# 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:

$\map f y = C e^{\map \phi x} - \map \phi x - 1$

## 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:

 $\displaystyle \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x$ $=$ $\displaystyle \dfrac {\map \phi x \, \map {\phi'} x} {\map {f'} y}$ $\displaystyle \leadsto \ \$ $\displaystyle \map {f'} y \dfrac {\d y} {\d x} - \map f y \, \map {\phi'} x$ $=$ $\displaystyle \map \phi x \, \map {\phi'} x$ multiplying $(1)$ by $\map {f'} y$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\d u} {\d x} - u \, \map {\phi'} x$ $=$ $\displaystyle \map \phi x \, \map {\phi'} x$ substituting $u$ for $\map f y$

This is a linear first order ordinary differential equation in the form:

$\dfrac {\d u} {\d x} + \map P x u = \map Q x$
$\displaystyle u e^{\int P \rd x} = \int Q e^{\int P \rd x} \rd x + C$

In this instance:

$\map P x = -\map {\phi'} x$

and:

$\map Q x = \map \phi x \, \map {\phi'} x$

Thus:

 $\displaystyle \int P \rd x$ $=$ $\displaystyle -\int \map {\phi'} x \rd x$ $\displaystyle$ $=$ $\displaystyle -\map \phi x + A$

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:

 $\displaystyle \int \map \phi x \, \map {\phi'} x e^{-\map \phi x + A} \rd x$ $=$ $\displaystyle \int v e^{-v + A} \rd v$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \int v e^{-v} e^A \rd v$ $\displaystyle$ $=$ $\displaystyle e^A \int v e^{-v} \rd v$ $\displaystyle$ $=$ $\displaystyle - e^A e^{-v} \paren {v + 1} + C'$ Primitive of $x e^{a x}$

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 $e^{\map \phi x - A}$:

 $\displaystyle \map f y$ $=$ $\displaystyle \paren {- \map \phi x - 1} + C e^{\map \phi x - A}$ $\displaystyle$ $=$ $\displaystyle C e^{- A} e^{\map \phi x} - \map \phi x - 1$ $\displaystyle$ $=$ $\displaystyle C e^{\map \phi x} - \map \phi x - 1$ By absorbing the constant factor $e^{- A}$ into $C$

Hence the result.

$\blacksquare$