First-Order Substitutions

Definition A Bernoulli equation is an equation of the form

where $n$ is a real number and $P$ and $Q$ are continuous on an open interval $I$. For $n = 0$ and $n = 1$ it is already linear, so the interesting cases are $n \neq 0, 1$. When $n$ is not an integer, $y^n$ is taken for $y > 0$.

Definition Let $n \neq 0, 1$ and let $y$ be a solution of the Bernoulli equation on an interval where $y \neq 0$ (and $y > 0$ when $n$ is not an integer). Then $v = y^{1-n}$ satisfies the linear equation

Conversely, a solution $v$ of this linear equation with $v > 0$ gives the solution $y = v^{1/(1-n)}$ of the Bernoulli equation, and when $1 - n$ is an odd integer any nonzero $v$ will do. When $n$ is an odd integer, $1 - n$ is even and $y = -v^{1/(1-n)}$ is a solution as well. The proof is one line, since $v' = (1 - n)y^{-n}y'$, and multiplying the Bernoulli equation by $(1 - n)y^{-n}$ gives the equation for $v$.

Definition A first order equation is called homogeneous (in the sense of degree zero) if it can be written as

with $F$ continuous on an interval. For $x \neq 0$, an equation $y' = f(x, y)$ has this form exactly when $f(tx, ty) = f(x, y)$ for all $t \neq 0$, and then $F(v) = f(1, v)$. On an interval where $x \neq 0$, the substitution $v = \frac{y}{x}$, that is $y = vx$ and $y' = v + xv'$, turns it into the separable equation

Definition The word homogeneous has two unrelated meanings. A linear equation $y' + P(x)y = Q(x)$ is homogeneous when $Q = 0$. An equation $y' = f(x, y)$ is homogeneous of degree zero when $f(tx, ty) = f(x, y)$. Neither property implies the other.

Definition A scaling substitution replaces the variables by constant multiples, $x = \alpha s$ and $y = \beta u$ with $\alpha, \beta \neq 0$. Then $\frac{dy}{dx} = \frac{\beta}{\alpha}\frac{du}{ds}$, so $y' = f(x, y)$ becomes

Choosing $\alpha$ and $\beta$ well removes constants from the equation. Degree zero homogeneity is exactly the statement that $y' = f(x, y)$ does not change under the scaling $x = ts$, $y = tu$, which is why $v = \frac{y}{x}$ works. The same idea goes further. Suppose the equation does not change under $x = ts$, $y = t^m u$ for some fixed power $m$ and every $t > 0$, which means $f(tx, t^m y) = t^{m-1}f(x, y)$. Then on $x > 0$ the substitution $y = vx^m$ turns it into the separable equation

Definition A substitution is valid only where it is defined and can be inverted. The substitution $v = y^{1-n}$ needs $y \neq 0$ (and $y > 0$ for noninteger $n$), and $v = \frac{y}{x}$ needs $x \neq 0$. Solutions outside these sets must be found separately. In particular, for $n > 0$ the function $y = 0$ solves every Bernoulli equation, and for $y' = F\left(\frac{y}{x}\right)$ every root $v_0$ of $F(v_0) = v_0$ gives the line $y = v_0 x$, which separation of $xv' = F(v) - v$ discards.