calling this #de_L_type1) if we assume $b(x)=a'(x)$ it kinda starts to look like a product rule $a(x)y'+a'(x)y=f(x)=(ay)'$ $ay=\int f(x) , dx$ <-yay! We can find the solutions to y." />
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|
||||
Linear equation: $$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$ (I’m calling this #de_L_type1) if we assume $b(x)=a'(x)$ it kinda starts to look like a product rule $a(x)y'+a'(x)y=f(x)=(ay)'$ $ay=\int f(x) , dx$ <-yay! We can find the solutions to y."/>
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<title>Linear DE (lec 2&3) - My New Hugo Site</title>
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display: block;
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#search {
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height: 100%;
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width: 0;
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position: fixed;
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background: var(--background-search);
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border-right: 1px solid var(--separator-color);
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overflow-x: hidden;
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scrollbar-width: none;
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#search::-webkit-scrollbar { display: none; }
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#search-header {
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padding: 12px;
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position: fixed;
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border-bottom: 1px solid var(--separator-color);
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#search .input-container {
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position: relative
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}
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<p>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.</p>
|
||||
<h1 id="linear-equation">Linear equation:</h1>
|
||||
<p>$$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$ (I’m calling this <a class="hashtag" onclick="focusTag(this)">de_L_type1)</a>
|
||||
if we assume $b(x)=a'(x)$ it kinda starts to look like a product rule
|
||||
$a(x)y'+a'(x)y=f(x)=(ay)'$
|
||||
$ay=\int f(x) , dx$ <svg width="11px" height="10px" viewBox="0 0 11 10" version="1.1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"> <g id="left-arrow" transform="translate(5.500000, 5.000000) scale(-1, 1) translate(-5.500000, -5.000000) "> <path d="M1.77635684e-14,5 L9,5" id="rod" stroke="#000000" stroke-width="2"></path> <path d="M11,5 L6,0.5 L6,9.5 L11,5 Z" id="point" fill="#000000"></path></g></svg>yay! We can find the solutions to y.</p>
|
||||
<p>we can rewrite the linear equation in what’s called standard form:
|
||||
$$\frac{ dy }{ dx }+P(x)y=Q(x)$$ (I’m calling this <a class="hashtag" onclick="focusTag(this)">de_L_type2)</a>
|
||||
we will define a function $\mu(x)$ called the integration factor, also expressed as $I(x)$
|
||||
Multiply both sides by $\mu(x)$
|
||||
$\mu(x) \frac{ dy }{ dx }+\underbrace{ \mu(x)P(x) }_{ \mu'(x) }y=\mu(x)Q(x)$
|
||||
Like shown above we imagine if $\mu(x) P(x)=\mu'(x)$ as it starts to look like the product rule again.
|
||||
$(\mu y)'=\mu(x)Q(x)$
|
||||
This is nice as now we can integrate both sides as usual and get a solution for y.
|
||||
$y=\frac{1}{\mu(x)}\int \mu(x)Q(x) , dx$ <a class="hashtag" onclick="focusTag(this)">remember</a>
|
||||
But what is $\mu(x)$? How do we find it?
|
||||
In order for $\mu(x) P(x)=\mu'(x)$ to be true, $\frac{ d\mu }{ dx }=\mu(x)P(x)\Rightarrow \frac{ d\mu }{ \mu }=P(x)dx\Rightarrow\int \frac{d\mu}{\mu}=\int P(x) , dx\Rightarrow\ \ln\mid \mu\mid=\int P(x) , dx$</p>
|
||||
<blockquote>
|
||||
<p>I’m not sure why the professor allows the absolute value to be dropped in the following step, I think he said that he argues all solutions can be found even if we focus only where $\mu$ is +, idk.</p>
|
||||
</blockquote>
|
||||
<p>finally we get that $\mu(x)=I(x)=e^{\int P(x) , dx}\quad \Box$ <a class="hashtag" onclick="focusTag(this)">remember</a></p>
|
||||
<hr>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">end</a> of lecture 2
|
||||
<a class="hashtag" onclick="focusTag(this)">start</a> of lecture 3</p>
|
||||
<h1 id="examples-of-linear-equations">Examples of linear equations:</h1>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
|
||||
<a class="hashtag" onclick="focusTag(this)">de_L_type2</a> Find the general solution to the equation:</p>
|
||||
<h2 id="1sinxy2cosxytanx">$$(1+\sin(x))y'+2\cos(x)y=\tan(x)$$</h2>
|
||||
<p>let $a(x)=1+sin(x)\qquad b(x)=2\cos(x)$
|
||||
we can see that $b(x)\ne a'(x)$ :( so we cant use
|
||||
<a class="hashtag" onclick="focusTag(this)">de_L_type1</a>
|
||||
let’s rearrange it into standard form:
|
||||
$y'+\frac{{2\cos(x)}}{1+\sin(x)}=\frac{\tan(x)}{1+\sin(x)}$
|
||||
$P(x):=\frac{2\cos(x)}{1+\sin(x)} \qquad Q(x)=\frac{\tan(x)}{1+\sin(x)}$
|
||||
then $I(x)=e^{\int {2\cos(x)}/(1+\sin(x)), dx}$
|
||||
let $u=1+\sin(x) \qquad du=\cos(x)dx$
|
||||
$I(x)=e^{\int \frac{2\cos(x)}{u} , \frac{du}{\cos(x)}}$
|
||||
$I(x)=e^{2\ln\mid u\mid}$
|
||||
$I(x)=\mid u\mid^2$
|
||||
$I(x)=\mid1+\sin(x)\mid^2$
|
||||
$I(x)=(1+\sin(x))^2$
|
||||
$y=\frac{1}{I(x)}\int I(x)Q(x) , dx$
|
||||
$y=\frac{1}{(1+\sin(x))^2}\int (\frac{(1+\sin(x))^2\tan(x)}{1+\sin(x)} , dx$
|
||||
$y=\frac{1}{(1+\sin(x))^2}\int \tan(x)+\frac{\sin^2(x)}{\cos(x)} , dx$
|
||||
$y=\frac{1}{(1+\sin(x))^2}(\ln\mid sec(x)\mid+\int \frac{\sin^2(x)}{\cos(x)} , dx)$</p>
|
||||
<blockquote>
|
||||
<p>using u substitution doesnt work for the second integral because I got $\int\frac{\sin(x)}{u}du \qquad u=\cos(x)$
|
||||
could try using $\sin^2(x)=\frac{{1-\cos(2x)}}{2}$ but looks hard with the 2x term,
|
||||
let’s try using $\sin^2(x)=1-\cos^2(x)$ instead. (also because I also remember this is what we used in class)</p>
|
||||
</blockquote>
|
||||
<p>$y=\frac{1}{(1+\sin(x))^2}(\ln\mid sec(x)\mid+\int \frac{1-\cos^2(x)}{\cos(x)} , dx)$</p>
|
||||
<p>$y=\frac{1}{(1+\sin(x))^2}(\ln\mid sec(x)\mid+\ln\mid sec(x)+\tan(x)\mid+\int -\cos(x) , dx)$
|
||||
Albeit a bit ugly, we have found the general solution to the DE:
|
||||
$$y=\frac{1}{(1+\sin(x))^2}(\ln\mid sec(x)\mid+\ln\mid sec(x)+\tan(x)\mid-\sin(x)+C)$$</p>
|
||||
<hr>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
|
||||
<a class="hashtag" onclick="focusTag(this)">IVP</a>
|
||||
<a class="hashtag" onclick="focusTag(this)">de_L_type2</a></p>
|
||||
<h2 id="ytanxycos2x-qquad-yleft-fracpi4-rightfrac12">$$y'+\tan(x)y=\cos^2(x) \qquad y\left( \frac{\pi}{4} \right)=\frac{1}{2}$$</h2>
|
||||
<p>Looks like a linear equation with an initial value.
|
||||
$P(x)=\tan(x) \qquad Q(x)=\cos^2(x) \qquad I(x)=e^{\int \tan(x) , dx}$
|
||||
$I(x)=e^{\ln\mid sec(x)\mid}$
|
||||
$I(x)=\mid sec(x)\mid$
|
||||
$I(x)=sec(x)$</p>
|
||||
<blockquote>
|
||||
<p>The prof simply drops the absolute value. I don’t understand why. Sigma asf tbh. I think it’s because he said linear DE are nice because their solutions are unique with an IVP, non linear equations are not necessarily unique. So if we find one solution we know that we found the only solution possible.</p>
|
||||
</blockquote>
|
||||
<p>$y=\cos(x)\int sec(x)\cos^2(x) , dx$
|
||||
$y=\cos(x)\int \cos(x) , dx$
|
||||
$y=cos(x)(sin(x)+C)$
|
||||
Now we issue the initial value:
|
||||
$\frac{1}{2}=\cos\left( \frac{\pi}{4} \right)\sin\left( \frac{\pi}{4}+C) \right)$
|
||||
$\frac{\frac{1}{2}}{\frac{1}{\sqrt{ 2 }}}=\frac{\sqrt{ 2 }}{2}=\frac{1}{\sqrt{ 2 }}=\sin\left( \frac{\pi}{4} \right)$
|
||||
$C=0$
|
||||
By plugging in C=0 in the general solution we get the solution to the IVP, as stated earlier, there can only be one solution to a linear IVP DE:</p>
|
||||
<h3 id="ycosxsinx">$$y=\cos(x)\sin(x)$$</h3>
|
||||
<hr>
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<h3>Referenced in</h3>
|
||||
|
||||
<ul>
|
||||
<li>No backlinks found</li>
|
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<meta name="description" content="#start of lecture 1
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Intro (Newton example):
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Newton example where we find the equations to describe a falling object using differential equations (DE’s)
|
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We know $F=ma$
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$F=m\frac{dv}{dt}=mg-kv$ <- we account for air resistance here. We can approximate the force of air resistance is …">
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<meta property="og:description" content="#start of lecture 1
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Intro (Newton example): Newton example where we find the equations to describe a falling object using differential equations (DE’s) We know $F=ma$ $F=m\frac{dv}{dt}=mg-kv$ <- we account for air resistance here. We can approximate the force of air resistance is proportional to the speed times a constant. We can rearrange and solve it as it is a separable DE: $\frac{dv}{mg-kv}=\frac{dt}{m}$ integrating both sides: $\int \frac{dv}{mg-kv}=\frac{t}{m}+C$ let $u=mg-kv \quad du=-kdv$ $\int \frac{dv}{mg-kv}=\int \frac{du}{-k*u}=\frac{1}{-k}\ln\mid mg-kv\mid=\frac{t}{m}+C$ Very cool, but I want the velocity as a function of time, isolate v $\ln\mid mg-kv\mid=-\frac{kt}{m}+C$ $\mid mg-kv\mid=e^{\frac{-kt}{m}+C}=e^{\frac{-kt}{m}}e^C$ $e^C$ is a + constant, the absolute value will multiply the inside expression by -1 when the inside is negative, so we can replace the $e^C$ constant with an arbitrary constant A that can be + or - $mg-kv=Ae^{\frac{-kt}{m}}$ so, the general solution is $v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$" />
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<meta name="twitter:description" content="#start of lecture 1
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Intro (Newton example): Newton example where we find the equations to describe a falling object using differential equations (DE’s) We know $F=ma$ $F=m\frac{dv}{dt}=mg-kv$ <- we account for air resistance here. We can approximate the force of air resistance is proportional to the speed times a constant. We can rearrange and solve it as it is a separable DE: $\frac{dv}{mg-kv}=\frac{dt}{m}$ integrating both sides: $\int \frac{dv}{mg-kv}=\frac{t}{m}+C$ let $u=mg-kv \quad du=-kdv$ $\int \frac{dv}{mg-kv}=\int \frac{du}{-k*u}=\frac{1}{-k}\ln\mid mg-kv\mid=\frac{t}{m}+C$ Very cool, but I want the velocity as a function of time, isolate v $\ln\mid mg-kv\mid=-\frac{kt}{m}+C$ $\mid mg-kv\mid=e^{\frac{-kt}{m}+C}=e^{\frac{-kt}{m}}e^C$ $e^C$ is a + constant, the absolute value will multiply the inside expression by -1 when the inside is negative, so we can replace the $e^C$ constant with an arbitrary constant A that can be + or - $mg-kv=Ae^{\frac{-kt}{m}}$ so, the general solution is $v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$"/>
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|
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|
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|
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|
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|
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<title>Seperable DE (lec 1) - My New Hugo Site</title>
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<main id="main">
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|
||||
<p>#start of lecture 1</p>
|
||||
<h3 id="intro-newton-example">Intro (Newton example):</h3>
|
||||
<p>Newton example where we find the equations to describe a falling object using differential equations (DE’s)
|
||||
We know $F=ma$
|
||||
$F=m\frac{dv}{dt}=mg-kv$ <svg width="11px" height="10px" viewBox="0 0 11 10" version="1.1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"> <g id="left-arrow" transform="translate(5.500000, 5.000000) scale(-1, 1) translate(-5.500000, -5.000000) "> <path d="M1.77635684e-14,5 L9,5" id="rod" stroke="#000000" stroke-width="2"></path> <path d="M11,5 L6,0.5 L6,9.5 L11,5 Z" id="point" fill="#000000"></path></g></svg> we account for air resistance here. We can approximate the force of air resistance is proportional to the speed times a constant.
|
||||
We can rearrange and solve it as it is a separable DE:
|
||||
$\frac{dv}{mg-kv}=\frac{dt}{m}$
|
||||
integrating both sides:
|
||||
$\int \frac{dv}{mg-kv}=\frac{t}{m}+C$
|
||||
let $u=mg-kv \quad du=-kdv$
|
||||
$\int \frac{dv}{mg-kv}=\int \frac{du}{-k*u}=\frac{1}{-k}\ln\mid mg-kv\mid=\frac{t}{m}+C$
|
||||
Very cool, but I want the velocity as a function of time, isolate v
|
||||
$\ln\mid mg-kv\mid=-\frac{kt}{m}+C$
|
||||
$\mid mg-kv\mid=e^{\frac{-kt}{m}+C}=e^{\frac{-kt}{m}}e^C$
|
||||
$e^C$ is a + constant, the absolute value will multiply the inside expression by -1 when the inside is negative, so we can replace the $e^C$ constant with an arbitrary constant A that can be + or -
|
||||
$mg-kv=Ae^{\frac{-kt}{m}}$
|
||||
so, the general solution is $v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$</p>
|
||||
<h3 id="separable-de">Separable DE:</h3>
|
||||
<pre><code>$\frac{dy}{dx}=f(y)g(x) \rightarrow \frac{dy}{f(y)}=g(x)dx\quad where\quad f(y)\ne0$
|
||||
(I'm calling this <a class="hashtag" onclick="focusTag(this)">de_s_type1)</a>
|
||||
</code></pre>
|
||||
<p>ex: $\frac{dy}{dt}=\frac{1-t^2}{y^2}$
|
||||
$y^2dy=dt(1-t^2)$
|
||||
integrating both sides yields:
|
||||
$\frac{y^3}{3}=t-\frac{t^3}{3}+C$
|
||||
$y=(3t-t^3+C)^\frac{1}{3}$</p>
|
||||
<h3 id="initial-value-problem-ivp">Initial value problem (IVP):</h3>
|
||||
<pre><code>A Differential equation with provided initial conditions.
|
||||
</code></pre>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
|
||||
<a class="hashtag" onclick="focusTag(this)">IVP</a> <a class="hashtag" onclick="focusTag(this)">de_s_type1</a>
|
||||
ex: $\frac{dy}{dx}=2x\cos^2(y), \quad y(0)=\frac{\pi}{4}$
|
||||
$\frac{dy}{\cos^2(y)}=2xdx$
|
||||
integrate both sides yields:
|
||||
$\int \frac{dy}{\cos^2(y)}=\tan(y)+C=x^2$
|
||||
plug in $y(0)=\frac{\pi}{4}$
|
||||
$\tan\left( \frac{\pi}{4} \right)+C=0$
|
||||
$1+C=0$
|
||||
$C=-1$
|
||||
So, the answer is: $y=\arctan(x^2+1)$</p>
|
||||
<p><a class="hashtag" onclick="focusTag(this)">end</a> of Lecture 1</p>
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<h3>Referenced in</h3>
|
||||
|
||||
<ul>
|
||||
<li>No backlinks found</li>
|
||||
|
||||
</ul>
|
||||
|
||||
|
||||
|
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</div>
|
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</main>
|
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|
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<script type="text/javascript">
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|
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|
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|
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|
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|
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|
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|
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</script>
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|
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</body>
|
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</html>
|
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|
|
@ -1,44 +0,0 @@
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<?xml version="1.0" encoding="utf-8" standalone="yes"?>
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<loc>/lec-4.html</loc>
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<loc>/bernoulli-de-lec-3.html</loc>
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<loc>/tags.html</loc>
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</urlset>
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405
public/tags.html
405
public/tags.html
|
|
@ -1,405 +0,0 @@
|
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<link rel="alternate" type="application/rss+xml" href="./tags/index.xml" title="My New Hugo Site" />
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<meta name="twitter:title" content="Tags"/>
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<meta name="twitter:description" content=""/>
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|
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|
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|
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<title> - My New Hugo Site</title>
|
||||
|
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|
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</head>
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<body>
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|
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<style>
|
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search-menu {
|
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|
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|
||||
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">
|
||||
|
||||
|
||||
<header>
|
||||
<h1>Tags</h1>
|
||||
</header>
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
</div>
|
||||
</main>
|
||||
|
||||
<script type="text/javascript">
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
</script>
|
||||
|
||||
</body>
|
||||
</html>
|
||||
|
|
@ -1,10 +0,0 @@
|
|||
<?xml version="1.0" encoding="utf-8" standalone="yes"?>
|
||||
<rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom">
|
||||
<channel>
|
||||
<title>Tags on My New Hugo Site</title>
|
||||
<link>/tags.html</link>
|
||||
<description>Recent content in Tags on My New Hugo Site</description>
|
||||
<generator>Hugo -- gohugo.io</generator>
|
||||
<language>en-us</language><atom:link href="/tags/index.xml" rel="self" type="application/rss+xml" />
|
||||
</channel>
|
||||
</rss>
|
||||
|
|
@ -1,504 +0,0 @@
|
|||
<!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="This is an incomplete list. I’m trying to add to it as I go.
|
||||
everything with the #remember tag as well as the representations of the various DE, they start with #de_
|
||||
Also remember the following:
|
||||
derivatives of trigs
|
||||
$\frac{d}{dx}\tan(x)=sec^2(x)$
|
||||
$\frac{d}{dx}sec(x)=sec(x)\tan(x)$
|
||||
… …">
|
||||
<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="This is an incomplete list. I’m trying to add to it as I go.
|
||||
everything with the #remember tag as well as the representations of the various DE, they start with #de_
|
||||
Also remember the following:
|
||||
derivatives of trigs $\frac{d}{dx}\tan(x)=sec^2(x)$ $\frac{d}{dx}sec(x)=sec(x)\tan(x)$ …
|
||||
integrals of trigs $\int \tan(x) , dx=\ln\mid \sec(x)\mid+C$ $\int sec(x) , dx=\ln\mid sec(x)+\tan(x)\mid+C$ …
|
||||
integration by parts LIATE -> log, inv trig, algebraic, trig, exp set u to the first in the list above $\int u(x)v'(x) , dx=uv-\int u'(x)v(x) , dx$" />
|
||||
<meta property="og:type" content="article" />
|
||||
<meta property="og:url" content="/things-to-remember.html" />
|
||||
|
||||
<meta name="twitter:card" content="summary"/>
|
||||
<meta name="twitter:title" content=""/>
|
||||
<meta name="twitter:description" content="This is an incomplete list. I’m trying to add to it as I go.
|
||||
everything with the #remember tag as well as the representations of the various DE, they start with #de_
|
||||
Also remember the following:
|
||||
derivatives of trigs $\frac{d}{dx}\tan(x)=sec^2(x)$ $\frac{d}{dx}sec(x)=sec(x)\tan(x)$ …
|
||||
integrals of trigs $\int \tan(x) , dx=\ln\mid \sec(x)\mid+C$ $\int sec(x) , dx=\ln\mid sec(x)+\tan(x)\mid+C$ …
|
||||
integration by parts LIATE -> log, inv trig, algebraic, trig, exp set u to the first in the list above $\int u(x)v'(x) , dx=uv-\int u'(x)v(x) , dx$"/>
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<title>Things to remember - 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;
|
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|
||||
transition: 1s;
|
||||
opacity: 0.5;
|
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|
||||
padding: 18px 0px 18px 0px;
|
||||
}
|
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|
||||
#toolbar:hover {
|
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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>This is an incomplete list. I’m trying to add to it as I go.</p>
|
||||
<p>everything with the
|
||||
<a class="hashtag" onclick="focusTag(this)">remember</a> tag as well as the representations of the various DE, they start with <a class=“hashtag” onclick=“focusTag(this)">de_</a></p>
|
||||
<p>Also remember the following:</p>
|
||||
<h2 id="derivatives-of-trigs">derivatives of trigs</h2>
|
||||
<p>$\frac{d}{dx}\tan(x)=sec^2(x)$
|
||||
$\frac{d}{dx}sec(x)=sec(x)\tan(x)$
|
||||
…</p>
|
||||
<h2 id="integrals-of-trigs">integrals of trigs</h2>
|
||||
<p>$\int \tan(x) , dx=\ln\mid \sec(x)\mid+C$
|
||||
$\int sec(x) , dx=\ln\mid sec(x)+\tan(x)\mid+C$
|
||||
…</p>
|
||||
<h2 id="integration-by-parts">integration by parts</h2>
|
||||
<p>LIATE <svg width="11px" height="10px" viewBox="0 0 11 10" version="1.1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" ><g id="right-arrow" ><path d="M1.77635684e-14,5 L9,5" id="rod" stroke="#000000" stroke-width="2" ></path><path d="M11,5 L6,0.5 L6,9.5 L11,5 Z" id="point" fill="#000000"></path></g></svg> log, inv trig, algebraic, trig, exp
|
||||
set u to the first in the list above
|
||||
$\int u(x)v'(x) , dx=uv-\int u'(x)v(x) , dx$</p>
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<h3>Referenced in</h3>
|
||||
|
||||
<ul>
|
||||
<li>No backlinks found</li>
|
||||
|
||||
</ul>
|
||||
|
||||
|
||||
|
||||
</div>
|
||||
</main>
|
||||
|
||||
<script type="text/javascript">
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
</script>
|
||||
|
||||
</body>
|
||||
</html>
|
||||
|
|
@ -0,0 +1,458 @@
|
|||
@charset "UTF-8";
|
||||
input {
|
||||
margin: 0;
|
||||
padding: 0;
|
||||
border: 0;
|
||||
font: inherit;
|
||||
color: inherit;
|
||||
font-size: 100%;
|
||||
vertical-align: baseline; }
|
||||
|
||||
*:focus {
|
||||
outline: none; }
|
||||
|
||||
textarea,
|
||||
input[type="search"],
|
||||
input[type="text"],
|
||||
input[type="button"],
|
||||
input[type="submit"] {
|
||||
-webkit-appearance: none;
|
||||
border-radius: 0; }
|
||||
|
||||
::selection {
|
||||
background: var(--selected-text-background-color); }
|
||||
|
||||
/**
|
||||
Bear base styles
|
||||
(Retrieved from official app, all credits belong to Bear Team)
|
||||
*/
|
||||
html, body, div, span, applet, object, iframe, h1, h2, h3, h4, h5, h6, p, blockquote, pre, a, abbr, acronym, address, big, cite, code, del, dfn, em, img, ins, kbd, q, s, samp, small, strike, strong, sub, sup, tt, var, b, u, i, center, dl, dt, dd, ol, ul, li, fieldset, form, label, legend, table, caption, tbody, tfoot, thead, tr, th, td, article, aside, canvas, details, embed, figure, figcaption, footer, header, hgroup, menu, nav, output, ruby, section, summary, time, mark, audio, video {
|
||||
margin: 0;
|
||||
padding: 0;
|
||||
border: 0;
|
||||
font: inherit;
|
||||
font-size: 100%;
|
||||
vertical-align: baseline; }
|
||||
|
||||
html {
|
||||
line-height: 1; }
|
||||
|
||||
ol, ul {
|
||||
list-style: none; }
|
||||
|
||||
table {
|
||||
border-collapse: collapse;
|
||||
border-spacing: 0; }
|
||||
|
||||
caption, th, td {
|
||||
text-align: left;
|
||||
font-weight: normal;
|
||||
vertical-align: middle; }
|
||||
|
||||
q, blockquote {
|
||||
quotes: none; }
|
||||
|
||||
q:before, q:after, blockquote:before, blockquote:after {
|
||||
content: "";
|
||||
content: none; }
|
||||
|
||||
a img {
|
||||
border: none; }
|
||||
|
||||
article, aside, details, figcaption, figure, footer, header, hgroup, main, menu, nav, section, summary {
|
||||
display: block; }
|
||||
|
||||
* {
|
||||
-moz-box-sizing: border-box;
|
||||
-webkit-box-sizing: border-box;
|
||||
box-sizing: border-box; }
|
||||
|
||||
html {
|
||||
font-size: 87.5%;
|
||||
line-height: 1.57143em; }
|
||||
|
||||
html {
|
||||
font-size: 14px;
|
||||
line-height: 1.6em;
|
||||
-webkit-text-size-adjust: 100%; }
|
||||
|
||||
body {
|
||||
background: var(--background);
|
||||
color: var(--text-base-color);
|
||||
text-rendering: optimizeLegibility;
|
||||
font-family: "AvenirNext-Regular"; }
|
||||
|
||||
a {
|
||||
color: var(--link-text-color);
|
||||
text-decoration: none; }
|
||||
|
||||
h1 {
|
||||
font-family: "AvenirNext-Medium";
|
||||
color: var(--title-text-color);
|
||||
font-size: 1.6em;
|
||||
line-height: 1.3em;
|
||||
margin-bottom: .78571em; }
|
||||
|
||||
h2 {
|
||||
font-family: "AvenirNext-Medium";
|
||||
color: var(--title-text-color);
|
||||
font-size: 1.3em;
|
||||
line-height: 1em;
|
||||
margin-bottom: .62857em; }
|
||||
|
||||
h3 {
|
||||
font-family: "AvenirNext-Medium";
|
||||
color: var(--title-text-color);
|
||||
font-size: 1.15em;
|
||||
line-height: 1em;
|
||||
margin-bottom: .47143em; }
|
||||
|
||||
p {
|
||||
margin-bottom: 1.57143em;
|
||||
hyphens: auto; }
|
||||
|
||||
hr {
|
||||
height: 1px;
|
||||
border: 0;
|
||||
background-color: #dedede;
|
||||
margin: -1px auto 1.57143em auto; }
|
||||
|
||||
ul, ol {
|
||||
margin-bottom: .31429em; }
|
||||
|
||||
ul ul, ul ol, ol ul, ol ol {
|
||||
margin-bottom: 0px; }
|
||||
|
||||
ol {
|
||||
counter-reset: ol_counter; }
|
||||
|
||||
ol li:before {
|
||||
content: counter(ol_counter) ".";
|
||||
counter-increment: ol_counter;
|
||||
color: var(--accent-text-color);
|
||||
text-align: right;
|
||||
display: inline-block;
|
||||
min-width: 1em;
|
||||
margin-right: 0.5em; }
|
||||
|
||||
b, strong {
|
||||
font-family: "AvenirNext-Bold"; }
|
||||
|
||||
i, em {
|
||||
font-family: "AvenirNext-Italic"; }
|
||||
|
||||
code {
|
||||
font-family: "Menlo-Regular"; }
|
||||
|
||||
.text-overflow-ellipsis {
|
||||
overflow: hidden;
|
||||
text-overflow: ellipsis;
|
||||
white-space: nowrap; }
|
||||
|
||||
.sf_code_string, .sf_code_selector, .sf_code_attr-name, .sf_code_char, .sf_code_builtin, .sf_code_inserted {
|
||||
color: #D33905; }
|
||||
|
||||
.sf_code_comment, .sf_code_prolog, .sf_code_doctype, .sf_code_cdata {
|
||||
color: #838383; }
|
||||
|
||||
.sf_code_number, .sf_code_boolean {
|
||||
color: #0E73A2; }
|
||||
|
||||
.sf_code_keyword, .sf_code_atrule, .sf_code_rule, .sf_code_attr-value, .sf_code_function, .sf_code_class-name, .sf_code_class, .sf_code_regex, .sf_code_important, .sf_code_variable, .sf_code_interpolation {
|
||||
color: #0E73A2; }
|
||||
|
||||
.sf_code_property, .sf_code_tag, .sf_code_constant, .sf_code_symbol, .sf_code_deleted {
|
||||
color: #1B00CE; }
|
||||
|
||||
.sf_code_macro, .sf_code_entity, .sf_code_operator, .sf_code_url {
|
||||
color: #920448; }
|
||||
|
||||
.note-wrapper {
|
||||
max-width: 75em;
|
||||
margin: 0px auto;
|
||||
padding: 1.57143em 3.14286em; }
|
||||
|
||||
.note-wrapper.spotlight-preview {
|
||||
overflow-x: hidden; }
|
||||
|
||||
u {
|
||||
text-decoration: none;
|
||||
background-image: linear-gradient(to bottom, rgba(0, 0, 0, 0) 50%, var(--accent-text-color) 50%);
|
||||
background-repeat: repeat-x;
|
||||
background-size: 2px 2px;
|
||||
background-position: 0 1.05em; }
|
||||
|
||||
s {
|
||||
color: #878787; }
|
||||
|
||||
p {
|
||||
margin-bottom: 0.1em; }
|
||||
|
||||
hr {
|
||||
margin-bottom: 0.7em;
|
||||
margin-top: 0.7em; }
|
||||
|
||||
ul li {
|
||||
text-indent: -0.35em; }
|
||||
|
||||
ul li:before {
|
||||
content: "•";
|
||||
color: var(--accent-text-color);
|
||||
display: inline-block;
|
||||
margin-right: 0.3em; }
|
||||
|
||||
ul ul {
|
||||
margin-left: 1.25714em; }
|
||||
|
||||
ol li {
|
||||
text-indent: -1.45em; }
|
||||
|
||||
ol ol {
|
||||
margin-left: 1.25714em; }
|
||||
|
||||
blockquote {
|
||||
display: block;
|
||||
margin-left: -1em;
|
||||
padding-left: 0.8em;
|
||||
border-left: 0.2em solid var(--accent-text-color); }
|
||||
|
||||
.todo-list ul {
|
||||
margin-left: 1.88571em; }
|
||||
|
||||
.todo-list li {
|
||||
text-indent: -1.75em; }
|
||||
|
||||
.todo-list li:before {
|
||||
content: "";
|
||||
display: static;
|
||||
margin-right: 0px; }
|
||||
|
||||
.todo-checkbox {
|
||||
text-indent: -1.7em; }
|
||||
|
||||
.todo-checkbox svg {
|
||||
margin-right: 0.3em;
|
||||
position: relative;
|
||||
top: 0.2em; }
|
||||
|
||||
.todo-checkbox svg #check {
|
||||
display: none; }
|
||||
|
||||
.todo-checkbox.todo-checked #check {
|
||||
display: inline; }
|
||||
|
||||
.todo-checkbox.todo-checked + .todo-text {
|
||||
text-decoration: line-through;
|
||||
color: #878787; }
|
||||
|
||||
.code-inline {
|
||||
display: inline;
|
||||
background: white;
|
||||
border: solid 1px #dedede;
|
||||
padding: 0.2em 0.5em;
|
||||
font-size: 0.9em; }
|
||||
|
||||
.code-multiline {
|
||||
display: block;
|
||||
background: white;
|
||||
border: solid 1px #dedede;
|
||||
padding: 0.7em 1em;
|
||||
font-size: 0.9em;
|
||||
overflow-x: auto; }
|
||||
|
||||
.hashtag {
|
||||
display: inline-block;
|
||||
color: var(--hashtag-text-color);
|
||||
background: var(--hashtag-background-color);
|
||||
padding: 0.0em 0.5em;
|
||||
border-radius: 1em;
|
||||
text-indent: 0; }
|
||||
|
||||
.hashtag a {
|
||||
color: var(--hashtag-text-color); }
|
||||
|
||||
.address a {
|
||||
color: #545454;
|
||||
background-image: linear-gradient(to bottom, rgba(0, 0, 0, 0) 50%, #0da35e 50%);
|
||||
background-repeat: repeat-x;
|
||||
background-size: 2px 2px;
|
||||
background-position: 0 1.05em; }
|
||||
|
||||
.address svg {
|
||||
position: relative;
|
||||
top: 0.2em;
|
||||
display: inline-block;
|
||||
margin-right: 0.2em; }
|
||||
|
||||
.color-preview {
|
||||
display: inline-block;
|
||||
width: 1em;
|
||||
height: 1em;
|
||||
border: solid 1px rgba(0, 0, 0, 0.3);
|
||||
border-radius: 50%;
|
||||
margin-right: 0.1em;
|
||||
position: relative;
|
||||
top: 0.2em;
|
||||
white-space: nowrap; }
|
||||
|
||||
.color-code {
|
||||
margin-right: 0.2em;
|
||||
font-family: "Menlo-Regular";
|
||||
font-size: 0.9em; }
|
||||
|
||||
.color-hash {
|
||||
opacity: 0.4; }
|
||||
|
||||
.ordered-list-number {
|
||||
color: var(--accent-text-color);
|
||||
text-align: right;
|
||||
display: inline-block;
|
||||
min-width: 1em; }
|
||||
|
||||
.arrow svg {
|
||||
position: relative;
|
||||
top: 0.08em;
|
||||
display: inline-block;
|
||||
margin-right: 0.15em;
|
||||
margin-left: 0.15em; }
|
||||
|
||||
.arrow svg #rod {
|
||||
stroke: #545454; }
|
||||
|
||||
.arrow svg #point {
|
||||
fill: #545454; }
|
||||
|
||||
mark {
|
||||
color: inherit;
|
||||
display: inline;
|
||||
padding: 0.2em 0.5em;
|
||||
background-color: var(--highlighter-marker-color); }
|
||||
|
||||
img {
|
||||
max-width: 100%;
|
||||
height: auto; }
|
||||
|
||||
/**
|
||||
Custom styles
|
||||
*/
|
||||
li > p {
|
||||
display: inline-block;
|
||||
margin-left: 16px; }
|
||||
|
||||
hr {
|
||||
background: var(--divider-color);
|
||||
margin-top: 2.6em;
|
||||
margin-bottom: 12px; }
|
||||
|
||||
h4 {
|
||||
font-family: "AvenirNext-Medium";
|
||||
color: var(--title-text-color);
|
||||
font-size: 1.05em;
|
||||
margin-bottom: .47143em; }
|
||||
|
||||
* + p,
|
||||
* + ul,
|
||||
* + ol,
|
||||
* + blockquote {
|
||||
/*margin-top: 1.6em;*/ }
|
||||
|
||||
svg + ul,
|
||||
svg + ol {
|
||||
margin-top: 0; }
|
||||
|
||||
* + h2,
|
||||
* + h3,
|
||||
* + h4 {
|
||||
margin-top: 2.8em; }
|
||||
|
||||
h1, h2, h3, h4, h5, h6 {
|
||||
position: relative; }
|
||||
|
||||
h1:before,
|
||||
h2:before,
|
||||
h3:before,
|
||||
h4:before,
|
||||
h5:before,
|
||||
h6:before {
|
||||
position: absolute;
|
||||
left: -2.2em;
|
||||
color: var(--heading-indicator);
|
||||
font-size: 12px; }
|
||||
|
||||
h1:before {
|
||||
content: "H1"; }
|
||||
|
||||
h2:before {
|
||||
content: "H2"; }
|
||||
|
||||
h3:before {
|
||||
content: "H3"; }
|
||||
|
||||
h4:before {
|
||||
content: "H4"; }
|
||||
|
||||
h5:before {
|
||||
content: "H5"; }
|
||||
|
||||
h6:before {
|
||||
content: "H6"; }
|
||||
|
||||
.note-wrapper {
|
||||
margin-top: 2em; }
|
||||
|
||||
.highlighted {
|
||||
background: var(--selected-text-background-color); }
|
||||
|
||||
.hashtag {
|
||||
cursor: pointer;
|
||||
margin-bottom: 4px;
|
||||
position: relative;
|
||||
margin-right: 2px; }
|
||||
|
||||
.hashtag:before {
|
||||
content: "#"; }
|
||||
|
||||
* + table {
|
||||
margin-top: 12px; }
|
||||
|
||||
table {
|
||||
border-radius: 4px;
|
||||
border: 1px solid var(--separator-color);
|
||||
border-collapse: inherit;
|
||||
overflow: hidden;
|
||||
width: 100%; }
|
||||
|
||||
table th {
|
||||
font-family: "Avenir-Medium", "AvenirLTStd-Medium"; }
|
||||
|
||||
table td,
|
||||
table th {
|
||||
padding: 0.3em 0.8em; }
|
||||
|
||||
table tr:nth-child(odd) {
|
||||
background-color: var(--selected-text-inactive-background-color); }
|
||||
|
||||
pre,
|
||||
p code,
|
||||
li code {
|
||||
border: 1px solid var(--separator-color);
|
||||
padding: 10px;
|
||||
font-size: 12px;
|
||||
background: white;
|
||||
overflow: hidden; }
|
||||
|
||||
li code,
|
||||
p code {
|
||||
padding: 2px; }
|
||||
|
||||
li img {
|
||||
width: 122px; }
|
||||
|
||||
.thumbnail {
|
||||
width: 122px;
|
||||
height: 76px;
|
||||
border: 1px solid var(--separator-color);
|
||||
object-fit: cover; }
|
||||
|
||||
.turbolinks-progress-bar {
|
||||
visibility: hidden; }
|
||||
|
|
@ -0,0 +1 @@
|
|||
{"Target":"css/style.css","MediaType":"text/css","Data":{}}
|
||||
|
|
@ -0,0 +1 @@
|
|||
<svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 640 640" style="enable-background:new 0 0 640 640" xml:space="preserve" width="32" height="32"><path style="fill:#fff" d="m395.9 484.2-126.9-61c-12.5-6-17.9-21.2-11.8-33.8l61-126.9c6-12.5 21.2-17.9 33.8-11.8 17.2 8.3 27.1 13 27.1 13l-.1-109.2 16.7-.1.1 117.1s57.4 24.2 83.1 40.1c3.7 2.3 10.2 6.8 12.9 14.4 2.1 6.1 2 13.1-1 19.3l-61 126.9c-6.2 12.7-21.4 18.1-33.9 12z"/><path style="fill:#609926" d="M622.7 149.8c-4.1-4.1-9.6-4-9.6-4s-117.2 6.6-177.9 8c-13.3.3-26.5.6-39.6.7v117.2c-5.5-2.6-11.1-5.3-16.6-7.9 0-36.4-.1-109.2-.1-109.2-29 .4-89.2-2.2-89.2-2.2s-141.4-7.1-156.8-8.5c-9.8-.6-22.5-2.1-39 1.5-8.7 1.8-33.5 7.4-53.8 26.9C-4.9 212.4 6.6 276.2 8 285.8c1.7 11.7 6.9 44.2 31.7 72.5 45.8 56.1 144.4 54.8 144.4 54.8s12.1 28.9 30.6 55.5c25 33.1 50.7 58.9 75.7 62 63 0 188.9-.1 188.9-.1s12 .1 28.3-10.3c14-8.5 26.5-23.4 26.5-23.4S547 483 565 451.5c5.5-9.7 10.1-19.1 14.1-28 0 0 55.2-117.1 55.2-231.1-1.1-34.5-9.6-40.6-11.6-42.6zM125.6 353.9c-25.9-8.5-36.9-18.7-36.9-18.7S69.6 321.8 60 295.4c-16.5-44.2-1.4-71.2-1.4-71.2s8.4-22.5 38.5-30c13.8-3.7 31-3.1 31-3.1s7.1 59.4 15.7 94.2c7.2 29.2 24.8 77.7 24.8 77.7s-26.1-3.1-43-9.1zm300.3 107.6s-6.1 14.5-19.6 15.4c-5.8.4-10.3-1.2-10.3-1.2s-.3-.1-5.3-2.1l-112.9-55s-10.9-5.7-12.8-15.6c-2.2-8.1 2.7-18.1 2.7-18.1L322 273s4.8-9.7 12.2-13c.6-.3 2.3-1 4.5-1.5 8.1-2.1 18 2.8 18 2.8L467.4 315s12.6 5.7 15.3 16.2c1.9 7.4-.5 14-1.8 17.2-6.3 15.4-55 113.1-55 113.1z"/><path style="fill:#609926" d="M326.8 380.1c-8.2.1-15.4 5.8-17.3 13.8-1.9 8 2 16.3 9.1 20 7.7 4 17.5 1.8 22.7-5.4 5.1-7.1 4.3-16.9-1.8-23.1l24-49.1c1.5.1 3.7.2 6.2-.5 4.1-.9 7.1-3.6 7.1-3.6 4.2 1.8 8.6 3.8 13.2 6.1 4.8 2.4 9.3 4.9 13.4 7.3.9.5 1.8 1.1 2.8 1.9 1.6 1.3 3.4 3.1 4.7 5.5 1.9 5.5-1.9 14.9-1.9 14.9-2.3 7.6-18.4 40.6-18.4 40.6-8.1-.2-15.3 5-17.7 12.5-2.6 8.1 1.1 17.3 8.9 21.3 7.8 4 17.4 1.7 22.5-5.3 5-6.8 4.6-16.3-1.1-22.6 1.9-3.7 3.7-7.4 5.6-11.3 5-10.4 13.5-30.4 13.5-30.4.9-1.7 5.7-10.3 2.7-21.3-2.5-11.4-12.6-16.7-12.6-16.7-12.2-7.9-29.2-15.2-29.2-15.2s0-4.1-1.1-7.1c-1.1-3.1-2.8-5.1-3.9-6.3 4.7-9.7 9.4-19.3 14.1-29-4.1-2-8.1-4-12.2-6.1-4.8 9.8-9.7 19.7-14.5 29.5-6.7-.1-12.9 3.5-16.1 9.4-3.4 6.3-2.7 14.1 1.9 19.8l-24.6 50.4z"/></svg>
|
||||
|
After Width: | Height: | Size: 2.2 KiB |
|
|
@ -2,7 +2,4 @@
|
|||
<p><a href="https://sasserisop.com">Back to Sasserisop homepage</a></p>
|
||||
<h1>{{ .Title }}</h1>
|
||||
{{ partial "content.html" . }}
|
||||
|
||||
|
||||
|
||||
{{ end }}
|
||||
|
|
|
|||
|
|
@ -1,4 +1,6 @@
|
|||
{{ define "main" }}
|
||||
<p style="text-align: center;"><a href="http://sasserisop.com">Back to Sasserisop homepage</a></p>
|
||||
{{ partial "content.html" . }}
|
||||
<p>Check out the source in this <a href="https:/git.sasserisop.com/Sasserisop/MATH201/src/branch/master">Gitea repository <img border="0" src="gitea-logo.svg" width="50" height="50"></a></p>
|
||||
|
||||
{{ end }}
|
||||
|
|
|
|||
Loading…
Reference in New Issue