Answer
$y_{p}=\frac{1}{15}x^4$
Work Step by Step
${y}'''+\frac{1}{x}{y}''-\frac{2}{x^2}{y}'+\frac{2}{x^3}y=2x$
$W(x,x^2,\frac{1}{x})=\begin{vmatrix}
x & x^2 & \frac{1}{x} \\
1 & 2x & -\frac{1}{x^2} \\
0 & 2 & \frac{2}{x^3} \\
\end{vmatrix}\;=\;\frac{6}{x}$
$W_{1}=\begin{vmatrix}
0 & x^2 & \frac{1}{x} \\
0 & 2x & -\frac{1}{x^2} \\
2x & 2 & \frac{2}{x^3} \\
\end{vmatrix}\;=\;-6x$
$W_{2}=\begin{vmatrix}
x & 0 & \frac{1}{x} \\
1 & 0 & -\frac{1}{x^2} \\
0 & 2x & \frac{2}{x^3} \\
\end{vmatrix}\;=\;4$
$W_{3}=\begin{vmatrix}
x & x^2 & 0 \\
1 & 2x & 0 \\
0 & 2 & 2x \\
\end{vmatrix}\;=\;2x^3$
${u}'_{1}=\frac{w_{1}}{w}=-x^2$
$u_{1}=\int -x^2=-\frac{1}{3}x^3$
${u}'_{2}=\frac{w_{2}}{w}=\frac{2}{3}x$
$u_{2}=\int \frac{2}{3}x= \frac{1}{3} x^3$
${u}'_{3}=\frac{w_{3}}{w}=\frac{1}{3}x^4$
$u_{3}=\int \frac{1}{3}x^4= \frac{1}{15} x^5 $
$y_{p}=y_{1}u_{1}+y_{2}u_{2}+y_{3}u_{3}$
$y_{p}=-\frac{1}{3}x^4+\frac{1}{3}x^4+\frac{1}{15}x^4=\frac{1}{15}x^4$
$y_{p}=\frac{1}{15}x^4$
