Answer¶
- 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
\[
\begin{aligned}
x^{2}\roundbr{m(m-1)x^{m-2}} - x\roundbr{mx^{m-1}} + x^{m} &= 0\newline
x^{m}\roundbr{m^{2} - m - m + 1} &= 0\newline
\implies m &= 1
\end{aligned}
\]
The solution is of the form
\[
\begin{aligned}
y_{1} &= x\newline
y_{2} &= x\ln x\newline
y_{h} &= c_{1}x + c_{2}x\ln x
\end{aligned}
\]
- Let's try to find \(y_{p}\) using variation of parameters. First we convert the cocefficient of \(\difftwo{y}\) to 1
\[
\begin{aligned}
\difftwo{y} - \frac{1}{x}\diffone{x} + \frac{1}{x^{2}}y &= \frac{\ln x}{x^{2}}\newline
W &= y_{1}\diffone{y_{2}} - y_{2}\diffone{y_{1}}\newline
&= x(\ln x + 1) - x\ln x(1) = x\newline
y_{p} &= -y_{1}\int\frac{y_{2}r}{W}dx + y_{2}\int\frac{y_{1}r}{W}dx\newline
&= -x\int\frac{\roundbr{\ln x}^{2}}{x^{2}}dx + x\ln x\int\frac{\ln x}{x^{2}} dx\newline
\end{aligned}
\]
Lets solve the two integrals one by one
\[
\begin{aligned}
\ln x &= t \implies dx = xdt = e^{t}dt\newline
\int\frac{\ln x}{x^{2}} dx &= \int te^{-t}dt\newline
&= -te^{-t} + \int e^{-t}dt = -te^{-t} - e^{-t}\newline
&= -\frac{\ln x + 1}{x}
\end{aligned}
\]
using integration by parts. Similarly,
\[
\begin{aligned}
\int\frac{\roundbr{\ln x}^{2}}{x^{2}}dx &= \int t^{2}e^{-t}dt\newline
&= -t^{2}e^{-t} + \int 2te^{-t}dt\newline
&= -t^{2}e^{-t} - 2te^{-t} + \int 2e^{-t}dt\newline
&= -t^{2}e^{-t} - 2te^{-t} - 2e^{-t}\newline
&= -\frac{\roundbr{\ln x}^{2} + 2\ln x + 2}{x}
\end{aligned}
\]
Hence,
\[
\begin{aligned}
y_{p} &= -x\roundbr{-\frac{\roundbr{\ln x}^{2} + 2\ln x + 2}{x}} + x\ln x\roundbr{-\frac{\ln x + 1}{x}}\newline
&= \roundbr{\ln x}^{2} + 2\ln x + 2 - \roundbr{\ln x}^{2} - \ln x\newline
&= \ln x + 2
\end{aligned}
\]
The general solution becomes
\[
\begin{aligned}
y = \roundbr{c_{1}x + c_{2}x}\ln x + \ln x + 2
\end{aligned}
\]