Finding a differentiable function with a particular property If $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable and satisfies $$2 \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=0,$$ show that there exists a differentiable function $\widetilde{f}:\mathbb{R} \to \mathbb{R}$ such that $f(x,y)=\widetilde{f}(x-2y)$ for every $(x,y) \in \mathbb{R}^2$.  
A hint to the problem says to show that for suitable $a,b,c,d \in \mathbb{R}$, the function $F(u,v)=f(au+bv,cu+dv)$ is independent of $u$.  By the chain rule, we have that $\frac{\partial F}{\partial u}=a\frac{\partial f}{\partial x}+c \frac{\partial f}{\partial y}$, so by the hypotheses, $\frac{\partial F}{\partial u}=(a-2c)\frac{\partial f}{\partial x}$.  Hence, $F$ as defined is independent of $u$ if $a=2c$.  
My question is, how can I use this information to obtain the function $\widetilde{f}$ with the required property?

Hint: As you observed, we may write $f(2u + v, u + v) = g(v)$ for some differentiable function $g$ (because it is independent of $u$). Now if you find $u,v$ such that $2u + v = x$ and $u + v = y$, then you can write $f(x,y) = g(v)$.   
