Answer
$y_{p}=\frac{1}{2} \int_{x_{0}}^{x} [\frac{x}{t^2}-\frac{2x^2}{t^3}+\frac{x^3}{t^4}]g(t)dt$
Work Step by Step
${y}'''-\frac{3}{x}{y}''+\frac{6}{x^2}{y}'-\frac{6}{x^3}y=\frac{g(x)}{x^3}$
$W(x,x^2,x^3)=\begin{vmatrix}
x & x^2 & x^3 \\
1 & 2x & 3x^2 \\
0 & 2 & 6x \\
\end{vmatrix}\;=\;2x^3$
$W_{1}=\begin{vmatrix}
0 & x^2 & x^3 \\
0 & 2x & 3x^2 \\
\frac{g(x)}{x^3} & 2 & 6x \\
\end{vmatrix}\;=\;xg(x)$
$W_{2}=\begin{vmatrix}
x & 0 & x^3 \\
1 & 0 & 3x^2 \\
0 & \frac{g(x)}{x^3} & 6x \\
\end{vmatrix}\;=\;-2g(x)$
$W_{3}=\begin{vmatrix}
x & x^2 & 0 \\
1 & 2x & 0 \\
0 & 2 & \frac{g(x)}{x^3} \\
\end{vmatrix}\;=\;\frac{1}{x}g(x)$
${u}'_{1}=\frac{w_{1}}{w}=\frac{g(x)}{2x^2}$
$u_{1}=\frac{1}{2} \int_{x_{0}}^{x} \frac{g(t)}{x^2}\;dt $
${u}'_{2}=\frac{w_{2}}{w}=-\frac{g(x)}{x^3}$
$u_{2}= - \int_{x_{0}}^{x} \frac{g(t)}{x^3}\;dt$
${u}'_{3}=\frac{w_{3}}{w}=\frac{g(x)}{2x^4}$
$u_{3}= \frac{1}{2} \int_{x_{0}}^{x} \frac{g(t)}{x^4} $
$y_{p}=y_{1}u_{1}+y_{2}u_{2}+y_{3}u_{3}$
$y_{p}=\frac{1}{2} \int_{x_{0}}^{x} \frac{g(t)}{x^2}\;dt x - \int_{x_{0}}^{x} \frac{g(t)}{x^3}\;dt x^2+ \frac{1}{2} \int_{x_{0}}^{x} \frac{g(t)}{x^4} x^3$
$y_{p}=\frac{1}{2} \int_{x_{0}}^{x} [\frac{x}{t^2}-\frac{2x^2}{t^3}+\frac{x^3}{t^4}]g(t)dt$
