How to find a solution "by inspection" w.r.t. the method of reduction of order for homogeneous linear ODE's? I'm currently studying ODE's and ran into an example problem that I'm having trouble understanding. The problem is from a section dealing with the method of reduction of order and is as follows:

Find a basis of solutions for the ODE
$$(x^2 - x)y'' - xy' + y = 0$$

I'm aware that reduction of order applies when we know one solution $y_1$ and we find the other by putting $y_2 = uy_1$ where $y_i$ and $u$ are both functions of $x$.
The solution for this ODE simply states that "by inspection" we can see that $y_1 = x$ is one solution. Indeed, if we put $y_2 = uy_1 = ux$ then we can solve the ODE, but I'm puzzled as to how one is to find out that $y_1 = x$ is supposedly such an obvious solution? Is there some sort of intuition that I should be accustomed to or are there actual methods? Thanks in advance.
 A: "By inspection" one put into the ODE some simple functions and check if the ODE is satisfied or not. If yes, we get a solution as expected. If not, we try another function.
For example in case of $$(x^2-x)y''-xy'+y=0$$
if we try $y=x$ which is one of the simplest function, we see that $y''=0$ and $y'=1$ thus $0-x+x=0$ which is OK. Obviously the function $y=x$ is a solution.
Of course, this is not successful each time. For example if we had tried $y=x^2$ that would be a failure.
Often, in elementary textbook exercises the ODE is chosen so that a solution be sufficiently simple to be obvious or to be found easily by trial and error. If  "inspection" is not sufficient, less simple methods have to be used.
A: First we have a solution which is $y_1 = c_1 x$: now to find other solution we make $y_2 = c_1(x) x$ and substitute into the DE obtaining
$$
x \left((x-1) x c_1''(x)+(x-2) c_1'(x)\right) = 0
$$
we follow with
$$
(x-1) x c_1''(x)+(x-2) c_1'(x) = 0
$$
and now making $C(x) = c_1'(x)$ we have a first order DE to solve
$$
(x-1) x C'(x)+(x-2) C(x) = 0
$$
which gives
$$
C(x) = c_2 \frac{1-x}{x^2}
$$
and then
$$
c_1'(x) = C(x),\ \ \ c_1(x) = c_3-c_2\left(\frac 1x+\ln x\right)
$$
and finally
$$
y_2(x) = \left(c_3-c_2\left(\frac 1x+\ln x\right)\right)x
$$
