On this page, we will look at integrating factors for firstorder, 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.
The integrating factor we will look at on this page applies to firstorder, 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 firstorder, 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.
video by MIT OCW 

Here are a couple of complete examples using integrating factors to solve firstorder, 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.
video by Dr Chris Tisdell 

video by Dr Chris Tisdell 

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.
\(\displaystyle{ x \frac{dy}{dx} + (x+1)y = 3 }\)
Problem Statement 

\(\displaystyle{ x \frac{dy}{dx} + (x+1)y = 3 }\)
Final Answer 

\(\displaystyle{ y(x) = \frac{3}{x} + \frac{C}{xe^x} }\)
Problem Statement 

\(\displaystyle{ x \frac{dy}{dx} + (x+1)y = 3 }\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\(\displaystyle{ y(x) = \frac{3}{x} + \frac{C}{xe^x} }\) 
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\(\displaystyle{ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4}{(1+x^2)^2} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4}{(1+x^2)^2} }\)
Final Answer 

\(\displaystyle{ y = \frac{4\arctan(x)+C}{1+x^2} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4}{(1+x^2)^2} }\)
Solution 

video by PatrickJMT 

Final Answer 

\(\displaystyle{ y = \frac{4\arctan(x)+C}{1+x^2} }\) 
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\(\displaystyle{ \frac{dy}{dx} + y/x = x }\); \(x > 0\); \(y(1)=0\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} + y/x = x }\); \(x > 0\); \(y(1)=0\)
Final Answer 

\(\displaystyle{ y(x) = \frac{1}{3x}(x^31) }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} + y/x = x }\); \(x > 0\); \(y(1)=0\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\(\displaystyle{ y(x) = \frac{1}{3x}(x^31) }\) 
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\(\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{3x} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{3x} }\)
Final Answer 

\(\displaystyle{ y = e^{3x}(x^2+C) }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{3x} }\)
Solution 

video by Krista King Math 

Final Answer 

\(\displaystyle{ y = e^{3x}(x^2+C) }\) 
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\(\displaystyle{ \frac{dy}{dx}  2xy = x }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx}  2xy = x }\)
Final Answer 

\(\displaystyle{ y = \frac{1}{2} + Ce^{x^2} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx}  2xy = x }\)
Solution 

video by PatrickJMT 

Final Answer 

\(\displaystyle{ y = \frac{1}{2} + Ce^{x^2} }\) 
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\(\displaystyle{ \frac{dy}{dx}  2y = e^{3x} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx}  2y = e^{3x} }\)
Final Answer 

\( y(x) = e^{2x}(e^x+C) \)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx}  2y = e^{3x} }\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y(x) = e^{2x}(e^x+C) \) 
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\(\displaystyle{ t^2\frac{dy}{dt} + 2ty = \sin(t) }\)
Problem Statement 

\(\displaystyle{ t^2\frac{dy}{dt} + 2ty = \sin(t) }\)
Final Answer 

\(\displaystyle{ y(t) = \frac{\cos(t)+C}{t^2} }\)
Problem Statement 

\(\displaystyle{ t^2\frac{dy}{dt} + 2ty = \sin(t) }\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\(\displaystyle{ y(t) = \frac{\cos(t)+C}{t^2} }\) 
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\(\displaystyle{ \frac{xy'y}{x^2} = 0 }\)
Problem Statement 

\(\displaystyle{ \frac{xy'y}{x^2} = 0 }\)
Final Answer 

\( y = cx \)
Problem Statement 

\(\displaystyle{ \frac{xy'y}{x^2} = 0 }\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y = cx \) 
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\( xy' = y + x^2\sin(x) \); \( y(\pi)=0 \)
Problem Statement 

\( xy' = y + x^2\sin(x) \); \( y(\pi)=0 \)
Final Answer 

\( y = x(\cos(x) + 1) \)
Problem Statement 

\( xy' = y + x^2\sin(x) \); \( y(\pi)=0 \)
Solution 

video by Krista King Math 

Final Answer 

\( y = x(\cos(x) + 1) \) 
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\(\displaystyle{ \frac{dy}{dx}  y = e^{3x} }\); \( y(0)=0 \)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx}  y = e^{3x} }\); \( y(0)=0 \)
Solution 

video by Dr Chris Tisdell 

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\(\displaystyle{ \frac{dy}{dx}  \frac{2}{x+1}y = 3 }\), \( y(0) = 2 \)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx}  \frac{2}{x+1}y = 3 }\), \( y(0) = 2 \)
Final Answer 

\( y = 3(x+1) + 5(x+1)^2 \)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx}  \frac{2}{x+1}y = 3 }\), \( y(0) = 2 \)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( y = 3(x+1) + 5(x+1)^2 \) 
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\(\displaystyle{ \frac{dy}{dt} = \frac{y}{t+1} + 4t^2 + 4t }\), \(y(1)=5\), \((t > 1)\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dt} = \frac{y}{t+1} + 4t^2 + 4t }\), \(y(1)=5\), \((t > 1)\)
Final Answer 

\( y(t) = (t+1)(2t^2+1/2) \)
Problem Statement 

\(\displaystyle{ \frac{dy}{dt} = \frac{y}{t+1} + 4t^2 + 4t }\), \(y(1)=5\), \((t > 1)\)
Solution 

video by MIP4U 

Final Answer 

\( y(t) = (t+1)(2t^2+1/2) \) 
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You CAN Ace Differential Equations
external links you may find helpful 

Cliff Notes  Integrating Factor For FirstOrder, Linear Equations 
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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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.