How to solve the differential equation ${\rm d}f_p(v) = A(p,v) f(p)$ in a manifold? Let $M$ be a smooth manifold and $f:M\to \Bbb R$ be smooth, such that  $${\rm d}f_p(v) = A(p,v) f(p), $$for all $v\in T_pM$, for some smooth coefficient  $A$. How to solve this equation rigorously? I know the answer should be of the form $f(p) = ce^{\lambda(p)} $ for some $c\in \Bbb R$ and $\lambda:M\to \Bbb R$ smooth.
I have a few ideas but they all feel like hand-waving. Sort as "let's solve for $f\circ \alpha$ where $\alpha$ is arbitrary" (say.. $(f\circ\alpha)'(t) = A(\alpha'(t)) (f\circ\alpha)(t)$ has solutions of the form $f(\alpha(t)) = ce^{\lambda(t)}$, then write $p$ instead of $t$), but needless to say, I'm unhappy with this.
 A: Write in local coordinates $(U;x^i)$, 
$$\frac{\partial f}{\partial x^i}=A_i(x)f(x),$$
where $A_i$'s are smooth. You can solve the system of PDEs if and only if Frobenius conditions hold.
Also, $A(p,v)$ should be linear in $v$.
A: An alternative approach is as follows: As mentioned in the answer of @EclipseSun, you have to assume that $A(p,v)$ is linear in $v$. Then this defines a one-form, let me call it $\alpha$, and your  equation becomes $df= f\alpha$. Assuming that $f$ is non-zero, you can write this as $\frac{df}{f}=\alpha$, i.e. as $d\log(f)=\alpha$. Thus you get a necessary condition for existence of a solution, namely $d\alpha=0$, i.e. $\frac{\partial A_i}{\partial x^j}=\frac{\partial A_j}{\partial x^i}$ for all $i$ and $j$. If this is satsified, then locally $\alpha=d\lambda$ for a smooth function $\lambda$ (which is determined up to an additive constant). This gives a solution of the expected form $f(x)=Ce^{\lambda(x)}$. If you are looking for global solutions, you need that in addition to $d\alpha=0$ the class of $\alpha$ in the first de-Rham cohomology is trivial. 
