How do I show that y = x is a solution for the differential equation x^2y'' + 2xy' - 2y = 0 in the boundary ]0, ∞[ and how do I find the general solution for the equation? I think that I can begin with y = x^r. then y' = rx^(r-1) and y'' = r(r-1)x^(r-2). Substituting in the original equation, then I would get r(r-1)x^r + 2x^r - 2x^r = 0. Then (r^
2 + r - 2)x^r = 0. Solving the quadratic equation for r, I get r = 1 and -2. But where do i go from here?

Question

How do I show that y = x is a solution for the differential equation x^2y'' + 2xy' - 2y = 0 in the boundary ]0,  ∞[ and how do I find the general solution for the equation? I think that I can begin with y = x^r. then y' = rx^(r-1) and y'' = r(r-1)x^(r-2). Substituting in the original equation, then I would get r(r-1)x^r + 2x^r - 2x^r = 0. Then (r^
2 + r - 2)x^r = 0. Solving the quadratic equation for r, I get r = 1 and -2. But where do i go from here?

sirm3d · Answer

you just proved that $$\large y=x^r=x^1=x$$ is a solution to the DE. you were also able to derive a second solution $$\large y=x^{-2}=\frac{1}{x^2}$$