Consider the functional
\begin{equation}\tag{1} S[y] = \int_a^bF(y, y', x)\ dx. \end{equation}
where $y(a) = A$ and $y(b) = B$. The Euler-Lagrange equation for finding the function $y(x)$ that makes the functional stationary is
$$\tag{2} \frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) = 0. $$
I was curious to find whether we can deduce the Euler-Lagrange equation by writing y as an infinite sum, $y(x) = \sum_{k=0}^{\infty} a_k x^k$ and then differentiating $S$ with respect to the parameters $a_k$. However, I'm having some trouble with this deduction.
My attempt has given me
$$\tag{3} \frac{\partial S}{\partial a_k} = \int_a^b \frac{\partial F(y, y', x)}{\partial a_k} \ dx = \int_a^b \frac{\partial F}{\partial y} \frac{\partial y}{\partial a_k} + \frac{\partial F}{\partial y'} \frac{\partial y'}{\partial a_k} = 0, $$
for all $k \geq 0$. Using the infinite sum, we find that
$$\tag{4} \frac{\partial y}{\partial a_k} = x^k. $$
And
$$\tag{5} y'(x) = \sum_{k=1}^{\infty} a_k k x^{k-1} \rightarrow \frac{\partial y'}{\partial a_k} = k x^{k-1} $$
for $k \gt 0$. For $k = 0$, $\frac{\partial y'}{\partial a_0} = 0$.
This gives the set of equations
$$\tag{6} \int_a^b \frac{\partial F}{\partial y} dx = 0, $$
for $k=0$ and, for $k > 0$,
$$\tag{7} \int_a^b \left( \frac{\partial F}{\partial y} x^k + \frac{\partial F}{\partial y'} k x^{k-1} \right)dx = 0. $$
I have then attempted to use the integration by parts trick on the second term of the left-hand side, which gives
$$\tag{8} \int_a^b \frac{\partial F}{\partial y'} k x^{k-1} dx = \left [\frac{\partial F}{\partial y'} x^{k} \right]_a^b - \int_a^b \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] x^k dx . $$
Therefore,
$$\tag{9} \int_a^b \left( \frac{\partial F}{\partial y} x^k - \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] x^k \right) dx = -\left [\frac{\partial F}{\partial y'} x^{k} \right]_a^b $$
From this point on I don’t know how to proceed.
Ideally, we would have
$$\tag{10} -\left [\frac{\partial F}{\partial y'} x^{k} \right]_a^b = 0 $$
so that we could write
$$\tag{11} \int_a^b \left( \frac{\partial F}{\partial y} x^k - \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] x^k \right) dx = \int_a^b \left( \frac{\partial F}{\partial y} - \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] \right) x^k dx = 0, $$
which would maybe imply that it must be true that
$$\tag{12} \left( \frac{\partial F}{\partial y} - \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] \right) = 0, $$
the Euler-Lagrange equation. So, how can I proceed with the derivation? Is this derivation valid? (I'm not very worried about mathematical rigour). Thanks