Finding all real solutions of a linear ODE.

Let $y(x) = z(\ln(x))$ for some unknown function $z(\cdot)$. Then,

$y'(x) = \frac{z'(\ln(x))}{x}$,   $y''(x) = \frac{z''(\ln(x))}{x^2} - \frac{z'(\ln(x))}{x^2}$ 

So that

$0 = y'' + \frac{4}{x}y' - \frac{10}{x^2}y = \frac{1}{x^2}[z''(\ln(x))+3z'(\ln(x))-10z(\ln(x))]$

Let $t:= \ln(x)$. Then, the previous equation is a linear ODE with constant coefficients:

$ 0 = z''(t)+3z'(t)-10z(t)$, where the characteristic equation is $(r+5)(r-2)$ so the solution is
$z(t):=c_1e^{-5t}+c_2e^{2t}$

Going back to our initial problem, $y(x) = z(\ln(x)) = c_1 x^{-5}+c_2x^2$, and this is a valid solution for all $x>0$.