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

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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:

$\dfrac {\mathrm d u}{\mathrm d x} + P \left({x}\right) u = Q \left({x}\right)$

whose general solution from Solution to Linear First Order Ordinary Differential Equation is:

$\displaystyle u e^{\int P \ \mathrm d x} = \int Q e^{\int P \ \mathrm d x} \ \mathrm d x + C$

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.

$\blacksquare$


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