We have the non-homogeneous differential equation $x^3y'''-3x^2y''+6xy'-6y=4x^2$ with conditions $y(1)=1, y'(1)=1, y''(1)=0$, and I have been tasked with finding its particular solution using Green's function. First, I found the homogeneous solution, $$y_h(x)=c_1x+c_2x^2+c_3x^3$$ Then, $y_1=x, y_2=x^2, y_3=x^3$. Then, after calculations, I obtain that $G(x,\xi)=\frac{x(\xi-x)^2}{2\xi}$. I know that $y_p(x)=\int_{x_0}^{x}G(x,\xi)f(\xi)d\xi$, and plugging in $1$ as the $x_0$ value, I obtain $y_p(x)=\frac13(x-1)^3x$. Finding the general solution after this is straight-forward; my question concerns the $x_0$ parameter.
Is it supposed to always be $x_0=0$, or does it depend on the initial values provided; I used $x_0=1$ because of this, but when I checked the general solution with WolframAlpha, it found $y_p(x)=\frac{x^4}{3}$. Is this because they do not take into account initial values, or set a default of $0$? My notes on this subject are a bit confusing; any help would be appreciated.