Differential Equations: Bernoulli's Equation $1 - 3rss' + r^2 s^2 s' = 0$

Bernoulli?

1- 3rss' + r^2 s^2 s' =0

Jhun Vert's picture

Yes, the given is a Bernoulli's equation
$1 - 3rs\,s' + r^2 s^2\,s' =0$

$1 - 3rs \, \dfrac{ds}{dr} + r^2 s^2 \, \dfrac{ds}{dr} = 0$

$dr - 3rs \, ds + r^2 s^2 \, ds = 0$

$dr - 3sr \, ds = -s^2 r^2 \, ds$
 

The equation is in the form
$dr + P(s) \, r \, ds = Q(s) \, r^n \, ds$