• The general solution is $y_{h} + y_{p}$. Lets find $y_{h}$ first.
  • This is a Cauchy-Euler equation whose solution can be assumed $x^{m}$. Substituiting \begin{align} x^{2}\roundbr{m(m-1)x^{m-2}} - x\roundbr{mx^{m-1}} + x^{m} &= 0\newline x^{m}\roundbr{m^{2} - m - m + 1} &= 0\newline \implies m &= 1 \end{align} The solution is of the form \begin{align} y_{1} &= x\newline y_{2} &= x\ln x\newline y_{h} &= c_{1}x + c_{2}x\ln x \end{align}

  • Let’s try to find $y_{p}$ using variation of parameters. First we convert the cocefficient of $\difftwo{y}$ to 1 \begin{align} \difftwo{y} - \frac{1}{x}\diffone{x} + \frac{1}{x^{2}}y &= \frac{\ln x}{x^{2}}\newline W &= y_{1}\diffone{y_{2}} - y_{2}\diffone{y_{1}}\newline &= x(\ln x + 1) - x\ln x(1) = x\newline y_{p} &= -y_{1}\int\frac{y_{2}r}{W}dx + y_{2}\int\frac{y_{1}r}{W}dx\newline &= -x\int\frac{\roundbr{\ln x}^{2}}{x^{2}}dx + x\ln x\int\frac{\ln x}{x^{2}} dx\newline \end{align} Lets solve the two integrals one by one \begin{align} \ln x &= t \implies dx = xdt = e^{t}dt\newline \int\frac{\ln x}{x^{2}} dx &= \int te^{-t}dt\newline &= -te^{-t} + \int e^{-t}dt = -te^{-t} - e^{-t}\newline &= -\frac{\ln x + 1}{x} \end{align} using integration by parts. Similarly, \begin{align} \int\frac{\roundbr{\ln x}^{2}}{x^{2}}dx &= \int t^{2}e^{-t}dt\newline &= -t^{2}e^{-t} + \int 2te^{-t}dt\newline &= -t^{2}e^{-t} - 2te^{-t} + \int 2e^{-t}dt\newline &= -t^{2}e^{-t} - 2te^{-t} - 2e^{-t}\newline &= -\frac{\roundbr{\ln x}^{2} + 2\ln x + 2}{x} \end{align}

    Hence, \begin{align} y_{p} &= -x\roundbr{-\frac{\roundbr{\ln x}^{2} + 2\ln x + 2}{x}} + x\ln x\roundbr{-\frac{\ln x + 1}{x}}\newline &= \roundbr{\ln x}^{2} + 2\ln x + 2 - \roundbr{\ln x}^{2} - \ln x\newline &= \ln x + 2 \end{align}

    The general solution becomes \begin{align} y = \roundbr{c_{1}x + c_{2}x}\ln x + \ln x + 2 \end{align}