\textbf{Bernoulli's differential equation:}

,,\qquad \boxed{y'  + Py = Qy^\alpha}

\textbf{General solution:}

Sbustitute \[y=z^r\] 

,,\qquad rz^{r-1}z' = Pz^r = Qz^{r\alpha}

Divide both sides with \[ rz^{r-1}\]

,,\qquad z' + \frac{P}{r} z = \frac{Q}{r}z^{r\alpha - r +1}

Setting \[r\] such that \[r\alpha - r + 1 = 0\] (i.e. \[r = \frac{1}{1-\alpha}\]) Bernoulli's equation becomes a linear differential equation of the first order that can be solved for \[z\]. 

,,\qquad z = e^{-(1-\alpha)\int P\mathrm{d}x}\left(C + (1-\alpha)\int Qe^{(1-\alpha)\int P\mathrm{d}x}\mathrm{d}x\right) 

Substituting back \[y =  z^r,\quad y=z^{\frac{1}{1-\alpha}},\quad z = y^{1-\alpha}\]:

,,\qquad \boxed{y = \left({e^{-(1-\alpha)\int P\mathrm{d}x}\left(C + (1-\alpha)\int Qe^{(1-\alpha)\int P\mathrm{d}x}\mathrm{d}x\right) }\right)^{\frac{1}{1-\alpha}}}


\textbf{Example:}

,,\qquad y' - y = xy^5

\textbf{Solution:}

Substituting \[y = z^{-1/4},\quad y' = -\frac{1}{4}z^{-5/4}z'\]:
,,\qquad -\frac{1}{4} z' - z = x

Solving this linear quation for \[z\] and substituting \[z = y^{-4}\] should yield

,,\qquad y^{-4} + x - \frac{1}{4} = Ce^{-4x}

,,\qquad