# Looking for the name of a class of ODE:

1. Aug 16, 2007

### Monochrome

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

$$M(x,y)y^{'}(x) + N(x,y) = 0$$
There exists:
$$\phi(x,y)$$
Such that
$$\frac{\partial\phi(x,y)}{\partial x}=N(x,y)$$

$$\frac{\partial\phi(x,y)}{\partial y}=M(x,y)$$

I'm not looking for a solution to anything particular to this but I can't find the type in my notes and I can't google it unless I know the name.

2. Aug 16, 2007

### HallsofIvy

That is an exact equation since
$$d\phi= M(x,y)dy+ NIx,y)dx[/itex] is an exact differential. Of course, since the differential equation says $d\phi= 0$, $\phi(x,y)= 0$ is the general solution. 3. Aug 16, 2007 ### HallsofIvy By the way, in physics, such a differential would correspond to a "conservative force field" and the function $\phi$ would be the "potential function". 4. Aug 17, 2007 ### Monochrome Thanks. Just a question: The I after the N is a typo yes? And how did you get $d\phi= 0$? Last edited: Aug 17, 2007 5. Aug 17, 2007 ### HallsofIvy Yex, that was a typo- my finger was aiming at "("! Since $\phi$ is a function of both x and y, $d\phi /dx$ would make no sense. By the chain rule, if x and y are functions of some third variable, t, [tex]\frac{d\phi}{dt}= \frac{\partial \phi}{\partial x}\frac{dx}{dt}+ \frac{\partial \phi}{\partial y}\frac{dy}{dt}$$
or, in differential notation,
$$d\phi= \frac{\partial \phi}{\partial x}dx+ \frac{\partial \phi}{\partial y}dy$$

6. Aug 17, 2007

### Monochrome

:rofl:

Thanks I forgot that this was dealing with partials. It makes sense now.