It is known that if $0<\epsilon<<1$ then differential equation with delay $$\frac{d}{dt}x(t)+\left(\frac{\pi}{2}+\epsilon\right)x(t-1)[1+x(t)]=0$$ has periodic solution.
In one book author introduces the algorithm to find this solution following:
First of all, let's normalize time by $t=(1+c)\tau, -1<c<1$ which gives us $$\frac{d}{d\tau}x(\tau)+\left(\frac{\pi}{2}+\epsilon\right)(1+c)x(\tau-1)[1+x(\tau)]=0.$$ The next step is to consider that $$x(τ,ζ)=ζcos\sigma_0\tau+ζ^2 x_2 (τ)+ζ^3 x_3 (τ)+⋯,$$ $$ε(ζ)=b_2 ζ^2+b_4 ζ^4+⋯,$$ $$c(ζ)=c_2 ζ^2+c_4 ζ^4+⋯,$$ assuming
$$\int_0^\frac{2\pi}{\sigma_0}x(τ,ζ)sin\sigma_0\tau d\tau\equiv0,$$ $$\int_0^\frac{2\pi}{\sigma_0}x(τ,ζ)cos\sigma_0\tau d\tau\equiv\frac{\pi}{\sigma_0}ζ,$$ $$x(τ,0)\equiv0,$$ $$\frac{d}{dζ}x(τ,ζ)|_{ζ=0}\equiv cos\sigma_0\tau,$$ where $\sigma_0=\frac{\pi}{2}$ in this case.
We substitute $x(τ,ζ), ε(ζ), c(ζ)$ in our differential equation, then equate coefficients to the same ζ degrees. For the first two ($ζ^2$ and $ζ^3$) we would get
$$ζ^2: \frac{d}{d\tau}x_2(\tau)+\frac{\pi}{2}x_2(τ-1)+\frac{\pi}{4}sinπτ=0,$$ $$ζ^3: \frac{d}{d\tau}x_3(τ)+\frac{\pi}{2}x_3(τ-1)+(b_2+\frac{\pi}{2}c_2) sin\frac{\pi}{2}τ+\frac{\pi}{2}x_2 (τ) sin\frac{\pi}{2}τ+\frac{\pi}{2}x_2(τ-1)cos\frac{\pi}{2}τ=0,$$ but the author gets $$ζ^2: \frac{d}{d\tau}x_2(\tau)+\frac{\pi}{2}x_2(τ-1)+\frac{\pi}{4}sinπτ=0,$$ $$ζ^3: \frac{d}{d\tau}x_3(τ)+\frac{\pi}{2}x_3(τ-1)+(b_2+\frac{\pi}{2}c_2) sin\frac{\pi}{2}τ+\frac{\pi}{2}x_2 (τ) sin\frac{\pi}{2}τ+\frac{\pi}{2}x_2(τ-1)cos\frac{\pi}{2}τ+\frac{\pi^2}{4}c_2cos\frac{\pi}{2}\tau=0.$$ As you can see in the $ζ^3$ case, there is other term $\frac{\pi^2}{4}c_2cos\frac{\pi}{2}\tau$ which I do not understand where it is coming from. In the algorithm's explanation author brings something about Fourier series and recursive functions, but I can not understand russian language very well so I'll show you pictures of that part. Hopefully someone gets it and could explain/show me how to get that extra term.
