Verify that the given functions form a basis for the space of solutions of the given differential equation:
x^2 y'' - 2xy' + 2y = 0, f_1(x) = x, f_2(x) = x^2, x > 0
Proof. First, we will verify that f_1(x)=x, f_2(x)=x^2 satisfy the differential equation x^2y''-2xy'+2y=0
It is verified that functions form a basis for the space of solutions of a differential equation.