# Exact de’s

1. Sep 8, 2004

### JonF

If you are testing for exactness in a d.e. that is in the form of Mdx + Ndy = 0

Do you compare M and N for the exactness test?

Or do you compare their partials?

Also when you are actually solving it, do you anti differentiate in respect to a partial the M term (or n). Or do you need to different the functions in front of the operators first?

2. Sep 8, 2004

### HallsofIvy

Staff Emeritus
Why in the world would you be asking this question? If you are learning about "exact differential equations" you surely must have a textbook that explains all that. Sit down with the textbook and work out the examples it gives. It is a really bad idea to think of mathematics in terms of "do this, then do that,...". You need to understand why you do those things.

The whole point of an exact differential is that there exist a function,F(x,y), such that dF= M(x,y)dx+ N(x,y)dy. You should have learned in calculus that
$df= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy$.

That means we must have $\frac{\partial F}{\partial x}= M(x,y)$ and
$\frac{\partial F}{\partial y}= N(x,y)$ for some function F. Assuming that's true look at the "mixed" derivatives of F: Differentiating F first with respect to x and then y, we get $\frac{\partial^2 F}{\partial y \partial x}= \frac{\partial M}{\partial y}[\itex] while differentiating F first with respect to y and then x, we get [itex]\frac{\partial^2 F}{\partial x \partial y}= \frac{\partial N}{\partial x}[\itex]. But, as long as the derivatives are continuous, those mixed derivatives must be equal- if such an F exists, then we must have [itex]\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}[\itex]. That's sometimes called the "cross-derivative" test. You "compare their partials" as you put it (but be sure to compare the right ones!). If have no idea what you mean by "different(iate) the functions in front of the operators first". Solving the equation is basically, finding F. You know that [itex]\fraction{\partial F}{\partial x}= M so if you anti-differentiate (treating y as a constant) you get F(x,y)= (anti-derivative of M wrt x)+ "constant" except that, since you are treating y as a constant, the "constant" may depend on y: F= (anti-derivative of M wrt x)+ g(y) where g is some unknown function of y. How do you find g? Differentiate F with respect to y: [itex]\frac{\partial F}{\partial y}= \frac{\partial (anti-derivative of M wrt x)}{\partial y}+ g'(y)$ and that must be equal to N.

3. Sep 8, 2004

### JonF

The text book assumes I’ve have completed calc3, which I am only currently taking.