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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The books answer was $\displaystyle y = \pm \sqrt{\frac{\epsilon}{\sigma + c\epsilon e^{-2\epsilon t}}}$

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