Question about exact diff eq

[vfisher05]

Hi, I'm new here and I have a question about exact differential equations and integrating factors.
I was looking at the following online pdf about special integrating factors:
http://www2.fiu.edu/~aladrog/SpecialIntegratingFactors.pdf

and I need help understanding the first example problem, which I'll print here.

1) (4xy + 3y² – x) dx + x(x + 2y) dy = 0

So I get it up to the point where you multiply by the integrating factor you find, x², to get:
(4x³y + 3x²y² – x³) dx + (x⁴ + 2x³y) dy = 0

But I don't understand why the example says this is exact, since when I find the partial derivative of M and N, they don't match.
I get:
M_x=12x²y + 6xy² - 3x²
N_y=2x³
Which don't equal, so I'm not really sure what to do from here.

And why did the example choose to group it like this? How does grouping it in specific ways help?
(4x³y dx + x⁴ dy) + (3x²y² dx + 2x³ydy) – x³ dx = 0

Thanks for any help in advance!

 

[Deleted member 4993]

Hi, I'm new here and I have a question about exact differential equations and integrating factors.
I was looking at the following online pdf about special integrating factors:
http://www2.fiu.edu/~aladrog/SpecialIntegratingFactors.pdf

and I need help understanding the first example problem, which I'll print here.

1) (4xy + 3y² – x) dx + x(x + 2y) dy = 0

So I get it up to the point where you multiply by the integrating factor you find, x², to get:
(4x³y + 3x²y² – x³) dx + (x⁴ + 2x³y) dy = 0

But I don't understand why the example says this is exact, since when I find the partial derivative of M and N, they don't match.
I get:
M_x=12x²y + 6xy² - 3x²
N_y=2x³
Which don't equal, so I'm not really sure what to do from here.

And why did the example choose to group it like this? How does grouping it in specific ways help?
(4x³y dx + x⁴ dy) + (3x²y² dx + 2x³ydy) – x³ dx = 0

Thanks for any help in advance!

Instead, if you had found:

M_y = 4x³ + 6x²y and .............................Corrected

N_x = 4x³ + 6x²y ....................................Corrected

a light-bulb would have gone off.

Remember - it is

\(\displaystyle \displaystyle{\frac{\partial}{\partial x}\left[ \frac{\partial f}{\partial y} \right ] \ = \ \frac{\partial}{\partial y}\left[ \frac{\partial f}{\partial x} \right ]}\)

 

[stapel]

...I need help understanding the first example problem, which I'll print here.

1) (4xy + 3y² – x) dx + x(x + 2y) dy = 0

So I get it up to the point where you multiply by the integrating factor you find, x², to get:
(4x³y + 3x²y² – x³) dx + (x⁴ + 2x³y) dy = 0

But I don't understand why the example says this is exact, since when I find the partial derivative of M and N, they don't match.
I get:
M_x=12x²y + 6xy² - 3x²
N_y=2x³

To be "exact", it has to be the other derivatives that match:

. . . . .\(\displaystyle \dfrac{\partial M}{\partial y}\, =\, 4x^3\, +\, 6x^2y\, -\, 0\, =\, 4x^3\, +\, 6x^2y\)

. . . . .\(\displaystyle \dfrac{\partial N}{\partial x}\, =\, 4x^3\, +\, 6x^2y\)

 

[vfisher05]

Ohhh ok thanks to you both!