First Order ODE/y' - f (y) phi' (x) over f' (y) = phi (x) phi' (x) over f' (y)

Theorem
Let $\map f y$ and $\map \phi x$ be known real functions of $y$ and $x$ respectively.

The general solution of:
 * $(1): \quad \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x = \dfrac {\map \phi x \, \map {\phi'} x} {\map {f'} y}$

is:
 * $\map f y = e^{2 \, \map \phi x} \paren {\map \phi x - 1} + C$

Proof
Let $u = \map f y$

Then by the Chain Rule for Derivatives:
 * $\dfrac {\d u} {\d x} = \map {f'} y \dfrac {\d y} {\d x}$

Then:

This is a linear first order ordinary differential equation in the form:
 * $\dfrac {\d u} {\d x} + \map P x u = \map Q x$

whose general solution from Solution to Linear First Order Ordinary Differential Equation is:
 * $\displaystyle u e^{\int P \rd x} = \int Q e^{\int P \rd x} \rd x + C$

In this instance:
 * $\map P x = -\map {\phi'} x$

and:
 * $\map Q x = \map \phi x \, \map {\phi'} x$

Thus:

Hence:


 * $\displaystyle u e^{-\map \phi x + A} = \int \map \phi x \, \map {\phi'} x e^{-\map \phi x + A} \rd x + C$

Let $v = \map \phi x$.

Then by Integration by Substitution:

This leaves us with:


 * $u e^{-\map \phi x + A} = e^A e^v \paren {v - 1} + C$

subsuming $C'$ into $C$.

Substituting back in, the general solution is seen to be:
 * $\map f y e^{-\map \phi x + A} = e^{\map \phi x + A} \paren {\map \phi x - 1} + C$

and so multiplying both sides by $e^{-\map \phi x + A}$:
 * $\map f y = e^{2 \map \phi x + A - A} \paren {\map \phi x - 1} + C$

and the constant disappears.

Hence the result.