Integrating Factor - Revision history

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Lila at 18:47, 22 September 2021

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Lila: Created page with "When solving first order linear differential equations of the form $\frac{dy}{dx} + p(x)y = g(x)$ , we can utilize the "integrating factor" $\mu$x$$ ..."

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Created page with "When solving first order linear differential equations of the form <math> \frac{dy}{dx} + p(x)y = g(x) </math>, we can utilize the "integrating factor" <math> \mu$x$</math>..."

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When solving first order linear differential equations of the form <math> \frac{dy}{dx} + p(x)y = g(x) </math>, we can utilize the "integrating factor" <math> \mu$x$</math>

==Resources==
* [https://tutorial.math.lamar.edu/classes/de/linear.aspx Solving Linear Equations], Paul's Online Notes

← Older revision		Revision as of 20:05, 16 November 2021
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	<p>Thus we have that for <span class="math-inline"><math>C</math></span> as a constant:</p>		<p>Thus we have that for <span class="math-inline"><math>C</math></span> as a constant:</p>
	<div style="text-align: center;"><math>\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{1}{t} \frac{dy}{dt} - \frac{1}{t^2} y = -e^{-t} \\ \quad \frac{d}{dt} \left ( \frac{y}{t} \right ) = -e^{-t} \\ \quad \int \frac{d}{dt} \left ( \frac{y}{t} \right ) \; dt = -\int e^{-t} \; dt \\ \quad \frac{y}{t} = e^{-t} + C \\ \quad y = te^{-t} + tC \end{align}</math></div>		<div style="text-align: center;"><math>\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{1}{t} \frac{dy}{dt} - \frac{1}{t^2} y = -e^{-t} \\ \quad \frac{d}{dt} \left ( \frac{y}{t} \right ) = -e^{-t} \\ \quad \int \frac{d}{dt} \left ( \frac{y}{t} \right ) \; dt = -\int e^{-t} \; dt \\ \quad \frac{y}{t} = e^{-t} + C \\ \quad y = te^{-t} + tC \end{align}</math></div>
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/the-method-of-integrating-factors The Method of Integrating Factors, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 7: / Line 7: @@
 <div style="text-align: center;"><math>\begin{align} \frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) g(t) \\ \end{align}</math></div>
 <p>The above differential equation can be solved by integrating both sides of the equation with respect to <span class="math-inline"><math>t</math></span> and isolating <span class="math-inline"><math>y</math></span>. The question now arises on how we can find such a function <span class="math-inline"><math>\mu (t)</math></span>.</p>
-<table class="wiki-content-table">
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 <td><strong>Definition:</strong> If <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span> is a first order differential equation, then <span class="math-inline"><math>\mu (t)</math></span> is called an <strong>Integrating Factor</strong> if for <span class="math-inline"><math>\mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t)</math></span> we have that <span class="math-inline"><math>\mu ' (t) = \mu (t) p(t)</math></span>.</td>
-</tr>
+</blockquote>
 <p>The following proposition will give us a formula for obtaining the integrating factor for differential equations in the form <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span>.</p>
-<table class="wiki-content-table">
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 <td><strong>Proposition 1:</strong> If <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span> is a differential equation, then an integrating factor <span class="math-inline"><math>\mu (t)</math></span> of this equation is given by the formula <span class="math-inline"><math>\mu (t) = e^{\int p(t) \; dt}</math></span>.</td>
-</tr>
+</blockquote>
 <ul>
 <li><strong>Proof:</strong> We want to find <span class="math-inline"><math>\mu (t)</math></span> such that <span class="math-inline"><math>\mu ' (t) = \mu (t) p(t)</math></span>. We can rewrite this equation as as <span class="math-inline"><math>\frac{\mu '(t)}{\mu (t)} = p(t)</math></span> and then:</li>

@@ Line 15: / Line 15: @@
 <table class="wiki-content-table">
 <tr>
-<td><strong>Proposition 1:</strong> If <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span> is a differential equation, then an integrating factor <span class="math-inline"><math>\mu (t)</math></span> of this equation is given by the formula <span class="math-inline"><math>\mu (t) = e^{\int p(t) \: dt}</math></span>.</td>
+<td><strong>Proposition 1:</strong> If <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span> is a differential equation, then an integrating factor <span class="math-inline"><math>\mu (t)</math></span> of this equation is given by the formula <span class="math-inline"><math>\mu (t) = e^{\int p(t) \; dt}</math></span>.</td>
 </tr>
 </table>
@@ Line 21: / Line 21: @@
 <li><strong>Proof:</strong> We want to find <span class="math-inline"><math>\mu (t)</math></span> such that <span class="math-inline"><math>\mu ' (t) = \mu (t) p(t)</math></span>. We can rewrite this equation as as <span class="math-inline"><math>\frac{\mu '(t)}{\mu (t)} = p(t)</math></span> and then:</li>
 </ul>
-<div style="text-align: center;"><math>\begin{align} \quad \frac{ \mu '(t)}{ \mu (t)} = p(t) \\ \quad \frac{d}{dt} \ln \mid \mu (t) \mid = p(t) \\ \quad \int \frac{d}{dt} \ln \mid \mu (t) \mid \: dt = \int p(t) \: dt \\ \quad \ln \mid \mu (t) \mid = \int p(t) \: dt \\ \quad \mu (t) = \pm e^{\int p(t) \: dt} \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \frac{ \mu '(t)}{ \mu (t)} = p(t) \\ \quad \frac{d}{dt} \ln \mid \mu (t) \mid = p(t) \\ \quad \int \frac{d}{dt} \ln \mid \mu (t) \mid \; dt = \int p(t) \; dt \\ \quad \ln \mid \mu (t) \mid = \int p(t) \; dt \\ \quad \mu (t) = \pm e^{\int p(t) \; dt} \end{align}</math></div>
 <ul>
-<li>Since we only need one integrating factor to solve differential equations in the form <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span>, we can more generally note that <span class="math-inline"><math>\mu (t) = e^{\int p(t) \: dt}</math></span> is an integrating factor of this differential equation. <span class="math-inline"><math>\blacksquare</math></span></li>
+<li>Since we only need one integrating factor to solve differential equations in the form <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span>, we can more generally note that <span class="math-inline"><math>\mu (t) = e^{\int p(t) \; dt}</math></span> is an integrating factor of this differential equation. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
-<p><em>Notice that from proposition 1 that integrating factors <span class="math-inline"><math>\mu (t)</math></span> are not unique. In fact, there are infinitely many integrating factors. This can be see when evaluating the indefinite integral in <span class="math-inline"><math>\mu (t) = e^{\int p(t) \: dt}</math></span> which will result in getting <span class="math-inline"><math>\mu (t) = e^{P(t) + C}</math></span> where <span class="math-inline"><math>P</math></span> is any antiderivative if <span class="math-inline"><math>p</math></span> and where <span class="math-inline"><math>C</math></span> is a constant. We will always use the simplest integrating factor in solving differential equations of this type.</em></p>
+<p><em>Notice that from proposition 1 that integrating factors <span class="math-inline"><math>\mu (t)</math></span> are not unique. In fact, there are infinitely many integrating factors. This can be see when evaluating the indefinite integral in <span class="math-inline"><math>\mu (t) = e^{\int p(t) \; dt}</math></span> which will result in getting <span class="math-inline"><math>\mu (t) = e^{P(t) + C}</math></span> where <span class="math-inline"><math>P</math></span> is any antiderivative if <span class="math-inline"><math>p</math></span> and where <span class="math-inline"><math>C</math></span> is a constant. We will always use the simplest integrating factor in solving differential equations of this type.</em></p>
 <p>Let's now look at some examples of applying the method of integrating factors.</p>
 ===Example 1===
 <p><strong>Find all solutions to the differential equation <span class="math-inline"><math>\frac{dy}{dt} + \frac{2y}{t} = \frac{\sin t}{t^2}</math></span>.</strong></p>
 <p>We first notice that our differential equation is in the appropriate form <span class="math-inline"><math>\frac{dy}{dt} + p(t) y = g(t)</math></span> where <span class="math-inline"><math>p(t) = \frac{2}{t}</math></span> and <span class="math-inline"><math>g(t) = \frac{\sin t}{t^2}</math></span>. We compute our integrating factor as:</p>
-<div style="text-align: center;"><math>\begin{align} \quad \mu (t) = e^{\int p(t) \: dt} = e^{\int \frac{2}{t} \: dt} = e^{2 \ln (t)} = e^{\ln (t^2)} = t^2 \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \mu (t) = e^{\int p(t) \; dt} = e^{\int \frac{2}{t} \; dt} = e^{2 \ln (t)} = e^{\ln (t^2)} = t^2 \end{align}</math></div>
 <p>Thus we have that for <span class="math-inline"><math>C</math></span> as a constant:</p>
-<div style="text-align: center;"><math>\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad t^2 \frac{dy}{dt} + 2t y = \sin t \\ \quad \frac{d}{dt} \left ( t^2 y \right ) = \sin t \\ \quad \int \frac{d}{dt} \left ( t^2 y \right ) \: dt = \int \sin t \: dt \\ \quad t^2 y = -\cos t + C \\ \quad y = \frac{-\cos t}{t^2} + \frac{C}{t^2} \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad t^2 \frac{dy}{dt} + 2t y = \sin t \\ \quad \frac{d}{dt} \left ( t^2 y \right ) = \sin t \\ \quad \int \frac{d}{dt} \left ( t^2 y \right ) \; dt = \int \sin t \; dt \\ \quad t^2 y = -\cos t + C \\ \quad y = \frac{-\cos t}{t^2} + \frac{C}{t^2} \end{align}</math></div>
 ===Example 2===
 <p><strong>Find all solutions to the differential equation <span class="math-inline"><math>\frac{dy}{dt} - \frac{y}{t} + te^{-t} = 0</math></span>.</strong></p>
 <p>We first rewrite our differential equation as <span class="math-inline"><math>\frac{dy}{dt} - \frac{y}{t} = - te^{-t}</math></span>. We note that in this form we have <span class="math-inline"><math>p(t) = - \frac{1}{t}</math></span> and <span class="math-inline"><math>g(t) = -t e^{-t}</math></span>. We now find an integrating factor:</p>
-<div style="text-align: center;"><math>\begin{align} \quad \mu (t) = e^{\int p(t) \: dt} = e^{\int -\frac{1}{t} \: dt} = e^{-\ln t} = e^{\ln \left ( \frac{1}{t} \right)} = \frac{1}{t} \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \mu (t) = e^{\int p(t) \; dt} = e^{\int -\frac{1}{t} \; dt} = e^{-\ln t} = e^{\ln \left ( \frac{1}{t} \right)} = \frac{1}{t} \end{align}</math></div>
 <p>Thus we have that for <span class="math-inline"><math>C</math></span> as a constant:</p>
-<div style="text-align: center;"><math>\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{1}{t} \frac{dy}{dt} - \frac{1}{t^2} y = -e^{-t} \\ \quad \frac{d}{dt} \left ( \frac{y}{t} \right ) = -e^{-t} \\ \quad \int \frac{d}{dt} \left ( \frac{y}{t} \right ) \: dt = -\int e^{-t} \: dt \\ \quad \frac{y}{t} = e^{-t} + C \\ \quad y = te^{-t} + tC \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{1}{t} \frac{dy}{dt} - \frac{1}{t^2} y = -e^{-t} \\ \quad \frac{d}{dt} \left ( \frac{y}{t} \right ) = -e^{-t} \\ \quad \int \frac{d}{dt} \left ( \frac{y}{t} \right ) \; dt = -\int e^{-t} \; dt \\ \quad \frac{y}{t} = e^{-t} + C \\ \quad y = te^{-t} + tC \end{align}</math></div>

@@ Line 1: / Line 1: @@
 When solving first order linear differential equations of the form <math> y' + p(x)y = g(x) </math>, we can utilize the "integrating factor" <math> \mu (x) = e^{\int p(x)dx}</math>.
-Steps to solving an equation of the form \frac{dy}{dx} + p(x)y = g(x):
+Steps to solving an equation of the form <math> \frac{dy}{dx} + p(x)y = g(x) </math>:
-# Find the integrating factor <math> \mu (x) = e^{\int p(x)dx}</math>, and note that <math> \mu '(x) = p(x)e^{\int p(x)dx} = p(x)\mu (x)</math>,
+# Find the integrating factor <math> \mu (x) = e^{\int p(x)dx} </math>, and note that <math> \mu '(x) = p(x)e^{\int p(x)dx} = p(x)\mu (x)</math>,
 # Multiply both sides of the equation by the integrating factor.
-# The left side of the equation, <math> y'\mu (x) + p(x)\mu (x)y </math>, can now be rewritten as <math> (\mu (x)y)' </math> since <math> y'\mu (x) + p(x)\mu (x)y = y'\mu (x) + \mu '(x)y</math>. Verify by taking the derivative of <math> \mu (x)y </math> with respect to x with the product rule.
+# The left side of the equation, <math> y'\mu (x) + p(x)\mu (x)y </math>, can now be rewritten as <math> (\mu (x)y)' </math> since <math> y'\mu (x) + p(x)\mu (x)y = y'\mu (x) + \mu '(x)y </math>. Verify by taking the derivative of <math> \mu (x)y </math> with respect to x with the product rule.
 # Now, integrate <math> (\mu (x)y)' = g(x)\mu (x)</math> to get <math> \mu (x)y = \int g(x)\mu (x)dx </math>.
 # Solve for y.
 ==Resources==
 * [https://tutorial.math.lamar.edu/classes/de/linear.aspx Solving Linear Equations], Paul's Online Notes

@@ Line 1: / Line 1: @@
-When solving first order linear differential equations of the form <math> \frac{dy}{dx} + p(x)y = g(x) </math>, we can utilize the "integrating factor" <math> \mu (x) = e^{\int p(x)dt}</math>.
+When solving first order linear differential equations of the form <math> y' + p(x)y = g(x) </math>, we can utilize the "integrating factor" <math> \mu (x) = e^{\int p(x)dx}</math>.
 ==Resources==
 * [https://tutorial.math.lamar.edu/classes/de/linear.aspx Solving Linear Equations], Paul's Online Notes

Integrating Factor - Revision history

Lila at 20:05, 16 November 2021

Lila at 20:04, 16 November 2021

Lila at 20:02, 16 November 2021

Lila at 20:02, 16 November 2021

Lila at 18:47, 22 September 2021

Lila at 17:50, 22 September 2021

Lila at 17:36, 22 September 2021

Lila at 17:34, 22 September 2021

Lila: Created page with "When solving first order linear differential equations of the form \frac{dy}{dx} + p(x)y = g(x) , we can utilize the "integrating factor" \mu\(x\)..."

Lila: Created page with "When solving first order linear differential equations of the form $\frac{dy}{dx} + p(x)y = g(x)$ , we can utilize the "integrating factor" $\mu\(x\)$ ..."