Since the degree of the dependent variable y is one it has to be reduced to the form

$\frac{dy}{dx}+Py=Q\ where\ P\ and\ Q\ are\ functions\ of\ x$

Dividing throughout by $x(x-1)$ we get

$\frac{dy}{dx}+\frac{x-2}{x(x-1)}y=\frac{x^3(2x-1)}{x(x-1)}$

$P=\frac{x-2}{x(x-1)}$ and $ Q=\frac{x^3(2x-1)}{x(x-1)}$

$ I.F=e^{\int\frac{x-2}{x(x-1)}}dx $

Consider

$\frac{x-2}{x(x-1)}$

$ \frac{x-2}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$ $\dots\dots{1}$

$ x-2=A(x-1)+Bx$

$ put\ x=0.\implies -2=A(-1)$

$\therefore A=2 $

$put\ x=1, \implies -1=B(1)$

$ \therefore B=-1$

Substituting the values of A and Bin eqn (1),we get

$\frac{x-2}{x(x-1)}=\frac{2}{x}-\frac{1}{x-1}$

$ I.F=e^{\int\left( \frac{2}{x}-\frac{1}{x-1}\right)dx}$

$=e^{2\log x-\log(x-1)}$

$ =e^{-\log{\frac{x^2}{x-1}}}=\frac{x-1}{x^2} $

The solution is

$y\frac{x-1}{x^2}=\int\frac{x-1}{x^2}.\frac{x^3(2x-1)}{x(x-1)}dy+c$

$=\int (2x-1)dx+c$

$=x^2-x+c$

$y\frac{x-1}{x^2}=x^2-x+c$