Test for Exactness, linearly dependent?

[ombudsmansect]

Hey guys was wondering if anyone knew what the go is with linearly dependent solutions to test for exactness, by that I mean
I have the differential equation (2x + y^2)dx + 4xydy = 0 (M,N)
So i test for exactness and

\partialM/\partialy = 2y \partialN/\partialx = 4y

So I wanted to ask if this does prove exactness, given that they are linearly dependent I cannot find an integrating factor, but I am not sure if I can just proceed to solve anyway

[HallsofIvy]

"linearly independent" or "dependent" has nothing to do with this problem. The given equation is NOT exact because 4y and 2y are not equal. Nor does "dependent" or "independent" have anything to do with you not being able to find an integrating factor.

However, the fact that N_x- M_y is a power of y only might suggest that you look for an integrating factor that is a power of x only: x^n.

[ombudsmansect]

hello again halls thanks for looking again,

I have run through the process to find an integrating factor and it gives me 1/sqrtx. This is also an intuitive guess at an integrating factor but it still yields a non exact solution, even though the formula specifically yields this answer. With integrating fact 1/sqrtx the test yields:

M/y = (2y)/sqrtx N/x=2y

there seems to be something about this relationship that says i cannot equate them. Perhaps there is another way I cannot see

[lurflurf]

1/sqrt(x) is an integrating factor test again

[HallsofIvy]

hello again halls thanks for looking again,

I have run through the process to find an integrating factor and it gives me 1/sqrtx. This is also an intuitive guess at an integrating factor but it still yields a non exact solution, even though the formula specifically yields this answer. With integrating fact 1/sqrtx the test yields:

M/y = (2y)/sqrtx N/x=2y

there seems to be something about this relationship that says i cannot equate them. Perhaps there is another way I cannot see
Then you have an error in your solution. x^{-1/2}= 1/\sqrt{x} certainly is an intgrating factor. Multiplying your equation by that,
(2x^{1/2} + x^{-1/2}y^2)dx + 4x^{1/2}ydy = 0

M_y= 2x^{-1/2}y
and
N_x= 4(1/2)x^{-1/2}y= 2x^{-1/2}y

Now, that means that there exist a function, F(x,y), such that
F_x= 2x^{1/2}+ x^{-1/2}y^3
so that
F= \frac{4}{3}x^{3/2}+ 2x^{1/2}y^3+ \phi(y)

Can you finish it?

[ombudsmansect]

hey guys thanks for your help, it most certainly was an integrating factor as you say, some careless differentiation there on my part. Worked out very nicely in the end with k(y) as 0 so the answer is just 4/3 x^1.5 + 2x^1.5y^3 as you have stated there hallsofivy. thanks again guys