# A non-exact nonlinear first ODE to solve

## Homework Statement

Solve the following equation.

## Homework Equations

( 3x2y4 + 2xy ) dx + ( 2x3y3 - x2 ) dy = 0

## The Attempt at a Solution

M = ( 3xy4 + 2xy )
N = ( 2x3y3 - x2 )

∂M/∂y = 12x2y3 + 2x
∂N/∂x = 6x2y3 - 2x

Then this equation looks like that the integrating factor is (xM-yN).
IF = x3y4 + 3x2y

then the new equation would be:

(3xy^3+2)/(x^2y^3+3x)dx + (2xy^3-1)/(xy^4+3y)dy = 0

but then if we find that this new equation is exact or not, it proves that this is not exact.
If so what should I do?

Note:
x3y2 + x2/y = c

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Mark44
Mentor

## Homework Statement

Solve the following equation.

## Homework Equations

( 3x2y4 + 2xy ) dx + ( 2x3y3 - x2 ) dy = 0

## The Attempt at a Solution

M = ( 3xy4 + 2xy )
The above should be $M = 3x^2y^4 + 2xy$
Nipuna Weerasekara said:
N = ( 2x3y3 - x2 )

∂M/∂y = 12x2y3 + 2x
∂N/∂x = 6x2y3 - 2x
You have a mistake above.
If $M = 3x^2y^4 + 2xy$ then $M_y = 12x^2y^3 + 2x$
Since My ≠ Nx, that's enough to show that the equation is not exact.

Nipuna Weerasekara said:
Then this equation looks like that the integrating factor is (xM-yN).
IF = x3y4 + 3x2y
I taught diff. equations a number of times, but this isn't a trick that I remember. Where does xM - yN come from?
Nipuna Weerasekara said:
then the new equation would be:

(3xy^3+2)/(x^2y^3+3x)dx + (2xy^3-1)/(xy^4+3y)dy = 0
I'm not following this at all. If you can find an integrating factor, you multiply both sides of the original diff. equation by it. If it really is an integrating factor, then the new equation will be exact, and you can get the solution.
Nipuna Weerasekara said:
but then if we find that this new equation is exact or not, it proves that this is not exact.
If so what should I do?

Note:
x3y2 + x2/y = c

LCKurtz
Homework Helper
Gold Member
Mark44
Mentor
I think the OP is referring, incorrectly, to the integrating factor method for non-exact equations.
I understand that he/she was trying to come up with an integrating factor. The link below presents the usual technique to start with when you're faced with an inexact equation: seeing if a function of x ($\mu(x)$) in the PDF below, and proceeding to a function of y alone ($\nu(y)$) if the previous try didn't work. What threw me was the immediate jump to xM - yN, which as you point out, is not the integrating factor.
LCKurtz said:
I haven't worked through this problem, but the method is described in the link http://users.math.msu.edu/users/sen/math_235/lectures/lec_5.pdf on pages 3 - 5.

There was a misinterpretation in the question by me... I solved it by myself. Thanks for the concern.