Homogeneous Differential Equations of First Order and First Degree

Homogeneous function. A function $f(x,y)$ is called a homogeneous function of degree $n$ if, for every $\lambda \neq 0$,

For example, $f(x,y) = x^2 + xy + y^2$ satisfies $f(\lambda x, \lambda y) = \lambda^2 x^2 + \lambda^2 xy + \lambda^2 y^2 = \lambda^2 f(x,y)$, so it is homogeneous of degree $2$.

Homogeneous differential equation. A first-order, first-degree differential equation $\dfrac{dy}{dx} = \dfrac{f(x,y)}{g(x,y)}$ is called homogeneous if $f$ and $g$ are both homogeneous of the same degree $n$. In that case the ratio $f(x,y)/g(x,y)$ is itself homogeneous of degree $0$, i.e. it depends on $x$ and $y$ only through their ratio $y/x$: writing $y = vx$,

a function of $v = y/x$ alone. This is the test used in practice: write the right-hand side in terms of $y/x$; if $x$ and $y$ cancel down to a function of $v = y/x$ only, the equation is homogeneous.

The substitution $y = vx$. Because the right-hand side is a pure function of $v = y/x$, the substitution $y = vx$ turns the equation into one in $v$ and $x$ alone -- and that new equation always separates. Differentiating $y = vx$ with respect to $x$ (product rule, since $v$ is now itself a function of $x$):

Substituting both this and $\dfrac{dy}{dx} = F(v)$ into the original equation gives

which is separable in $v$ and $x$:

Carrying out the integration (Section 4's method) and finally substituting $v = y/x$ back gives the general solution in terms of $x$ and $y$.