17Calculus - Integrating Factors for Linear Differential Equations
On this page, we will look at integrating factors for first-order, linear equations. (See the exact differential equations page for discussion of integrating factors for inexact equations.) The integrating factor converts the differential equation into a form that can be solved by direct integration.
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The integrating factor we will look at on this page applies to first-order, linear differential equations of the form
\(\displaystyle{ \frac{dy}{dt} + p(t)y = g(t) }\).
The idea of the technique of integrating factors is deceptively simple, yet quite powerful. When you have a first-order, linear differential equation of the form
\(\displaystyle{ \frac{dy}{dt} + p(t)y = g(t) }\)
and you multiply this equation by the generated integrating factor
\( \mu(t) = \exp \int{ p(t)~dt }\)
[ what does exp mean? ]
This converts the differential equation into the form
\(\displaystyle{ \frac{d}{dt}[ \mu(t)y] = \mu(t)g(t) }\)
You can then integrate to get \(\displaystyle{ \mu(t)y }\) and divide by \(\mu(t)\) to solve for \(y\).
(Note: We will not go through the derivation of the integrating factor here at this time. However, going through the derivation in your textbook will really help you understand what is going on here.)
Okay, let's watch some videos, so we can see how this works.
Here is a good introduction to integrating factors that is not too long.
MIT OCW - integrating factors introduction [27mins-8secs]
video by MIT OCW |
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Here are a couple of complete examples using integrating factors to solve first-order, linear differential equations. In the first video, he works an example and explains integrating factors. In the second video, he expands on his discussion, giving more detail, while working a second example. In both videos, he explains how the product rule from calculus relates to this technique.
Dr Chris Tisdell - integrating factors (1) [8mins-18secs]
video by Dr Chris Tisdell |
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Dr Chris Tisdell - integrating factors (2) [11mins-34secs]
video by Dr Chris Tisdell |
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Okay, time for some practice problems.
Practice
Unless otherwise instructed, find the general solutions to these differential equations using the method of integrating factors. If initial condition(s) are given, find the particular solution also. Give your answers in exact form.
\(\displaystyle{ x \frac{dy}{dx} + (x+1)y = 3 }\)
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\(\displaystyle{ x \frac{dy}{dx} + (x+1)y = 3 }\)
Final Answer |
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\(\displaystyle{ y(x) = \frac{3}{x} + \frac{C}{xe^x} }\)
Problem Statement |
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\(\displaystyle{ x \frac{dy}{dx} + (x+1)y = 3 }\)
Solution |
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581 video
video by Dr Chris Tisdell |
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Final Answer |
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\(\displaystyle{ y(x) = \frac{3}{x} + \frac{C}{xe^x} }\) |
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\(\displaystyle{ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4}{(1+x^2)^2} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4}{(1+x^2)^2} }\)
Final Answer |
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\(\displaystyle{ y = \frac{4\arctan(x)+C}{1+x^2} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4}{(1+x^2)^2} }\)
Solution |
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582 video
video by PatrickJMT |
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Final Answer |
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\(\displaystyle{ y = \frac{4\arctan(x)+C}{1+x^2} }\) |
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\(\displaystyle{ \frac{dy}{dx} + y/x = x }\); \(x > 0\); \(y(1)=0\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} + y/x = x }\); \(x > 0\); \(y(1)=0\)
Final Answer |
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\(\displaystyle{ y(x) = \frac{1}{3x}(x^3-1) }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} + y/x = x }\); \(x > 0\); \(y(1)=0\)
Solution |
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590 video
video by Dr Chris Tisdell |
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Final Answer |
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\(\displaystyle{ y(x) = \frac{1}{3x}(x^3-1) }\) |
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\(\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{-3x} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{-3x} }\)
Final Answer |
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\(\displaystyle{ y = e^{-3x}(x^2+C) }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{-3x} }\)
Solution |
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583 video
video by Krista King Math |
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Final Answer |
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\(\displaystyle{ y = e^{-3x}(x^2+C) }\) |
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\(\displaystyle{ \frac{dy}{dx} - 2xy = x }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} - 2xy = x }\)
Final Answer |
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\(\displaystyle{ y = -\frac{1}{2} + Ce^{x^2} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} - 2xy = x }\)
Solution |
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584 video
video by PatrickJMT |
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Final Answer |
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\(\displaystyle{ y = -\frac{1}{2} + Ce^{x^2} }\) |
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\(\displaystyle{ \frac{dy}{dx} - 2y = e^{3x} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} - 2y = e^{3x} }\)
Final Answer |
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\( y(x) = e^{2x}(e^x+C) \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} - 2y = e^{3x} }\)
Solution |
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589 video
video by Dr Chris Tisdell |
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Final Answer |
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\( y(x) = e^{2x}(e^x+C) \) |
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\(\displaystyle{ t^2\frac{dy}{dt} + 2ty = \sin(t) }\)
Problem Statement |
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\(\displaystyle{ t^2\frac{dy}{dt} + 2ty = \sin(t) }\)
Final Answer |
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\(\displaystyle{ y(t) = \frac{-\cos(t)+C}{t^2} }\)
Problem Statement |
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\(\displaystyle{ t^2\frac{dy}{dt} + 2ty = \sin(t) }\)
Solution |
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591 video
video by Dr Chris Tisdell |
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Final Answer |
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\(\displaystyle{ y(t) = \frac{-\cos(t)+C}{t^2} }\) |
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\(\displaystyle{ \frac{xy'-y}{x^2} = 0 }\)
Problem Statement |
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\(\displaystyle{ \frac{xy'-y}{x^2} = 0 }\)
Final Answer |
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\( y = cx \)
Problem Statement |
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\(\displaystyle{ \frac{xy'-y}{x^2} = 0 }\)
Solution |
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592 video
video by Dr Chris Tisdell |
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Final Answer |
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\( y = cx \) |
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\( xy' = y + x^2\sin(x) \); \( y(\pi)=0 \)
Problem Statement |
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\( xy' = y + x^2\sin(x) \); \( y(\pi)=0 \)
Final Answer |
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\( y = -x(\cos(x) + 1) \)
Problem Statement |
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\( xy' = y + x^2\sin(x) \); \( y(\pi)=0 \)
Solution |
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619 video
video by Krista King Math |
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Final Answer |
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\( y = -x(\cos(x) + 1) \) |
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\(\displaystyle{ \frac{dy}{dx} - y = e^{3x} }\); \( y(0)=0 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} - y = e^{3x} }\); \( y(0)=0 \)
Solution |
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1844 video
video by Dr Chris Tisdell |
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\(\displaystyle{ \frac{dy}{dx} - \frac{2}{x+1}y = 3 }\), \( y(0) = 2 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} - \frac{2}{x+1}y = 3 }\), \( y(0) = 2 \)
Final Answer |
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\( y = -3(x+1) + 5(x+1)^2 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} - \frac{2}{x+1}y = 3 }\), \( y(0) = 2 \)
Solution |
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1949 video
video by Dr Chris Tisdell |
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Final Answer |
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\( y = -3(x+1) + 5(x+1)^2 \) |
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\(\displaystyle{ \frac{dy}{dt} = \frac{y}{t+1} + 4t^2 + 4t }\), \(y(1)=5\), \((t > -1)\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dt} = \frac{y}{t+1} + 4t^2 + 4t }\), \(y(1)=5\), \((t > -1)\)
Final Answer |
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\( y(t) = (t+1)(2t^2+1/2) \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dt} = \frac{y}{t+1} + 4t^2 + 4t }\), \(y(1)=5\), \((t > -1)\)
Solution |
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2081 video
video by MIP4U |
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Final Answer |
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\( y(t) = (t+1)(2t^2+1/2) \) |
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You CAN Ace Differential Equations
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Cliff Notes - Integrating Factor For First-Order, Linear Equations |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Topics Listed Alphabetically
Single Variable Calculus |
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Multi-Variable Calculus |
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Differential Equations |
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Precalculus |
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Engineering |
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Circuits |
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Semiconductors |
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Practice Instructions
Unless otherwise instructed, find the general solutions to these differential equations using the method of integrating factors. If initial condition(s) are given, find the particular solution also. Give your answers in exact form.
