I was asked to prove the following statement:
Suppose $P(x,y),Q(x,y)$ are both $n$-th homogeneous functions ($n\ne-1$), i.e., $$P(tx,ty)=t^nP(x,y),\, Q(tx,ty)=t^nQ(x,y)\quad\forall t\in\Bbb R$$ and that $$P(x,y)dx+Q(x,y)dy=0$$ is a total differentiation equation, prove that its solution is $$xP(x,y)+yQ(x,y)=C$$ where $C$ is an arbitrary constant.
My approach, however, seemed to disagree with this conclusion:
Since $Pdx+Qdy=0$ is a total differentiation equation, we have $P_y=Q_x$. Suppose $dU=Pdx+Qdy$, then the solution is clearly $U=C$. Therefore $U=xP+yQ$ and
$$U_x=P+xP_x+yQ_x=P+xP_x+yP_y$$
But $U_x=P$ according to $dU=Pdx+Qdy$, so all we have to do is to prove
$$xP_x+yP_y=0$$
However, according to Euler's Homogeneous Function Theorem, we ought to have
$$xP_x+yP_y=nP$$
which does not make sense to me..
Is there anything wrong in my approach? Or the statement itself is just not correct?