Integrating Factor Technique: Example

Find all the solutions to


Solution: Note that this equation is in fact homogeneous. But let us use the technique of exact and nonexact to solve it. Let us follow these steps:

We rewrite the equation to get


Hence, tex2html_wrap_inline58 and tex2html_wrap_inline60 .

We have


which clearly implies that the equation is not exact.

Let us find an integrating factor. We have


Therefore, an integrating factor u(x) exists and is given by


The new equation is


which is exact. (Check it!)

Let us find F(x,y). Consider the system:


Let us integrate the first equation. We get


Differentiate with respect to y and use the second equation of the system to get


which implies tex2html_wrap_inline82 , that is, tex2html_wrap_inline84 is constant. Therefore, the function F(x,y) is given by


We don't have to keep the constant C due to the nature of the solutions (see next step).

All the solutions are given by the implicit equation


Remark: Note that if you consider the function


then we get another integrating factor for the same equation. That is, the new equation


is exact. So, from this example, we see that we may not have uniqueness of the integrating factor. Also, you may learn that if the integrating factor is given to you, the only thing you have to do is multiply your equation and check that the new one is exact.

[Differential Equations] [First Order D.E.] [Integrating Factor Technique]
[Geometry] [Algebra] [Trigonometry ]
[Calculus] [Complex Variables] [Matrix Algebra]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Author: Mohamed Amine Khamsi

Copyright 1999-2018 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour