Solve ODE with perturbation $\frac{dx}{dt}=1+\epsilon f(t)g(x), x(0)=0$

Question

Consider the ODE: $\frac{dx}{dt}=1+\epsilon f(t)g(x), x(0)=0$, where $f,g$ smooth.I want to find the time $t$ s.t. $x(t)=1$ up to order $\epsilon$.

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To expand $x=x_0+x_1\epsilon+x_2\epsilon^2+\dots$ and $g=g_1x+g_2x^2+\cdots$. I don't know how to continue

Answer 1

In zeroth order, that is considering the equation disregarding terms $O(ϵ)$, you get $$ x(t)=t+O(ϵ) $$ Thus $x(t)=1$ for $t=1+O(ϵ)$. You could want one order higher. Then you have to evaluate $$ 1+ϵ\dot x_1(t)+O(ϵ^2) = 1+ϵf(t)(g(t)+ϵg'(t)x_1+O(ϵ^2))\\ \implies \dot x_1=f(t)g(t)\implies x_1(t)=\int_0^t f(s)g(s)ds $$ Thus $$x(1-ϵu)=1-ϵu+ϵ\int_0^1 f(s)g(s)ds+O(ϵ^2)$$ so that $x(1-ϵu)=1+O(ϵ^2)$ for $$ u=\int_0^1 f(s)g(s)ds $$