A Regular Perturbation Expansion
for

Solution of with
y(0)=1.
We seek a solution of form

Logan 1.4 on page 50 (1st edition).

Here is the proposed series solution:

$ys = Sum[y _ n[t] ϵ^n, {n, 0, 2}] + O[ϵ]^3$

Here is the equation with the proposed perturbation solution:

y _ 0^'[t] + y _ 1^'[t] ϵ + y _ 2^'[t] ϵ^2 + O[ϵ]^3 == (1 + y _ 0[t]^2) + (y _ 0[t]^2 + 2 y _ 0[t] y _ 1[t]) ϵ + (2 y _ 0[t] y _ 1[t] + y _ 1[t]^2 + 2 y _ 0[t] y _ 2[t]) ϵ^2 + O[ϵ]^3

With the O[ϵ] symbol, the LogicalExpand does a powerful step:

Here is another way to look at it:

Here is the ODE for :

...and it's solution:

$sol0 = DSolve[{eqns[[1]], y _ 0[0] == 1}, y _ 0[t], t]$

${{y _ 0[t] -> Tan[π/4 + t]}}$

Here is the ODE for , making use of the known solution for :

Here is the solution for :

$sol1 = DSolve[{eqn1, y _ 1[0] == 0}, y _ 1[t], t]$

${{y _ 1[t] -> (-1 - 2 t + Cos[2 t])/(2 (-1 + Sin[2 t]))}}$

${y _ 0[t] -> Tan[π/4 + t]}$

-(-1 - 2 t + Cos[2 t])^2/(4 (-1 + Sin[2 t])^2) - ((-1 - 2 t + Cos[2 t]) Tan[π/4 + t])/(-1 + Sin[2 t]) - 2 Tan[π/4 + t] y _ 2[t] + y _ 2^'[t] == 0

$sol2 = DSolve[{eqn2, y _ 2[0] == 0}, y _ 2[t], t]$

${{y _ 2[t] -> 1/8 (Cos[t] - Sin[t])^(-1 - (2 Cos[2 t])/((Cos[t] - Sin[t])^3 (Cos[t] + Sin[t])) + Sin[4 t]/((Cos[t] - Sin[t])^3 (Cos[t] + Sin[t]))) (-3 Cos[t] - 2 t Cos[t] + 4 t^2 Cos[t] + 3 Cos[t] Cos[2 t] + 5 Sin[t] + 10 t Sin[t] + 4 t^2 Sin[t] - 3 Cos[2 t] Sin[t])}}$

Now convert these replacement rules into approximate solutions, of increasing O[ϵ],
that we can plot:

Tan[π/4 + t] + ((-1 - 2 t + Cos[2 t]) ϵ)/(2 (-1 + Sin[2 t])) + 1/8 (Cos[t] - Sin[t])^(-1 - (2 Cos[2 t])/((Cos[t] - Sin[t])^3 (Cos[t] + Sin[t])) + Sin[4 t]/((Cos[t] - Sin[t])^3 (Cos[t] + Sin[t]))) (-3 Cos[t] - 2 t Cos[t] + 4 t^2 Cos[t] + 3 Cos[t] Cos[2 t] + 5 Sin[t] + 10 t Sin[t] + 4 t^2 Sin[t] - 3 Cos[2 t] Sin[t]) ϵ^2 + O[ϵ]^3

1/8 ϵ^2 (Cos[t] - Sin[t])^(-1 - (2 Cos[2 t])/((Cos[t] - Sin[t])^3 (Cos[t] + Sin[t])) + Sin[4 t]/((Cos[t] - Sin[t])^3 (Cos[t] + Sin[t]))) (-3 Cos[t] - 2 t Cos[t] + 4 t^2 Cos[t] + 3 Cos[t] Cos[2 t] + 5 Sin[t] + 10 t Sin[t] + 4 t^2 Sin[t] - 3 Cos[2 t] Sin[t]) + (ϵ (-1 - 2 t + Cos[2 t]))/(2 (-1 + Sin[2 t])) + Tan[π/4 + t]