17Calculus - Integrating Factors for Linear Differential Equations

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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.

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
$$\mu(t) = \exp \int{ p(t)~dt }$$

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

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

Dr Chris Tisdell - integrating factors (2) [11mins-34secs]

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

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

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

Problem Statement

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

Solution

581 video

video by Dr Chris Tisdell

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

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

Problem Statement

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

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

Problem Statement

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

Solution

582 video

video by PatrickJMT

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

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

Problem Statement

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

$$\displaystyle{ y(x) = \frac{1}{3x}(x^3-1) }$$

Problem Statement

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

Solution

590 video

video by Dr Chris Tisdell

$$\displaystyle{ y(x) = \frac{1}{3x}(x^3-1) }$$

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$$\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{-3x} }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{-3x} }$$

$$\displaystyle{ y = e^{-3x}(x^2+C) }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} + 3y = 2xe^{-3x} }$$

Solution

583 video

video by Krista King Math

$$\displaystyle{ y = e^{-3x}(x^2+C) }$$

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$$\displaystyle{ \frac{dy}{dx} - 2xy = x }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} - 2xy = x }$$

$$\displaystyle{ y = -\frac{1}{2} + Ce^{x^2} }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} - 2xy = x }$$

Solution

584 video

video by PatrickJMT

$$\displaystyle{ y = -\frac{1}{2} + Ce^{x^2} }$$

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$$\displaystyle{ \frac{dy}{dx} - 2y = e^{3x} }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} - 2y = e^{3x} }$$

$$y(x) = e^{2x}(e^x+C)$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} - 2y = e^{3x} }$$

Solution

589 video

video by Dr Chris Tisdell

$$y(x) = e^{2x}(e^x+C)$$

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

Problem Statement

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

$$\displaystyle{ y(t) = \frac{-\cos(t)+C}{t^2} }$$

Problem Statement

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

Solution

591 video

video by Dr Chris Tisdell

$$\displaystyle{ y(t) = \frac{-\cos(t)+C}{t^2} }$$

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$$\displaystyle{ \frac{xy'-y}{x^2} = 0 }$$

Problem Statement

$$\displaystyle{ \frac{xy'-y}{x^2} = 0 }$$

$$y = cx$$

Problem Statement

$$\displaystyle{ \frac{xy'-y}{x^2} = 0 }$$

Solution

592 video

video by Dr Chris Tisdell

$$y = cx$$

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

Problem Statement

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

$$y = -x(\cos(x) + 1)$$

Problem Statement

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

Solution

619 video

video by Krista King Math

$$y = -x(\cos(x) + 1)$$

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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

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

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

$$y = -3(x+1) + 5(x+1)^2$$

Problem Statement

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

Solution

1949 video

video by Dr Chris Tisdell

$$y = -3(x+1) + 5(x+1)^2$$

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

Problem Statement

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

$$y(t) = (t+1)(2t^2+1/2)$$

Problem Statement

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

Solution

2081 video

video by MIP4U

$$y(t) = (t+1)(2t^2+1/2)$$

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You CAN Ace Differential Equations

Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

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

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

Engineering

Circuits

Semiconductors

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