Looking for the name of a class of ODE:

[Monochrome]

Homework Statement

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.

 

[HallsofIvy]

That is an exact equation since
d\phi= M(x,y)dy+ NIx,y)dx[/itex]<br /> is an exact differential.<br /> <br /> Of course, since the differential equation says d\phi= 0, \phi(x,y)= 0 is the general solution.

 

[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".

 

[Monochrome]

Thanks.
Just a question: The I after the N is a typo yes? And how did you get d\phi= 0?

 

[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,
\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

 

[Monochrome]

Yex, that was a typo-

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