Originally Posted by

**chisigma** The DE...

$\displaystyle x^{2}\ y^{''} - x\ y^{'} + 2\ y=0$ (1)

... is 'Euler's type' and a particular solution of it has the form $\displaystyle x^{\alpha}$. Substituting $\displaystyle x^{\alpha}$ in (1) You arrive at the algebraic equation...

$\displaystyle \alpha^{2}-2\ \alpha +2 =0$ (2)

...the solution of which are $\displaystyle 1 \pm i$, so that the general solution of (1) is...

$\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (3)

Of course if You want a real solution it must be $\displaystyle c_{1}=c_{2}=c$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$