Answer
See below.
Work Step by Step
Take the derivatives of the function. $$y(x)=c_1x^4+c_2x^{-2}$$ $$y'(x)=4c_1x^3-2c_2x^{-3}$$ $$y''(x)=12c_1x^2+6c_2x^{-4}$$ Substitute these functions into the differential equation to get $$x^2y''-xy'-8y=0$$ $$x^2(12c_1x^2+6c_2x^{-4})-x(4c_1x^3-2c_2x^{-3})-8(c_1x^4+c_2x^{-2})=0$$ $$12c_1x^4+6c_2x^{-2}-4c_1x^4+2c_2x^{-2}-8c_1x^4-8c_2x^{-2}=0$$ $$0=0$$ This equation is always true when $x>0$, so the equation is a solution to the differential equation.
