(x^2 - 2x)y'' - (x^2 - 2)y' + (2x - 2)y = 0
Let's first notice that $c y_1(x)= c e^x$ is also a solution.
To find other solutions let's suppose that $c$ depends of $x$ (this method is named 'variation of constants') :
If $y(x)= c(x) e^x$ then your O.D.E. becomes : (x^2 - 2x)(c''+c'+c'+c)e^x - (x^2 - 2)(c'+c)e^x + (2x - 2)ce^x = 0 (x^2 - 2x)(c''+2c'+c) - (x^2 - 2)(c'+c) + (2x - 2)c = 0 Of course the $c$ terms disappear and we get :
(x^2 - 2x)(c''+2c') - (x^2 - 2)c' = 0 Let's set d(x)=c'(x) then : (x^2 - 2x)d' = (x^2 - 2)d-(x^2 - 2x)2d (x^2 - 2x)d' = (-x^2 +4x- 2)d
\frac{d'}d = \frac{-x^2 +4x- 2}{x^2 - 2x} I'll let search the integral at the right, the answer should be ($C_0$, $C_1$, $C_2$ are constants) : $ \ln(d)=\ln(x^2-2x)-x+C_0 $
$ d=(x^2-2x)e^{-x}C_1 $
but c'=d so that $ c=C_2+C_1\int (x^2-2x)e^{-x} dx $
$ c=C_2-C_1x^2e^{-x} $ And we got the wished general solution : $ y(x)=c(x)e^x=C_2e^x-C_1x^2 $