# Confusion about notation and convention regarding differentials as objects and operators

1. Jun 9, 2015

### hideelo

I am currently reading "Differential Equatons with Applications" by Ritger and Rose, and I need some clarification about some notation and convention that they are using. I think it all stems from a lack of clarity of the difference between the operator d/dx and the "object" (I dont know what to call it, I dont really know what it is) dx

If I have an equation of the following form

d(f(x,y)) = h(x,y)dx

the approach they then take is to "integrate both sides" and they remain with the following equation

f(x,y) = g(x,y) + c where g(x,y) = ∫h(x,y)dx

the integration on the right is clearly with respect to x, I am not so clear as to what they are doing on the left however. It seems like they are asserting that the integral of the derivative of a function is the function itself. But with respect to what are they integrating it? Is this only with respect to x? If so then the answer is wrong since ∫ d(f(x,y))dx is in general not f(x,y) . Is this some sort of line integral and we are using gradient theorem? If so are we doing the same thing on the other side of the equation? I dont think so, as we are only integrating with respect to x

The other confusion comes from the following, if have the same initial equation as above, d(f(x,y)) = h(x,y)dx is that the same as (d/dx)(f(x,y)) = h(x,y)? If yes, how so? How about if I have something of the following form ∂/∂x (f(x,y)) = h(x,y)?

TIA

2. Jun 9, 2015

### haruspex

I'm not sure that is meaningful in isolation. It certainly looks like it might mean ∂/∂x (f(x,y)) = h(x,y), but I can imagine other circumstances. E.g. we might be considering stepping a small distance along a path in the XY plane, so that in general d(f(x,y)) = h(x,y)dx+k(x,y).dy, but that it happens in this case that k(x,y) (at some point) is zero.
Assuming it does mean ∂/∂x (f(x,y)) = h(x,y), we can integrate wrt x to get f(x,y)=∫h(x,y).dx+k(y), where k is a 'function of integration', analogous to a constant of integration.
∫ d(f(x,y))dx does not parse. You mean just ∫ d(f(x,y)).