# Diff. Eq. Show that the following equation is not exact.

1. Mar 2, 2014

### Jtechguy21

1. The problem statement, all variables and given/known data

Show that the following equation is not exact.
Then find and simplify the integrating factor that makes the equation exact.(You do not have to solve the equation)

(Y^2 - x)dx + (4xy)dy=0

2. Relevant equations

3. The attempt at a solution

M(x,y) =(Y^2 - x)dx N(x,y)=(4xy)dy

Partial Derivative of m with respect to y is 2y

Partial Derivative of N with respect to x is xy

^shows they are not exact.
how exactly do i find this integrating factor that makes them exact?

2. Mar 2, 2014

### LCKurtz

If your text doesn't tell you how, look here:

http://www.cliffsnotes.com/math/differential-equations/first-order-equations/integrating-factors

3. Mar 2, 2014

### Jtechguy21

yes!!! thank you so much. I figured it out with the link you gave me.
My integrating factor was x^-(1/2) yay :)
they are exact now.

I found it really strange my book does not show me how to find the IF for exact equations, and the homework examples already have the IF given. hmmm. anyways.

can you answer another question for me. In the link I used case 1. but i am unsure when i need to use case2. the conditions are nearly identical.

Consider the differential equation M dx + N dy = 0. If this equation is not exact, then M y will not equal N x ; that is, M y – N x ≠ 0. However, if

case1 is a function of x only
(My-Nx)/(N)

Case2 is a function of y only
(My-Nx)/(-M)

Last edited: Mar 2, 2014
4. Mar 3, 2014

### LCKurtz

If the equation is exact so $M_y=N_x$, you don't need an integrating factor. It's ready to go as it is.

You don't have a choice of which, if any, case to use. If you are lucky and get either the function of x or function of y in that test, you can find an integrating factor which will make the equation exact. But an equation might not be exact and neither of the above cases work either. Then you have to resort to other methods to solve it, presuming it can be solved analytically at all. These techniques don't always work.

5. Mar 3, 2014

### Jtechguy21

thank you for explaining this to me :) It all makes sense to me now!