Integrating Factor Technique: Example

Find all the solutions to

displaymath54

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:

(1)
We rewrite the equation to get

displaymath56

Hence, tex2html_wrap_inline58 and tex2html_wrap_inline60 .

(2)
We have

displaymath62,

which clearly implies that the equation is not exact.

(3)
Let us find an integrating factor. We have

displaymath64.

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

displaymath68

(4)
The new equation is

displaymath70,

which is exact. (Check it!)

(5)
Let us find F(x,y). Consider the system:

displaymath74

(6)
Let us integrate the first equation. We get

displaymath76

(7)
Differentiate with respect to y and use the second equation of the system to get

displaymath80,

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

displaymath88

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

(8)
All the solutions are given by the implicit equation

displaymath92

Remark: Note that if you consider the function

displaymath94,

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

displaymath96

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]
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Author: Mohamed Amine Khamsi

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