Question

find the general solution (x^2 - 1)y'' - 2xy' + 2y = x^2 + 1

Answer 1

I'm not sure where to start, it looks like euler-cauchy but the -1 in the leading coefficient is messing me up

Answer 2

By inspection it looks like y=x is the solution to the complementary equation

Answer 3

y(x) = x*_C2+(x^2+1)*_C1+(1/2)*(x-1)^2*ln(x-1)+(1/2*(x+1))*((x+1)*ln(x+1)-2*x+2)

Maybe?

Answer 4

you mean this

(x^2-1)*(diff(y(x), x, x))-2*x*(diff(y(x), x))+2*y(x) = x^2+1