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:

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


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:

\(\ds \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x\) \(=\) \(\ds \dfrac {\map \phi x \map {\phi'} x} {\map {f'} y}\)
\(\ds \leadsto \ \ \) \(\ds \map {f'} y \dfrac {\d y} {\d x} - \map f y \map {\phi'} x\) \(=\) \(\ds \map \phi x \map {\phi'} x\) multiplying $(1)$ by $\map {f'} y$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d u} {\d x} - u \map {\phi'} x\) \(=\) \(\ds \map \phi x \map {\phi'} x\) 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:

$\ds u e^{\int P \rd x} = \int Q e^{\int P \rd x} \rd x + C$

In this instance:

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

and:

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

Thus:

\(\ds \int P \rd x\) \(=\) \(\ds -\int \map {\phi'} x \rd x\)
\(\ds \) \(=\) \(\ds -\map \phi x + A\)


Hence:

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


Let $v = \map \phi x$.

Then by Integration by Substitution:

\(\ds \int \map \phi x \map {\phi'} x e^{-\map \phi x + A} \rd x\) \(=\) \(\ds \int v e^{-v + A} \rd v\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int v e^{-v} e^A \rd v\)
\(\ds \) \(=\) \(\ds e^A \int v e^{-v} \rd v\)
\(\ds \) \(=\) \(\ds - e^A e^{-v} \paren {v + 1} + C'\) 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 $e^{\map \phi x - A}$:

\(\ds \map f y\) \(=\) \(\ds \paren {-\map \phi x - 1} + C e^{\map \phi x - A}\)
\(\ds \) \(=\) \(\ds C e^{-A} e^{\map \phi x} - \map \phi x - 1\)
\(\ds \) \(=\) \(\ds C e^{\map \phi x} - \map \phi x - 1\) by absorbing the constant factor $e^{-A}$ into $C$

Hence the result.

$\blacksquare$


Sources

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