r(r-1)xrr+ xr= xr(r2+ 1)= 0.
In order that that be true for all x, you must have r2+ 1= 0 which does NOT have 1 and -1 as roots.
A different way of solving an equation like this is to make the change of variable x= et which changes the d.e. into a "constant coefficients" equation that has the same characteristic equation.
No, neither satisfies the equation. If y= x then y'= 1, y''= 0 so your equation becomes x+ x= 2x, not 0.y1 satisfies the solutions provided by y2 does not,
The y2 solutions look like they're for ln(x)
Does anyone know how I would get ln(x) as a solutions from x-1? (I know ln(x) is just the integrate of x-1 but I don't know why you would integrate)