## Integrating Factors

### Video List

A01  integrating factor 1

A05  A06  A07  A08  integrating factor 2

PatrickJMT Videos

A02  integ factor 1

A04  integ factor 2

A03  integrating factor 1

A09  integrating factor 2

Classify

Form

first order

$$\displaystyle{ y' + p(t)y = g(t) }$$

linear

$$\displaystyle{ M(x,y)dx + N(x,y)dy = 0 }$$ not exact

The first type of integrating factor we will look at 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
$$\displaystyle{ \mu(t) = exp \int{ p(t)~dt } }$$       [ 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.

Video 1 - - Here is a good introduction to integrating factors that is not too long.   open video

These next two videos are related and it is best to watch them in order, since the second one builds on the first.
Video 2 - - In this video, he works an example first and then explains integrating factors.   open video

Video 3 - - In this video, he expands on his discussion giving more detail. In both videos, he explains how the product rule from calculus relates to this technique.   open video

In the above discussion, we required that the differential equation be in the form $$\displaystyle{ \frac{dy}{dt} + p(t)y = g(t) }$$ in order for the integrating factor $$\displaystyle{ \mu(t) = exp \int{ p(t)~dt } }$$ to work. There is another type of equation where an integrating factor can be used to convert an equation to an exact equation.

Integrating Factors For Inexact Equations

Note - - If you haven't studied exact differential equations yet, not all of this discussion will make sense. However, it may help you to scan this and then come back later, after you study exact equations, and read this more thoroughly.

If you have an equation of the form $$M(x,y)dx + N(x,y)dy = 0$$, and this is NOT an exact equation, i.e. $$M_y \neq N_x$$, then there are two possible integrating factors that convert the differential equation to an exact equation.

 $$\displaystyle{ \mu_1 = exp \int{ \frac{M_y - N_x}{N} dx } }$$ $$\displaystyle{ \mu_2 = exp \int{ \frac{M_y - N_x}{-M} dx } }$$

The trick comes in when you are asked to evaluate these integrals. The evaluation is not always possible and can get quite messy. Regardless, these integrating factors can be quite useful.
Okay, it's time for some videos.

These two videos are sequential, so it is best to watch them in this order.
Video 4 - - In this video, he starts an in-depth example where he calculates an integrating factor for a specific inexact equation. He finishes the example in the next video.   open video

Video 5 - - Here is the completion of the example he starts in the previous video. He calculated an integrating factor in the above video and he finishes the problem, which is now exact, in this video.   open video

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Integrating Factors Practice Problems

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. Give your answers in exact form.

 Level A - Basic

Practice A01

$$\displaystyle{ x \frac{dy}{dx} + (x+1)y = 3 }$$;   $$x > 0$$

Practice A02

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

Practice A03

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

Practice Problem [ A01 ] Solution

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

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Practice Problem [ A02 ] Solution

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

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Practice Problem [ A03 ] Solution

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

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Practice A04

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

Practice A05

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

Practice A06

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

Practice A07

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

Practice Problem [ A04 ] Solution

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

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Practice Problem [ A05 ] Solution

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

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Practice Problem [ A06 ] Solution

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

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Practice Problem [ A07 ] Solution

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

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Practice A08

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

Practice A09

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

Practice Problem [ A08 ] Solution

$$\displaystyle{ y = cx }$$

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Practice Problem [ A09 ] Solution

$$\displaystyle{ y = -x(\cos(x)+1) }$$