In the book of Int. to the Ordinary Differential Equation by Coddington, at page 130, in question 130, it is asked that
The equation $$y'' + e^x y = 0$$ has a solution $\phi$ of the form $$\phi(x) = \sum_{k=0}^\infty c_k x^k$$ which satisfies $\phi (0) = 1$, $\phi'(0) = 0$. Find $c_k$.
I have plugged the Power series solution $\phi$ into the ODE, and found that every
$$\sum_{k_0}^\infty [(k+2)(k+1)a_{k+2} + a_k e^x] x^k = 0$$.
However, we can interpret this result in two different way.
First; For a given $x$, the expression $$[(k+2)(k+1)a_{k+2} + a_k e^x]$$ has to be zero for all $k=0,1....$,
Second; we can write this equation as
$$\sum_{k_0}^\infty [ (k+2)(k+1)a_{k+2}x^k + a_k e^x x^k] = 0 $$, and since each $x^j$ and $x^ie^x$ are linearly independent provided that $x\not = 0$, each coefficient $a_k$ has to be zero. Since in particular, we are dealing with $x = 0$, we can still reach the same conclusion as in the first case, but what if we were dealing with $x=x_0 \not = 0$, then can we apply the first logic ?
Making $y = \sum_{k=0}^\infty a_k x^k$ and $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$
$$ y e^x = \sum_{k=0}^\infty a_k x^k \sum_{j=0}^\infty\frac{x^j}{j!} = \sum_{k=0}^\infty\sum_{\nu=0}^{\nu=k}\frac{a_{k-\nu}}{\nu !}x^k $$
and then
$$ y''+e^x y \Rightarrow \sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k + \sum_{k=0}^\infty\sum_{\nu=0}^{\nu=k}\frac{a_{k-\nu}}{\nu !}x^k = 0 $$
and finally
$$ (k+2)(k+1)a_{k+2} + \sum_{\nu=0}^{\nu= k}\frac{1}{\nu!}a_{k-\nu} = 0 $$