I have a 1-form in the $(x,y)$ plane, and I can write it as: $$ \tilde\omega= f \,dA = g \,dB $$ With $f, g,dA,dB \neq 0$.
I want to prove that if the following equality holds then the functions only differ by a multiple or a constant. For example $A= B+ 3$ or $f= \frac{1}{3}g$ and $A=3B$.
But I have problems with the proof. I know that $A$ and $B$ have the same contour lines. But I don't know if that is enough. And for that condition it follows that: $$ dA= \frac{g}{f} \, dB $$ Using the exterior derivative $$ d(dA)=0= d(\frac{g}{f}) \wedge dB $$
And then I am stuck, I think that since I have a wedge product equal to zero, then one is a multiple of the other. $$ d(\frac{g}{f}) =c \, dB $$
$$ \frac{g}{f}= cB + k $$
Where $c, k $ are constants. Now I would like to prove that $c=0$. By contradiction if $c \neq 0 $ Then $$ dA=\frac{1}{c} \frac{g}{f}d(\frac{g}{f}) \frac{2}{2} $$
$$ dA=\frac{1}{2c} \,d((\frac{g}{f})^2)=d(\frac{1}{2c}(\frac{g}{f})^2) $$
$$ A= \frac{1}{2c}(\frac{g}{f})^2 +r $$ where $r$ is a constant. And that is all I got. I think that the very special form of $A$ when $c \neq 0$ will make a contradiction, but I don't know how. Could anyone help me finish the proof or to give a counter-example. Thank you (:
$A=x^2$, $f=x$, $f\,dA=x(2x\,dx)=2x^2\,dx$, $B=x$, $g=2x^2$, $g\,dB=2x^2\,dx$ is a counterexample.