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

From ProofWiki
Jump to navigation Jump to search

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:

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

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

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

Hence the result.

$\blacksquare$


Sources