Answer
See below
Work Step by Step
Given: $y''+e^xy=0$
Let $y=\sum a_nx^n$, obtain:
$y''+e^xy$$=\sum n(n-1)a_{n}x^{n-2}-\sum \frac{x^n}{n!}\sum a_mx^m\\
=\sum [(n+2)(n+1)a_{n+2}x^n-\sum (\sum \frac{a_m}{(n-m)!})x^n\\
=\sum [(n+2)(n+1)a_{n+2}-\sum \frac{a_m}{(n-m)!}]x^n\\
=0$
Comparing $x^k$ on both sides:
$a_{n+2}=\frac{1}{(n+1)(n+2)}\sum \frac{a_m}{(n-m)!}$
It gives:
$a_2=\frac{1}{2}a_0\\
a_3=\frac{a_0+a_1}{6}\\
a_4=\frac{a_0+a_1}{12}\\
a_5=\frac{1}{120}(5a_0+4a_1)$
Therefore: $f(x)=a_0 (1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{24}+...)+a_1(x+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{30}...)$
The independent solutions are:
$y_1(x)=1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{24}+...\\
y_2(x)=x+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{30}...$
Therefore, the radius of convergence of the series solution is $\frac{1}{R}=0 \rightarrow R=\infty$
