Originally Posted by

**arbolis** Thanks for the tips.

I propose a solution of the form $\displaystyle \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}$. After derivating once and twice and plugging into the DE the indicial equation gives me $\displaystyle c=0$ or $\displaystyle c=1$. And also $\displaystyle a_{n+1}=\frac{a_n}{n+c}$. But the denominator blows up if $\displaystyle c=0$ and $\displaystyle n=0$, thus $\displaystyle c=1$ only. Thus $\displaystyle a_{n+1}=\frac{a_n}{n+1}$.

$\displaystyle a_0$ is arbitrary but different from 0. I'm guessing I have to choose it so that $\displaystyle y (0)=1$ as requested by the statement of the problem. When I choose$\displaystyle a_0=1$, I get $\displaystyle a_1=1$, $\displaystyle a_2=\frac{1}{2}$ and in general, $\displaystyle a_n=\frac{1}{n!}$. So that $\displaystyle \phi (x)=\sum _{n=0}^\infty \frac{x^{n+1}}{n!} \Rightarrow y(x)=\sum _{n=0}^{\infty} \frac{x^n}{n!}=e^x$. It seems like I got lucky this time when I chose the arbitrary constant $\displaystyle a_0$ since $\displaystyle y(0)=1$ as required.

Is what I've done correct so far?

P.S.:This indeed satisfies the original DE. Wow! Thank you very much!

Post Scriptum 2: I have no idea how to find A(x) and B(x) for a second solution. I tried to write them as infinite series and plug into the DE but I get an enormous expression that doesn't look easily solvable. I'm missing a trick here.