# Thread: Bernoulli's Equation - Fluid Flow Problem

1. ## Bernoulli's Equation - Fluid Flow Problem

$y' = \epsilon y - \sigma y^3$ ; $\epsilon > 0$ and $\sigma > 0$

Show that if $n$ does not equal $0$ or $1$, then the substitution $v = y^{1 - n}$ reduces Bernoulli's equation to a linear equation.

Bernoulli's Equation Standard Form $y' + p(t)y = q(t)y^n$

$y' - \epsilon y = -\sigma y^3 \rightarrow y'y^{-3} - \epsilon y^{-2} = -\sigma$

$n = 3$ ; $v = y^{1-3} = y^{-2} \rightarrow dv = -2y^{-3} dy \rightarrow -\frac{1}{2}dv = y^{-3}dy$

$-\frac{1}{2}v' - \epsilon v = -\sigma$ (Linear)

$v' + 2\epsilon v = 2\sigma$

$\mu = e^{\int 2\epsilon} = e^{2\epsilon t}$

$e^{2\epsilon t}[v' + 2\epsilon v] = 2\sigma e^{2\epsilon t}$

$\frac{dv}{dt}[e^{2\epsilon t}v] = 2\sigma e^{2\epsilon t}$

$\int dv[e^{2\epsilon t}v] = \int [2\sigma e^{2\epsilon t}]dt$

$e^{2\epsilon t}v = 2\sigma \int e^{2\epsilon t}dt \rightarrow e^{2\epsilon t}v = 2\sigma \left[\frac{e^{2\epsilon t}}{2\epsilon}}\right]$

$v = \frac{\sigma e^{2\epsilon t}}{\epsilon e^{2\epsilon t}} \rightarrow v = \frac{\sigma}{\epsilon}$

$y^{-2} = \frac{\sigma}{\epsilon}$

$y = \pm \sqrt{\frac{\epsilon}{\sigma}$

$y = \pm \sqrt{\frac{\epsilon}{\sigma + c\epsilon e^{-2\epsilon t}}}$

My answer's not even close, can someone walk me through this problem so I can see what I did wrong?

3. ## Re: Bernoulli's Equation - Fluid Flow Problem

Can't believe forgetting the constant of integration had such an effect on the final answer. Well anyways, I got $y = \pm \sqrt{\frac{\epsilon e^{2\epsion t}}{\sigma e^{2\epsilon t} + c\epsilon}}$

Is my answer equivalent to there's? It's kind of hard to see. $y = \pm \sqrt{\frac{\epsilon}{\sigma + c\epsilon e^{-2\epsilon t}}}$

4. ## Re: Bernoulli's Equation - Fluid Flow Problem

Multiply top and bottom of the fraction inside the square root sign with $e^{-2\epsilon t}$. I think the exponential in your numerator might be off.

5. ## Re: Bernoulli's Equation - Fluid Flow Problem

Oh forgot the epsilon in it.

6. ## Re: Bernoulli's Equation - Fluid Flow Problem

So, you're all set now?