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

## Theorem

Let $f \left({y}\right)$ and $\phi \left({x}\right)$ be known real functions of $y$ and $x$ respectively.

The general solution of:

$(1): \quad \dfrac {\mathrm d y} {\mathrm d x} - \dfrac {f \left({y}\right)} {f' \left({y}\right)} \phi' \left({x}\right) = \dfrac {\phi \left({x}\right) \phi' \left({x}\right)} {f' \left({y}\right)}$

is:

$\displaystyle f \left({y}\right) = e^{2 \phi \left({x}\right)} \left({\phi \left({x}\right) - 1}\right) + C$

## Proof

Let $u = f \left({y}\right)$

Then by the Chain Rule:

$\dfrac {\mathrm d u} {\mathrm d x} = f' \left({y}\right) \dfrac {\mathrm d y} {\mathrm d x}$

Then:

 $\displaystyle \dfrac {\mathrm d y} {\mathrm d x} - \dfrac {f \left({y}\right)} {f' \left({y}\right)} \phi' \left({x}\right)$ $=$ $\displaystyle \dfrac {\phi \left({x}\right) \phi' \left({x}\right)} {f' \left({y}\right)}$ $\displaystyle \implies \ \$ $\displaystyle f' \left({y}\right) \dfrac {\mathrm d y} {\mathrm d x} - f \left({y}\right) \phi' \left({x}\right)$ $=$ $\displaystyle \phi \left({x}\right) \phi' \left({x}\right)$ multiplying $(1)$ by $f' \left({y}\right)$ $\displaystyle \implies \ \$ $\displaystyle \dfrac {\mathrm d u} {\mathrm d x} - u \phi' \left({x}\right)$ $=$ $\displaystyle \phi \left({x}\right) \phi' \left({x}\right)$ substituting $u$ for $f \left({y}\right)$

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

$\dfrac {\mathrm d u}{\mathrm d x} + P \left({x}\right) u = Q \left({x}\right)$
$\displaystyle u e^{\int P \ \mathrm d x} = \int Q e^{\int P \ \mathrm d x} \ \mathrm d x + C$

In this instance:

$P \left({x}\right) = -\phi' \left({x}\right)$

and:

$Q \left({x}\right) = \phi \left({x}\right) \phi' \left({x}\right)$

Thus:

 $\displaystyle \int P \ \mathrm d x$ $=$ $\displaystyle -\int \phi' \left({x}\right) \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle -\phi \left({x}\right) + A$

Hence:

$\displaystyle u e^{-\phi \left({x}\right) + A} = \int \phi \left({x}\right) \phi' \left({x}\right) e^{-\phi \left({x}\right) + A} \ \mathrm d x + C$

Let $v = \phi \left({x}\right)$.

Then by Integration by Substitution:

$\displaystyle \int \phi \left({x}\right) \phi' \left({x}\right) e^{-\phi \left({x}\right) + A} \ \mathrm d x = \int v e^{-v + A} \ \mathrm d v$

 $\displaystyle \int \phi \left({x}\right) \phi' \left({x}\right) e^{-\phi \left({x}\right) + A} \ \mathrm d x$ $=$ $\displaystyle \int v e^{-v + A} \ \mathrm d v$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \int v e^{-v} e^A \ \mathrm d v$ $\displaystyle$ $=$ $\displaystyle e^A \int v e^{-v} \ \mathrm d v$ $\displaystyle$ $=$ $\displaystyle e^A e^v \left({v - 1}\right) + C'$ Primitive of $x e^{a x}$

This leaves us with:

$\displaystyle u e^{-\phi \left({x}\right) + A} = e^A e^v \left({v - 1}\right) + C$

subsuming $C'$ into $C$.

Substituting back in, the general solution is seen to be:

$\displaystyle f \left({y}\right) e^{-\phi \left({x}\right) + A} = e^{\phi \left({x}\right) + A} \left({\phi \left({x}\right) - 1}\right) + C$

and so multiplying both sides by $e^{-\phi \left({x}\right) + A}$:

$\displaystyle f \left({y}\right) = e^{2 \phi \left({x}\right) + A - A} \left({\phi \left({x}\right) - 1}\right) + C$

and the constant disappears.

Hence the result.

$\blacksquare$