Assume y=x^z, where z is a constant, so y'=zx^(z-1), y''=z(z-1)x^(z-2), y'''=z(z-1)(z-2)x^(z-3).
Substitute for y and its derivatives:
3zx^z(z-1)(z-2)+10zx^z(z-1)-8zx^z-4x^z=0;
x^z(3z(z-1)(z-2)+10z(z-1)-8z-4)=0. Trivial case is x=0.
Auxiliary equation is: 3z(z^2-3z+2)+10z^2-10z-8z-4=0;
3z^3-9z^2+6z+10z^2-10z-8z-4=0=3z^3+z^2-12z-4=3z(z^2-4)+z^2-4=(3z+1)(z-2)(z+2).
So we have z=-1/3, 2 or -2. Therefore y=x^(-1/3), y=x^2, y=x^-2.
Let's check the simplest of these y=x^2: y'=2x, y''=2, y'''=0:
0+20x^2-16x^2-4x^2=0 OK.
Now y=x^-2: y'=-2x^-3, y''=6x^-4, y'''=-24x^-5:
-72x^-2+60x^-2+16x^-2-4x^2=0 OK. Looking good!