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

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## 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:

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

## Proof

Let $u = \map f y$

Then by the Chain Rule:

- $\dfrac {\d u} {\d x} = \map {f'} y \dfrac {\d y} {\d x}$

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:

- $\dfrac {\d u} {\d x} + \map P x u = \map Q x$

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:

\(\displaystyle \int P \rd x\) | \(=\) | \(\displaystyle -\int \map {\phi'} x \rd x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\map \phi x + A\) |

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:

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

\(\displaystyle \int \map \phi x \, \map {\phi'} x e^{-\map \phi x + A} \rd x\) | \(=\) | \(\displaystyle \int v e^{-v + A} \rd v\) | 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:

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

subsuming $C'$ into $C$.

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

- $\displaystyle \map f y e^{-\map \phi x + A} = e^{\map \phi x + A} \paren {\map \phi x - 1} + C$

and so multiplying both sides by $e^{-\map \phi x + A}$:

- $\displaystyle \map f y = e^{2 \map \phi x + A - A} \paren {\map \phi x - 1} + C$

and the constant disappears.

Hence the result.

$\blacksquare$

## Sources

- 1896: Joseph Edwards:
*Integral Calculus for Beginners with an Introduction to Differential Equations*: Exercise $18 \ \text{(d)}$