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General Integrating Factor

Metronome

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Jun 12, 2018
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Consider the differential equation 2y dx + 3x dy = 0. Determine conditions on a and b so that u(x, y) = (x^a)(y^b) is an integrating factor. Find a particular integrating factor and use it to solve the differential equation.
There is a remote possibility that this is not an intentional exercise, as my book actually states the equation as 2y dx + 3x dx = 0. I have assumed this to be a misprint. This exercise appears in the section on exact differential equations.

The scalar curl of the equation is (3x)/x - (2y)/y = 1. Multiplying by u(x, y) gives 2(x^a)(y^(b + 1)) dx + 3(x^(a + 1))(y^b) dy = 0, which has a scalar curl of (3a - 2b + 5)(x^a)(y^b). (3a - 2b + 5)(x^a)(y^b) = 0 could maybe sorta qualify as conditions on a and b that yield an integrating factor, but only in a very implicit and not very insightful way, such that I don't think this is the answer to the first part of the question.

The equation is homogeneous, so it might be appropriate to make the substitution y = xv; dy = dx v + x dv. This transforms the equation into 5xv dx + 3x^2 dv = 0, which has a scalar curl of x. Dividing through by x to get 5v dx + 3x dv = 0 yields a scalar curl of -2. These two substituted equations multiplied by u(x, y) have scalar curls of (3a - 2b + 1)(x^(a + b + 1))(v^b) and (3a - 2b - 2)(x^(a + b))(v^b), respectively. The integrating factor u(x, y) doesn't seem to simplify things in any case.

Obviously (x^-1)(y^-1) is a particular integrating factor, as it separates the variables, but I'm not sure how to arrive at this conclusion without that insight, given a general integrating factor, as implied by the problem.
 

HallsofIvy

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Jan 27, 2012
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What you are doing seems overly complicated- as if you were trying to follow some "formula" rather than thinking about it.

IF the equation really is "2ydx+ 3xdy= 0" then that is the same as \(\displaystyle \frac{2dx}{x}+ \frac{3dy}{y}= 0\) and each part can now be integrated and so this is an "exact" equation. In other words, an "integrating factor" is \(\displaystyle \frac{1}{xy}=x^{-1}y^{-1}\).

Integrating then gives \(\displaystyle 2 ln(x)+ 3ln(y)= ln(x^2)+ ln(y^3)= ln(x^2y^3)= c\). Taking the exponential of both sides, \(\displaystyle x^3y^3=C\) where \(\displaystyle C= e^c\)
 

Metronome

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Jun 12, 2018
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32
What you are doing seems overly complicated- as if you were trying to follow some "formula" rather than thinking about it.

IF the equation really is "2ydx+ 3xdy= 0" then that is the same as \(\displaystyle \frac{2dx}{x}+ \frac{3dy}{y}= 0\) and each part can now be integrated and so this is an "exact" equation. In other words, an "integrating factor" is \(\displaystyle \frac{1}{xy}=x^{-1}y^{-1}\).

Integrating then gives \(\displaystyle 2 ln(x)+ 3ln(y)= ln(x^2)+ ln(y^3)= ln(x^2y^3)= c\). Taking the exponential of both sides, \(\displaystyle x^3y^3=C\) where \(\displaystyle C= e^c\)
This skips the first part of the problem, at least as I'm reading and interpreting it. They seem to be asking for a general expression for all integrating factors of the form (x^a)(y^b) that would work. Then in the second part of the problem, they ask for a particular integrating factor and a solution, as you gave via separation of variables. It's really the first part that I'm stuck on though.
 

Dr.Peterson

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Nov 12, 2017
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2,769
Think about the equation you got, 2(x^a)(y^(b + 1)) dx + 3(x^(a + 1))(y^b) dy = 0. What is the test for this being an exact differential? Apply that test, and you have a condition for x^a y^b being an integrating factor.

Then you can choose any value of a and determine what b has to be; if you choose a=-1, you'll get HallsofIvy's answer, but they let you find any particular factor you want.
 

Metronome

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Joined
Jun 12, 2018
Messages
32
Think about the equation you got, 2(x^a)(y^(b + 1)) dx + 3(x^(a + 1))(y^b) dy = 0. What is the test for this being an exact differential? Apply that test, and you have a condition for x^a y^b being an integrating factor.

Then you can choose any value of a and determine what b has to be; if you choose a=-1, you'll get HallsofIvy's answer, but they let you find any particular factor you want.
Ahh, looks like one of my attempts was on the right track (with an arithmetic error), but I was focusing on the wrong factors when equating it to 0. I understand it now, thanks!
 
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