Difference between revisions of "Integrating Factor"

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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>.
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==The Method of Integrating Factors==
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<p>Let <span class="math-inline"><math>p</math></span> and <span class="math-inline"><math>g</math></span> be functions of <span class="math-inline"><math>t</math></span> and consider the following first order differential equation:</p>
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<div style="text-align: center;"><math>\begin{align} \quad \frac{dy}{dt} + p(t) y = g(t) \end{align}</math></div>
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<p>If we multiply the both sides of the equation above by the function <span class="math-inline"><math>\mu (t)</math></span> we get that:</p>
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<div style="text-align: center;"><math>\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \end{align}</math></div>
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<p>If we can guarantee that <span class="math-inline"><math>\mu ' (t) = \mu (t) p(t)</math></span>, then notice that by applying the product rule for differentiation that we get: <span class="math-inline"><math>\frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) \frac{dy}{dt} + \mu ' (t) y</math></span> and substituting <span class="math-inline"><math>\mu ' (t) = \mu (t) y</math></span> and we get that <span class="math-inline"><math>\frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) \frac{dy}{dt} + \mu (t) p(t) y</math></span> which is exactly the lefthand side of the equation above. Thus we get that:</p>
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<div style="text-align: center;"><math>\begin{align} \frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) g(t) \\ \end{align}</math></div>
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<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>
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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<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>
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</blockquote>
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<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>
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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<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>
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</blockquote>
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<ul>
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<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>
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</ul>
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<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>
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<ul>
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<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>
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</ul>
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<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>
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<p>Let's now look at some examples of applying the method of integrating factors.</p>
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===Example 1===
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<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>
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<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>
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<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>
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<p>Thus we have that for <span class="math-inline"><math>C</math></span> as a constant:</p>
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<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>
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===Example 2===
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<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>
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<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>
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<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>
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<p>Thus we have that for <span class="math-inline"><math>C</math></span> as a constant:</p>
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<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>
  
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Steps to solving an equation of the form <math> \frac{dy}{dx} + p(x)y = g(x) </math>:
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==Licensing==
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# 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>,
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Content obtained and/or adapted from:
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# Multiply both sides of the equation by the integrating factor.
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* [http://mathonline.wikidot.com/the-method-of-integrating-factors The Method of Integrating Factors, mathonline.wikidot.com] under a CC BY-SA license
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# 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.
 
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# Now, integrate <math> (\mu (x)y)' = g(x)\mu (x)</math> to get <math> \mu (x)y = \int g(x)\mu (x)dx </math>.
 
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# Solve for y.
 
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Example problem: <math> y' + \frac{2}{t}y = t - 1 + \frac{1}{t} </math>
 
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# <math> \mu (x) = e^{\int \frac{2}{t}dt}  = e^{2\ln{|t|}} = e^{\ln{|t|^2}} = t^2 </math>
 
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# <math> t^2y' + 2ty = t^3 - t^2 + t </math>
 
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# <math> (t^2y)' = t^3 - t^2 + t </math>
 
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# <math> t^2y = \int (t^3 - t^2 + t)dt  = \frac{1}{4}t^4 - \frac{1}{3}t^3 + \frac{1}{2}t^2 + C </math>
 
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# <math> y = \frac{1}{4}t^2 - \frac{1}{3}t + \frac{1}{2} + \frac{C}{t^2} </math>
 
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==Resources==
 
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* [https://tutorial.math.lamar.edu/classes/de/linear.aspx Solving Linear Equations], Paul's Online Notes
 

Latest revision as of 14:05, 16 November 2021

The Method of Integrating Factors

Let and be functions of and consider the following first order differential equation:

If we multiply the both sides of the equation above by the function we get that:

If we can guarantee that , then notice that by applying the product rule for differentiation that we get: and substituting and we get that which is exactly the lefthand side of the equation above. Thus we get that:

The above differential equation can be solved by integrating both sides of the equation with respect to and isolating . The question now arises on how we can find such a function .

Definition: If is a first order differential equation, then is called an Integrating Factor if for we have that .

The following proposition will give us a formula for obtaining the integrating factor for differential equations in the form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dt} + p(t) y = g(t)} .

Proposition 1: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dt} + p(t) y = g(t)} is a differential equation, then an integrating factor Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t)} of this equation is given by the formula Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t) = e^{\int p(t) \; dt}} .

  • Proof: We want to find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t)} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu ' (t) = \mu (t) p(t)} . We can rewrite this equation as as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mu '(t)}{\mu (t)} = p(t)} and then:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}
  • Since we only need one integrating factor to solve differential equations in the form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dt} + p(t) y = g(t)} , we can more generally note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t) = e^{\int p(t) \; dt}} is an integrating factor of this differential equation. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Notice that from proposition 1 that integrating factors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t)} are not unique. In fact, there are infinitely many integrating factors. This can be see when evaluating the indefinite integral in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t) = e^{\int p(t) \; dt}} which will result in getting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t) = e^{P(t) + C}} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P} is any antiderivative if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} and where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} is a constant. We will always use the simplest integrating factor in solving differential equations of this type.

Let's now look at some examples of applying the method of integrating factors.

Example 1

Find all solutions to the differential equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dt} + \frac{2y}{t} = \frac{\sin t}{t^2}} .

We first notice that our differential equation is in the appropriate form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dt} + p(t) y = g(t)} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(t) = \frac{2}{t}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(t) = \frac{\sin t}{t^2}} . We compute our integrating factor as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Thus we have that for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} as a constant:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Example 2

Find all solutions to the differential equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dt} - \frac{y}{t} + te^{-t} = 0} .

We first rewrite our differential equation as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dt} - \frac{y}{t} = - te^{-t}} . We note that in this form we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(t) = - \frac{1}{t}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(t) = -t e^{-t}} . We now find an integrating factor:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Thus we have that for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} as a constant:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Licensing

Content obtained and/or adapted from:

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