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:

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


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:

\(\displaystyle \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x\) \(=\) \(\displaystyle \dfrac {\map \phi x \, \map {\phi'} x} {\map {f'} y}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {f'} y \dfrac {\d y} {\d x} - \map f y \, \map {\phi'} x\) \(=\) \(\displaystyle \map \phi x \, \map {\phi'} x\) 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:

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

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

and the constant disappears.

Hence the result.

$\blacksquare$


Sources