Answer
See answer below
Work Step by Step
Take the derivatives of the function.
$$y(x)=c_1e^x+c_2e^{-x}+(2x^2+2x+1)$$
$$y'(x)=c_1e^x-c_2e^{-x}+(2x^2+2x+1)+c_2e^{-x}(4x+2)$$
$$y''(x)=c_1e^x+c_2e^{-x}+(2x^2+2x+1)-2c_2e^{-x}(4x+2)+4c_2e^{-x}$$
Substitute these functions into the differential equation to get $$xy''-2y'+(2-x)y=0$$
$$x[c_1e^x+c_2e^{-x}+(2x^2+2x+1)-2c_2e^{-x}(4x+2)+4c_2e^{-x}]-2[c_1e^x-c_2e^{-x}+(2x^2+2x+1)+c_2e^{-x}(4x+2)]+(2-x)[c_1e^x+c_2e^{-x}+(2x^2+2x+1)]=0$$
$$0=0$$
This equation is always true, so the equation is a solution to the differential equation.
