This is from example 2 of chapter seven, consider the IVP
\begin{equation} \begin{cases} y''=f(x)y, y(0)=1, y'(0)=1 \end{cases} \end{equation} we then make a perturbation expansion $y(x)=\sum_{n=0}^{\infty}\epsilon^ny_n(x)$
I get \begin{equation} f(x)\sum_{n=0}^{\infty}\epsilon^{n+1}y_n(x)=\sum_{n=0}^{\infty}\epsilon^ny''_n(x) \end{equation}
I don't see how we deduce both \begin{equation} f(x)y_{n-1}(x)=y''_n(x) \end{equation}
and the solution \begin{equation} y_n=\int_{0}^{x}dt\int_{0}^{t}dsf(s)y_{n-1}(s) \end{equation}
for all $n\geq 1$ Sorry if I'm asking to be spoon fed here but I feel like I am being blind to something very simple...