Perturbative solution to an initial-value problem

Question

Could anyone help me with the part marked in red? I have trouble understanding the reasoning behind the technique used, and also the steps to go from (7.1.8) and (7.1.9) to (7.1.10). Thank you!

Answer 1

Using the expansion in (7.1.9) and plugging into (7.1.8) we have $$ \sum \epsilon^n y_n'' = f(x) \sum \epsilon^{n+1} y_n \\ \sum_{n=0}^{\infty} \epsilon^n y_n'' = f(x) \sum_{n=1}^{\infty} \epsilon^n y_{n-1} $$ We assume $\epsilon \ll 1$ and so the leading order is $y_0'' = 0$. Which leaves the higher order solutions $$ \underbrace{y_0''}_{\text{0th order}} + \sum_{n=1}^{\infty} \epsilon^n y_n'' = f(x) \sum_{n=1}^{\infty} \epsilon^n y_{n-1} $$ Which gives (7.1.10) $$ y_n'' = y_{n-1} f(x) $$