Using Regular Perturbation, obtain approximate solution:

Question

Consider the following ODE: $y'+ y = \epsilon y^2, y(0,\epsilon)=1$

Assuming the perturbation amsats gives:

$y'_0 + y_0 =0, y_0(0)=1,$

$y'_1 + y_1 = y^2_0$, $y_1(0)=0$

How is: $y_0(x)=e^{(-x)}$ and $y_1(x) = e^{(-x)}-e^{(-2x)}$

Answer 1

Here you can, to confirm the perturbartion computations, solve the problem directly as a Bernoulli equation to find that $$ \frac{d}{dx}(y^{-1})=-y^{-2}y'=-ϵ+y^{-1}\implies y^{-1}(x)=ϵ+(1-ϵ)e^{x} $$ and then $$ y(x)=\frac{e^{-x}}{1-ϵ(1-e^{-x})} $$ which you can expand as geometric series to find the perturbation series.

Answer 2

In both cases you have an equation of the form $$ z'+z=f(x),\quad z(0)=a. $$ Multiply by $e^x$ to get $$ \bigl(e^x\,z\bigr)'=e^x\,f(x). $$ Now integrate to find the solution $$ z(x)=a\,e^{-x}+e^{-x}\int_0^xe^t\,f(t)\,dt. $$