Derivative of homogenous polynomial of two variables

Question

Claim:

If $F(x, y)$ is a polynomial that satisfies $F(tx, ty) = t^n F(x, y)$, for all $t > 0$, then for $F(x, y) = 0$ $$\frac{dy}{dx} = \frac yx$$

Proof:

Let $y = vx$

Now as the polynomial is homogenous, it can be written in the form \begin{align} F(x, y)& =0\\ \sum_{}^{} a_k x^k y^{n-k} &=0 \\ \sum a_k v^{n-k}x^n &=0\\ x^n\sum a_k v^{n-k} &=0\\ \sum a_k v^{n-k} &=0 \end{align}

Now this is a polynomial in $v$, and therefore $v$ is a constant. \begin{align} \frac{dv}{dx} &= 0\\ \frac{dy}{dx} &= v+x \frac{dv}{dx}\\ \frac{dy}{dx} &= v\\ \frac{dy}{dx} &= \frac yx \end{align}

Questions:

1. Is this proof valid? I object because $v$ is multi-valued.
2. How do I guarantee such a $v$ exists? I think that the implicit function theorem may be applied, but I am not sure.
3. Does anyone know of a more direct proof?

Motivation:

If $x^{13}y^7=(x+y)^{20}$, prove that $\frac {dy}{ dx}=\frac yx$

Answer 1

Since $F$ is homogeneous of degree $n$, Euler's Theorem applied to $F$ yields, for all $x,y$ such that $F(x,y)=0$, $x\frac{\partial F}{\partial x}$ +$y\frac{\partial F}{\partial y}$=$nF(x,y)$. Hence, $x\frac{\partial F}{\partial x}$ +$y\frac{\partial F}{\partial y}$=$0$.

Along the level set the $F(x,y)=0$, $\frac{dy}{dx} = -\frac{\partial F}{\partial x}/\frac{\partial F}{\partial y}=\frac{y}{x}$ (the last term follows from Euler's Theorem, as needed).

Answer 2

You can certainly write $v=\dfrac yx$ whenever $x\ne 0$. This defines $v$ as a function on the curve $F(x,y)=0$ and it will surely not be a function of $x$. The sentence "Now this is a polynomial in $v$, and therefore $v$ is a constant" requires thought. Away from $x=0$, the curve has the equation $\sum a_kv^{n-k} = 0$ in the $(x,v)$-plane. You're correct that this does not define $v$ as a function of $x$; indeed, the solution set consists of (up to) $n$ lines $v=\text{constant}$ in the $xv$-plane. But restricting to any one of these lines, $v$ is a bona-fide constant function of $x$ and, so implicit differentiation gives $$\frac{dy}{dx} = v + x \frac{dv}{dx} = v = \frac yx$$ at any point with $(x,v)$ with $x=0$ on this curve.