# 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?

• Did you mean $3 x^2 y''$? Feb 18, 2015 at 17:22
• Yes, my apologies, I will fix that. Feb 18, 2015 at 17:23

• Thanks! Might I ask, however, why I must forget about the second root that is caused by the $\pm$ in the quadratic equation? Is it because it forms a second set of exponentials/sines and cosines that are the same as the first? Feb 18, 2015 at 17:25
• Well that's easy: Because they give the same solutions! The second root only gives a minus in front of the $\sin$ but this doesn't matter since $C_3$ is arbitrary. Feb 18, 2015 at 18:16