17calculus 17calculus
First Order Second Order Laplace Transforms Additional Topics Applications, Practice
Separation of Variables
Linear
Integrating Factors (Linear)
Substitution
Exact Equations
Integrating Factors (Exact)
Linear
Constant Coefficients
Substitution
Reduction of Order
Undetermined Coefficients
Variation of Parameters
Polynomial Coefficients
Cauchy-Euler Equations
Chebyshev Equations
Laplace Transforms
Unit Step Function
Unit Impulse Function
Square Wave
Shifting Theorems
Solve Initial Value Problems
Classify Differential Equations
Fourier Series
Slope Fields
Wronskian
Existence and Uniqueness
Boundary Value Problems
Euler's Method
Inhomogeneous ODE's
Resonance
Partial Differential Equations
Linear Systems
Exponential Growth/Decay
Population Dynamics
Projectile Motion
Chemical Concentration
Fluids (Mixing)
Practice Problems
Practice Exam List
Exam A1
Exam A3
Exam B2

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Integrating Factors for Linear 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.

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

Here are a couple of complete examples using integrating factors to solve first-order, linear differential equations.

Dr Chris Tisdell - integrating factors

Okay, time for some practice problems.

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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 also. Give your answers in exact form.

Level A - Basic

Practice A01

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

answer

solution

Practice A02

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

answer

solution

Practice A03

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

answer

solution

Practice A04

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

answer

solution

Practice A05

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

answer

solution

Practice A06

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

answer

solution

Practice A07

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

answer

solution

Practice A08

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

answer

solution

Practice A09

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

answer

solution

Practice A10

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

solution

Practice A11

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

answer

solution

Practice A12

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

answer

solution

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