A non-exact nonlinear first ODE to solve

[Nipuna Weerasekara]

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?
Please help...

Note:
The given answer is:
x³y² + x²/y = c

 

[Mark44]

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##

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 M_y ≠ N_x, that's enough to show that the equation is not exact.

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?

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.

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?
Please help...

Note:
The given answer is:
x³y² + x²/y = c

 

[LCKurtz]

I think the OP is referring, incorrectly, to the integrating factor method for non-exact equations. 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.

 

[Mark44]

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.

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.

 

[Nipuna Weerasekara]

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