Differential Equations: Bernoulli's Equation $1 - 3rss' + r^2 s^2 s' = 0$ Submitted by agentcollins on Sun, 07/17/2016 - 10:18 Bernoulli? 1- 3rss' + r^2 s^2 s' =0 Tags Bernoulli's Equation (Differential Equation) Log in to post comments Yes, the given is a Bernoulli Jhun Vert Sun, 07/17/2016 - 13:10 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 $dr + P(s) \, r \, ds = Q(s) \, r^n \, ds$ Log in to post comments
Yes, the given is a Bernoulli Jhun Vert Sun, 07/17/2016 - 13:10 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 $dr + P(s) \, r \, ds = Q(s) \, r^n \, ds$ Log in to post comments
Yes, the given is a Bernoulli
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
$dr + P(s) \, r \, ds = Q(s) \, r^n \, ds$