Answer

We first solve the homogenous equation by substituiting \(x^{m}\) as the solution (this is an Euler-Cauchy equation)

\[ \begin{aligned} x^{3}\diffthree{y} - 3x^{2}\difftwo{y} + 6x\diffone{y} - 6y &= 0\newline x^{3}m(m-1)(m-2)x^{m-3} - 3x^{2}m(m-1)x^{m-2} + 6xmx^{m-1} - 6x^{m} &= 0\newline \implies m(m-1)(m-2) -3m(m-1) + 6m - 6 &= 0\newline m^{3} - 6m^{2} + 11m - 6 &= 0\newline m &= 1, 2, 3 \end{aligned} \]

Root 1 is easily identifiable by inspection. Remaining roots can be obtained by factorization. Hence, \(y_{h}\) is

\[ \begin{aligned} y_{h} &= c_{1}x + c_{2}x^{2} + c_{3}x^{3} \end{aligned} \]

with \(x\), \(x^{2}\) and \(x^{3}\) as the bases. Since \(r(x)\) is not a standard function (for method of undetermined coefficients), we will try to use variation of parameters to solve for \(y_{p}\)

\[ \begin{aligned} W &= \detm{\begin{matrix} x &x^{2} &x^{3}\newline 1 &2x &3x^{2}\newline 0 &2 &6x \end{matrix}} = 2x^{3} \end{aligned} \]

Hence, all \(W_{i}\) are

\[ \begin{aligned} W_{1} &= \detm{\begin{matrix} 0 &x^{2} &x^{3}\newline 0 &2x &3x^{2}\newline 1 &2 &6x \end{matrix}} = x^{4}\newline W_{2} &= \detm{\begin{matrix} x &0 &x^{3}\newline 1 &0 &3x^{2}\newline 0 &1 &6x \end{matrix}} = -2x^{3}\newline W_{3} &= \detm{\begin{matrix} x &x^{2} &0\newline 1 &2x &0\newline 0 &2 &1 \end{matrix}} = x^{2} \end{aligned} \]

Now, to apply the method of undetermined, we also convert our equation to the standard form

\[ \begin{aligned} \diffthree{y} - \frac{3}{x}\difftwo{y} + \frac{6}{x^{2}}\diffone{y} - \frac{6}{x^{3}}y = x\ln x \end{aligned} \]

\[ \begin{aligned} y_{p}(x) &= \sum_{k=1}^{n}y_{k}(x)\int \frac{W_{k}(x)}{W(x)}r(x)dx\newline &= x\int\frac{x^{4}}{2x^{3}}x\ln x dx + x^{2}\int\frac{-2x^{3}}{2x^{3}}x\ln x dx + x^{3}\int\frac{x^{2}}{2x^{3}}x\ln x dx\newline &= \frac{x}{2}\roundbr{\frac{x^{3}}{3}\ln x - \frac{x^{3}}{9}} - x^{2}\roundbr{\frac{x^{2}}{2}\ln x - \frac{x^{2}}{4}} + \frac{x^{3}}{2}\roundbr{x\ln x - x}\newline &= \frac{x^{4}}{6}\ln x - \frac{11}{36}x^{4} \end{aligned} \]

Thus, the general solution is

\[ \begin{aligned} y = c_{1}x + c_{2}x^{2} + c_{3}x^{3} + \frac{x^{4}}{6}\roundbr{\ln x - \frac{11}{6}} \end{aligned} \]