<s>[INST] Cauchy-Euler Equidimensional DE $x^3y'''\left(x\right)+3x^2y''\left(x\right)+5xy'\left(x\right)-5y\left(x\right)=0$ I'm studying Cauchy-Euler equations from Fundamentals of Differential Equations by Nagle et. al. and I've come across the following problem (pg. 454, #9).

\begin{align}
 x^3y'''\left(x\right)+3x^2y''\left(x\right)+5xy'\left(x\right)-5y\left(x\right)=0,\:\:r>0,\tag{1}
 \end{align}

and this is what I have done so far. I began with the substitution of $y=x^r$ to get
\begin{align}
x^3r\left(r-1\right)\left(r-2\right)x^{r-3}+3x^2r\left(r-1\right)x^{r-2}+5xrx^{r-1}-5x^r=0\tag{2}
\end{align}
which simplifies to
\begin{align}
r^3+4r-5=0.\tag{3}
\end{align}
(3) has a factor of $\left(r-1\right)$, so taking that out we have
\begin{align}
\left(r-1\right)\left(r^2+r+5\right)=0,\tag{4}
\end{align}
and this gives us one root of $r=\left\{1\right\}$, but then to get the others I had to resort to the well-known quadratic equation, resulting in
\begin{align}
r&=\frac{-1\pm\sqrt{1-20}}{2}\\
&=\frac{-1}{2}\pm \frac{i\sqrt{19}}{2},\tag{5}
\end{align}
and therefore I have roots of
\begin{align}
r=\left\{1,\:\frac{-1}{2}+\frac{i\sqrt{19}}{2},\:\frac{-1}{2}-\frac{i\sqrt{19}}{2}\right\}.\tag{6}
\end{align}
Using this I'm going to have to use the fact that
\begin{align}
y&=x^{\alpha+i\beta}=e^{\left(\alpha+i\beta\right)\log\left(x\right)}\tag{7}\\
&=e^{\alpha\log\left(x\right)}\cos\left(\beta\log\left(x\right)\right)+ie^{\alpha\log\left(x\right)}\sin\left(\beta\log\left(x\right)\right)\tag{8}\\
&=x^\alpha\cos\left(\beta\log\left(x\right)\right)+ix^\alpha\sin\left(\beta\log\left(x\right)\right).\tag{9}
\end{align}
This therefore (should) give us a solution of
\begin{align}
y=C_1x+C_2x^{-1/2}\cos\left(\frac{\sqrt{19}}{2}\log\left(x\right)\right)+C_3x^{-1/2}\sin\left(\frac{\sqrt{19}}{2}\log\left(x\right)\right),\:\:r>0,\tag{10}
\end{align}
if I have done everything correctly. 
Have I made a mistake anywhere that you can see?
 [/INST] Seems absolutely correct, except for the typo pointed out by Amzoti.
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