I'm trying to solve the Bernoulli's equation via perturbation method but I need some help understanding how its done:
We start off with $y'=-y+\epsilon y^2$ with $y(0)=1$.
Then how is it possible to get the transformation $y_{ex}= e^{-t}/1+\epsilon(e^{-t}-1)$ and then change it to $y'=-y$ and $y(0)=1$ and solve it from there?
And how are you supposed to even come up with the $y_{ex}$ out of the blue?
On
Or treat it as a Bernoulli ODE by setting $y=u^{-1}$ and $$ -u'=-u+ϵ \implies (u(t)-ϵ)=e^{t}·(u(0)-ϵ)\\[0.7em] \implies y(t)=\frac{y(0)}{ϵ·y(0)+e^t·(1-ϵ·y(0))}=\frac{e^{-t}·y(0)}{1+ϵ·(e^{-t}-1)·y(0)} $$
However, from the task description you are to use this as an example for the perturbation method, so setting $y=y_0+ϵ·y_1+ϵ^2·y_2+…$ one gets $$ y'_0+ϵ·y'_1+ϵ^2·y'_2+…=-(y_0+ϵ·y_1+ϵ^2·y_2+…)+ϵ·(y_0+ϵ·y_1+ϵ^2·y_2+…)^2 $$ and by comparing the terms of equal power of $ϵ$ one successively obtains equations \begin{align} y'_0&=-y_0\\ y'_1&=-y_1+y_0^2\\ y'_2&=-y_2+2y_0y_1\\ y'_3&=-y_3+y_1^2+2y_0y_2\\ &… \end{align} which all have the homogeneous solution $e^{-t}$.
$$y' = -y + \epsilon y^2 \implies \frac{dy}{\epsilon y^2 - y} = dt$$ Using partial fractions: $$\frac{1}{\epsilon y^2 - y} = \frac{\epsilon}{\epsilon y - 1} - \frac{1}{y}$$ From this, we have: $$\ln(\epsilon y - 1) - \ln(y) = t + C$$ $$\ln\bigg(\frac{\epsilon y - 1}{y}\bigg) = \cdot t+ C$$ $$y^{-1} = \epsilon - A\cdot e^t$$