You CAN Ace Differential Equations

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

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

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

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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.

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

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

581 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

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

582 solution video

video by PatrickJMT

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

590 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

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

583 solution video

video by Krista King Math

Final Answer

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

close solution

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

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

584 solution video

video by PatrickJMT

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

589 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

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

591 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

\(\displaystyle{ y = cx }\)

Problem Statement

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

Solution

592 solution video

video by Dr Chris Tisdell

Final Answer

\(\displaystyle{ y = cx }\)

close solution

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

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

619 solution video

video by Krista King Math

Final Answer

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

close solution

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

Problem Statement

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

Solution

1844 solution video

video by Dr Chris Tisdell

close solution

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

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

1949 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

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

2081 solution video

video by MIP4U

Final Answer

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

close solution
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