In a homework question, we were asked to find the first order perturbation correction to $f(x)$ of the function defined by $$f'(x)-f(x)=\epsilon\log(f(x))$$ where $\epsilon$ is a small parameter. The boundary condition is $f(0)=1$.
In the solution, $\log[f(x)]$ were expanded as a polynomial series (which in this case I assume to be the Maclaurin series) giving: $$\log(f(x))=\sum_{n=0}^{\infty} a_n f^n.$$
It is known that $\log(x)$ cannot be expanded around $x=0$, it looks like the solution expanded it around $f(x)=0$.
f(x) was also expanded in ε. $$f=f_0+\epsilon f_1+\epsilon^2 f_2 ...$$ Thus $$f'_0+\epsilon f'_1-f_0-\epsilon f_1 =\epsilon\sum_{n=0}^{\infty} a_n (f_0+\epsilon f_1 ...)^n$$ $$ f'_1- f_1 =\sum_{n=0}^{\infty} a_n (f_0)^n$$ Is the solution correct?