Not a mathematician. I have got a problem of finding by integration a field $F$ from the expression of its gradient. It is a physical problem in continuous space $x,y$ and time $t$ and with real-valued smooth functions.
I first show the vanilla case I coped with; please do correct where applicable. The gradient relationship has the simple form $$ \textrm{grad} F =a \{\cos (\omega t), \sin(\omega t)\} $$ I consider the differential form for the spatial variation: $$ dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy $$ Since the gradient components are independent of space, and of $F$, in my case the formula above becomes the relationship $$ dF = a \cos(\omega t)\, dx + a \sin(\omega t)\,dy $$ I integrate both sides along an arbitrary line on the plane $(x,y)$ and the result is $$ F -F_0 = a \cos(\omega t)\, (x-x_0) + a \sin(\omega t)\,(y-y_0) $$ where $F_0 = F(x_0,y_0)$, which I can set. The surface of $F$ is a rotating plane.
The case that I found difficult is when the gradient depends on $F$, say as simply as this form $$ \textrm{grad} F =aF \{\cos (\omega t), \sin(\omega t)\} $$ The same differential form as above becomes $$ dF = aF \cos(\omega t)\, dx + aF \sin(\omega t)\,dy $$ If I tread the same steps as in the vanilla case above, I do not know how to derive a closed form for $F$ any longer.
Would someone please show and explain the way to derive $F$ from this second gradient expression?
$$\frac{dF}{F} = A \cos( \omega t) dx + A \sin(\omega t) dy $$
$$\int \frac{dF}{F} = a \cos( \omega t) \int dx + a \sin(\omega t)\int dy $$
$$\ln\frac{F}{F_0} = a \cos( \omega t) (x - x_0) + a \sin(\omega t) (y-y_0)$$
If you expand the exponential in a Taylor series around $x_0$, the first two terms of it return the expression for the simpler equation $\textrm{grad} F = a\{\cos(\omega t), \sin(\omega t)\}$.
With thanks to @TedSchifrin for the nudge.