Second order linear homogeneous ODE with variable coefficients We have a second order linear homogeneous ODE with variable coefficients of the form:
$$y''(x)+ \frac{b(x)}{a(x)}y'(x)+\frac{c(x)}{a(x)} y(x)=0.$$


*

*For $y=c_1\eta(x)+c_2\eta'(x)$, find $a(x)$, $b(x)$ and $c(x)$.

*Use the result when $y = \log x$.

*Check the result obtained.

I'm having trouble with the first part, mainly it seems like such a general form of an ODE that not much can be done. I was thinking of getting a system of three equations and solving that, two of them being immediate:
$$a(x)\eta''(x)+b(x)\eta'(x)+c(x)\eta(x)=0$$
$$a(x)\eta'''(x)+b(x)\eta''(x)+c(x)\eta'(x)=0$$ 
I don't know where a third linearly independent equation could come from. Also, are $c_1$ and $c_2$ important enough that I should work with them?
Thanks.
 A: The phrasing of the question is singularly vague; if this is indeed the exact way the question was put to you, I cannot blame you the least for not knowing how to proceed.
I interpret the first bullet as follows:

Assume that there exists a function $\eta(x)$ such that every solution to the above ODE can be written as a linear combination of $\eta(x)$ and $\eta'(x)$, i.e. $y(x) = c_1 \eta(x) + c_2 \eta'(x)$ is a solution to the ODE for every choice of $c_{1,2}$.

Call $q(x) = \frac{b(x)}{a(x)}$ and $r(x) = \frac{c(x)}{a(x)}$. Then, by substituting $\eta(x)$ resp. $\eta'(x)$ in the original ODE, you obtain expressions for $q(x)$ and $r(x)$ in terms of $\eta$ and its derivatives.
To obtain the third equation you're looking for, an idea would be to take the derivative of the entire ODE, and use the fact that $\eta'(x)$ solves the original ODE. After some cancellations, this yields
\begin{equation}
 r'(x) \eta(x) + q'(x) \eta'(x) = 0. \tag{1}
\end{equation}
The first two expressions obtained enable you to write
\begin{align}
 q(x) &= \frac{\eta(x) \eta'''(x) - \eta'(x) \eta''(x)}{\eta'(x)^2-\eta(x) \eta''(x)},\tag{2}\\ 
r(x) &= \frac{\eta''(x)^2 - \eta'(x) \eta'''(x)}{\eta'(x)^2-\eta(x) \eta''(x)}. \tag{3}
\end{align}
Note that $q(x) \eta'(x) + r(x) \eta(x) = -\eta''(x)$. Taking the derivative of this equation, and using the fact that $\eta'(x)$ solves the original ODE, you end up with $(1)$. In other words: there are only two independent equations, so you can only solve for $q(x)$ and $r(x)$, and not for $a,b,c$.
Note that an obvious choice for $a,b,c$ based on $(2)$ and $(3)$ would be
\begin{align}
a(x) &= \eta'(x)^2-\eta(x) \eta''(x),\\
b(x) &= \eta(x) \eta'''(x) - \eta'(x) \eta''(x),\\
c(x) &= \eta''(x)^2 - \eta'(x) \eta'''(x),
\end{align}
but as you can see, this choice is not uniquely determined: you could multiply these choices of $a,b,c$ with an arbitrary function of $x$ and still satisfy $(2)$ and $(3)$.
For the rest: I sincerely hope you inadvertently made a typo, and that it should read

For $\eta(x) = \log(x)$, apply the above results.

