Suppose we have the equation in terms of differential forms
$$ d\mathcal{Y}=f(C)dC,$$ here $d\mathcal{Y}(x^i)$ is one form, $C(x^i)$ is a scalar, $f(C)$ is a function of $C$ only.
Can it be integrated? Is the most general solution given by $$ \mathcal{Y}=g(C)$$ with some arbitrary function $g(C)$.
I feel like it can be done and simply substituting $\mathcal{Y}=g(C)$ back into the equation we get $$ d\mathcal{Y}=\partial_C g(C) dC,$$ so $\partial_C g(C)=f(C)$. Does it make sense?
We can look at the equation $d\mathcal{Y}=f(C)dC$ as the definition of the exact differential. Then obviously $\mathcal{Y}$ depends only on $C$, so the solution is $\mathcal{Y}=g(C)$.