Solving an ODE by deriving it to get a nice form I came across this ODE:
$$2xy'(x)-y(x)=\log(x),\quad x>0 \tag{1}$$
Now one sees that the inhomogeneous part is an Euler-Differential-Equation. In fact, the nice way to solve this would probably be to use the substitution $x=e^t$ and $h(t)=y(e^t)$ since the problem here seems to be the inhomogeneity, which is $\log(x)$.
When I initially saw the problem, I didn't like the $\log(x)$ inhomogeneity, since I don't have any tools to solve such a thing. So I tried to get rid and saw that if I differentiate the ODE and multiply it with $x$, I'll get a very nice form:
Solution:
We differentiate $(1)$ and multiply with $x$.
$$2y'-2xy''-y'=1/x\tag{2}\quad\Rightarrow\quad xy'-2x^2y''=1$$
Homogeneous:
We use the ansatz $y=x^\lambda$ to get
$$P(\lambda)=\lambda-\lambda(\lambda-1)=2\lambda-\lambda^2=\lambda(2-\lambda)=0 \quad \Rightarrow \quad \lambda_1=0, \lambda_2=\frac{1}{2} \tag{3}$$
So we get
$$y_h=A+B\sqrt{x} \tag{4}$$
Particular: Using a theorem from the Book Analysis 1 - Königsberger (see below) we get the particular solution:
$$y_p=\frac{1}{2}x\tag{5}$$
General Solution:
So we find:
$$y=A+B\sqrt{x}+\frac{1}{2}x$$
Question: Now I am very unsure if the whole "I differentiate the original equation and solve the resulting" actually works. I'd assume I now should divide by $x$ and integrate, but I just can't get it to work so I guess the whole idea is flawed?

Theorem: Let $P(D)y=q$ (this is just the operator notaton) be our diff. eq. With $q$ having the form: $$q(x)=(b_0+b_1x+\dots+b_mx^m)e^{\mu x}$$ and let $\mu$ be a k-th root
  of $P$ (think: the char. poly. found in the homogeneous part) whereas
  $k=0$ means $P(\mu)\neq 0$. Then, $P(D)y=q$ has a solution of the form
$$y_p(x)=(c_0+c_1x+\dots +c_mx^m)x^ke^{\mu x}$$
If $m=0$ we can further say:
$$y_p=\frac{b_0}{P^{(k)}(\mu)}x^k e^{\mu x}$$


After substitution $x=e^t,\;t=\log{x}$ we have ODE with constant coefficients: 
$$2y'-y=t$$
Solution is
$$y=Ce^{t/2}-t-2=C\sqrt{x}-\log{x}-2$$
