$\frac{dy}{dx}=\frac{y}{x}$ for a homogeneous equation in $x,y$?

Question

A problem book I use mentions the following:

...since the curve $32x^3y^2=(x+y)^5$ is homogeneous, $\frac{dy}{dx}=\frac{y}{x}$

I have indeed verified the derivative. Is this a known theorem? I generalized it to $$cx^ay^b=(x+y)^{a+b},$$

but does there exist a more generalized form?

Answer 1

In fact, the rather surprizing relationship

$$\dfrac{dy}{dx}=\dfrac{y}{x}=\text{slope in point } \ (x,y)$$

is easy to understand on the curves of certain of these equations because these "curves" are in fact union of straight lines.

Let us take the case of the general equation you mention;

$$cx^{a}y^{b}=(x+y)^{a+b} \ \ \iff \ \ \left(1+\frac{y}{x}\right)^{a}\left(1+\frac{x}{y}\right)^{b}=c\tag{1}$$

Its curve, when non void, is plainly the union of lines passing through the origin (equations $y/x=t$), where $t$ is any real root of the following equation:

$$\left(1+t\right)^{a}\left(1+\frac{1}{t}\right)^{b}=c \tag{2}$$

which may or may not have real roots.

Let us take the particular case $a=1, \ b=2, \ c=8$ (see figure) where equation (2), a third degree polynomial equation, has $3$ real solutions, among them $t=1$. The resulting "curve" is made of the union of $3$ straight lines ; among them the line with equation $y/x=t=1$).

Answer 2

Hint

For the general problem $$f(x,y)=c\,x^a\,y^b-(x+y)^{a+b}=0$$ $$\frac{\partial f(x,y)}{\partial x}=a c x^{a-1} y^b-(a+b) (x+y)^{a+b-1}$$ $$\frac{\partial f(x,y)}{\partial y}=b c x^a y^{b-1}-(a+b) (x+y)^{a+b-1}$$

Now, write $$(x+y)^{a+b-1}=\frac 1 {x+y}(x+y)^{a+b}=\frac{c\,x^a\,y^b }{x+y}$$ Replace and simplify : the result is in three characters.

This is a direct application of Euler's identity for an homogeneous function as @Jean Marie already commented.